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8.3: Solutions for Chapter 3


Exercise 3.3

There are five non-ID columns in Eq. (3.1)and five arrows in Eq. (3.2). This is not a coincidence: there is always one arrow for every non-ID column.

Exercise 3.9

To do this precisely, we should define concatenation technically.

If G = (V, A, s, t) is a graph, define a path in G to be a tuple of the form (v, a(_{1}), . , a(_{n})) where v (in) V is a vertex, s(a(_{1})) = v, and t(a(_{i})) = s(a(_{i+1})) for all i (in) {1, . , n − 1}; the length of this path is n, and this definition makes sense for any n (in) (mathbb{N}). We say that the source of p is s(p) := v and the target of p is defined to be

Two paths p = (v, a(_{1}), . , a(_{m})) and q = (w, b(_{1}), . , b(_{n})) can be concatenated if t(p) = s(q), in which case the concatenated path p ; q is defined to be

(p ; q) := (v, a(_{1}), . , a(_{m}), b(_{1}), . , b(_{n})).

We are now ready to check unitality and associativity. A path p is an identity in Free(G) iff p = (v) for some v (in) V.

It is easy to see from the above that (v) and (w, b(_{1}), ..., b(_{n})) can be concatenated iff v = w, in which case the result is (w,b(_{1}), ..., b(_{n})). Similarly(v, a(_{1}), ..., a(_{m})) and (w) can be concatenated iff w = t(a(_{m})), in which case the result is (v, a(_{1}), . , a(_{m})). Finally, for associativity with p and q as above and r = (x, c(_{1}), . , co), the formula readily reads that whichever way they are concatenated, (p ; q) ; r or p ; (q ; r), the result is

(v, a(_{1}), ..., a(_{m}), b(_{1}), ..., b(_{n}), c(_{1}), ..., c(_{o})).

Exercise 3.10

We often like to name identity morphisms by the objects they’re on, and we do that here: v(_{2}) means id(_{v2}) . We write ☒ when the composite does not make sense (i.e. when the target of the first morphism does not agree with the source of the second).

Exercise 3.12

1. The category 1 has one object v(_{1}) and one morphism, the identity id(_{v1}).

2. The category 0 is empty; it has no objects and nomorphisms.

3. The pattern for number of morphisms in 0, 1, 2, 3 is 0, 1, 3, 6; does this pattern look familiar? These are the first few ‘triangle numbers,’ so we could guess that the number of morphisms in n, the free category on the following graph

is 1 + 2 + · · · + n. This makes sense because (and the proof strategy would be to verify that) the above graph has n paths of length 0, it has n − 1 paths of length 1, and so on : it has n i paths of length i for every 0 ≤ i n.

Exercise 3.15

The correspondence was given by sending a path to its length.

Concatenating a path of length m with a path of length n results in a path of length m + n.

Exercise 3.16

1. The ten paths are as follows

A, A ; f, A ; g, A ; f ; h, A ; g ; i, B, B ; h, C, C ; i, D

2. A ; f ; h is parallel to A ; g ; i, in that they both have the same domain and both have the same codomain.

3. A is not parallel to any of the other nine paths.

Exercise 3.17

The morphisms in the given diagram are as follows:

A, A ; f, A ; g, A ; j, B, B ; h, C, C ; i, D

Note that A ; f ; h = j = A ; g ; i.

Exercise 3.19

There are four morphisms in D, shown below, namely z, s, s ; s, and s ; s ; s:

Exercise 3.21

The equations that make the graphs into preorders are shown below

Exercise 3.22

The preorder reflection of a category C has the same objects and either one morphism or none between two objects, depending on whether or not a morphism between them exists in C. So the preorder reflection of (mathbb{N}) has one object and one morphism from it to itself, which must be the identity. In other words, the preorder reflection of (mathbb{N}) is 1.

Exercise 3.25

A function f : (underline{2}) → (underline{3}) can be described as an ordered pair (f (1), f (2)). The nine such functions are given by the following ordered pairs, which we arrange into a 2-dimensional grid with 3 entries in each dimension, just for “funzies”:(^{1})

(1, 1) (2, 1) (3, 1)

(1, 2) (2, 2) (3, 2)

(1, 3) (2, 3) (3, 3)

Exercise 3.30

1. The inverse to f (a) = 2, f (b) = 1, f (c) = 3 is given by

f (^{−1})(1) = b, f (^{−1})2) = a, f (^{−1})(3) = c.

2. There are 6 distinct isomorphisms. In general, if A and B are sets, each with n elements, then the number of isomorphisms between them is n factorial, often denoted n!. So for example there are 5 ∗ 4 ∗ 3 ∗ 2 ∗ 1 = 120 isomorphisms between {1, 2, 3, 4, 5} and {a, b, c, d, e}.

Exercise 3.31

We have to show that for any object c (in) C, the identity idc has an inverse, i.e. a morphism f : c c such that f ; id(_{c}) = id(_{c}) and id(_{c}) ; f = id(_{c}). Take f = id(_{c}); this works.

Exercise 3.32

1. The monoid in Example 3.13 is not a group, because the morphism s has no inverse.

Indeed each morphism is of the form s(^{n}) for some n (in) (mathbb{N}) and composing it with s gives s(^{n+1}), which is never s(^{0}).p

2. C from Example 3.18 is a group: the identity is always an isomorphism, and the other morphism s has inverse s.

Exercise 3.33

You may have found a person whose mathematical claims you can trust! Whenever you compose two morphisms in Free(G), their lengths add, and the identities are exactly those morphisms whose length is 0. In order forp to be an isomorphism, there must be some q such that p ; q = id and q ; p = id, in which case the length of p (or q) must be 0.

Exercise 3.37

The other three functors 2 3 are shown here:

Exercise 3.39

There are nine morphisms in F; as usual we denote identities by the object they’re on. These morphisms are sent to the following morphisms in C:

A' (longmapsto) A, f' (longmapsto) f, g' (longmapsto) g, f' ; h' (longmapsto) f ; h, g' ; i' (longmapsto) f ; h,

B′ (longmapsto) B, h′ (longmapsto) h, C ′ (longmapsto) C, i′ (longmapsto) i, D′ (longmapsto) D.

If one of these seems different from the rest, it’s probably g' ; i ' (longmapsto) f ; h.

But note that in fact also g' ; i '(longmapsto) g ; i because g ; i = f ; h, so it’s not an outlier after all.

Exercise 3.40

We need to give two functors F, G from (stackrel{a}{ullet} stackrel{f}{longrightarrow} stackrel{b}{ullet}) to (stackrel{a'}{ullet} underset{f_{2}}{stackrel{f_{1}}{longrightarrow}} stackrel{b'}{ullet}) whose on-objects parts are the same and whose on-morphisms parts are different. There are only two ways to do this, and we choose one of them:

F(a) := a′, G(a) := a′, F(b) := b′, G(b) := b′, F (f ) := f(_{1}), and G (f ) := f(_{2}).

The other way reverses f(_{1}) and f(_{2}).

Exercise 3.43

1. Let C be a category.

Then defining id(_{C}) : C → C by id(_{C})(x) = x for every object and morphism in C is a functor because it preserves identities id(_{C})(id(_{c})) = id(_{c}) = id(_{id_{C}})(c) for each object c (in) Ob(C), and it preserves composition id(_{C})(f ; g) = f ; g = id(_{C})(f) ; id(_{C})((_{g})) for each pair of composable morphisms f , g in C.

2. Given functors F : C → D and G : D → E, we need to show that F ; G is a functor, i.e. that it preserves preserves identities and compositions. If c (in) C is an object then (F ; G)(id(_{c})) = G(F(id(_{c}))) = G(id(_{F(c)})) = id(_{G(F(c)})) because F and G preserve identities. If f, g are composable morphisms in C then

(F ; G)(f ; g) = G(F(f ) ; F(g)) = G(F(f )) ; G(F(g))

because F and G preserve composition

3. We have proposed objects, morphisms, identities, and a composition formula for a category Cat: they are categories, functors, and the identities and compositions given above. We need to check that the two properties, unitality and associativity, hold. So suppose F: C → D is a functor and we pre-compose it as above with id(_{C}); it is easy to see that the result will again be F, and similarly if we post-compose F with id(_{D}). This gives unitality, and associativity is just as easy, though more wordy. Given F as above and G : D → (mathcal{E}) and H : (mathcal{E}) → F, we need to show that (F ; G) ; H = F ; (G ; H). It’s a simple application of the definition: for any x (in) C, be it an object or morphism, we have

((F ; G) ; H)(c) = H ((F ; G))(c)) = H (G (F (c))) = (G ; H)(F (c)) = (F ; (G ; H))(c).

Exercise 3.45

Let S (in) Set be a set. Define F(\_{S}): 1 Set by F(\_{S})(1) = S and F(\_{S})(id(\_{1})) = id(\_{S}). With this definition, F(\_{S}) preserves identities and compositions (the only compositions in 1 is the composite of the identity with itself), so F(\_{S}) is a functor with F(\_{S})(1) = S as desired.

Exercise 3.48

We are asked what sort of data “makes sense” for the schemas below?

This is a subjective question, so we propose an answer for your consideration.
1. Data on this schema, i.e. a set-valued functor, assigns a set D(z) and a function D(s): D(z) → D(z), such that applying that function twice is the identity. This sort of function is called an involution

of the set Dz:

It’s a do-si-do, a “partner move,” where everyone picks a partner (possibly themselves) and

exchanges with them. One example one could take D to be the set of pixels in a photograph, and take s to be the function sending each pixel to its mirror image across the vertical center line of the photograph.

2. WecouldmakeD(c)thesetofpeopleata“secretSanta”Christmasparty,where everyone gives a gift to someone, possibly themselves. Take D(b) to be the set of gifts, g the giver function (each gift is given by a person), and h the receiver function (each gift is received by a person), D(a) is the set of people who give a gift to themselves, and d( f ) : D(a) → D(b) is the inclusion.

Exercise 3.5

1. The expert packs so much information in so little space! Suppose given three objects F, G, H (in) D(^{C}); these are functors F, G, H : C → D. Morphisms α: F G and β: G H are natural transfor- mations. Most beginners seem to think about a natural transformation in terms of its naturality squares, but the main thing to keep in mind is its components; the naturality squares constitute a check that comes later.

So for each c (in) C, α has a component α(_{c}) : F(c) → G(c) and β has a component β(_{c}) : G(c) → H(c) in D. The expert has told us to define (α ; β)(_{c}) := (α(_{c}) ; β(_{c})), and indeed that is a morphism F(c) → H(c).
Now we do the check. For any f : c c′ in C, the inner squares of the following diagram commute because α and β are natural; hence the outer rectangle does too:

2. We propose that the identity natural transformation id(_{F}) on a functor F : C → D has as its c-component the morphism (id(_{F}))(_{c}) := id(_{F})((_{c})) in D, for any c. The naturality square

obviously commutes for any f : c c′. And it is unital: post-composing idF with any β : F G (and similarly for precomposing with any α : E F) results in a natural transformation id(_{F}) ; β with components (id(_{F}))(_{c}) ; β(_{c}) = (id(_{F})((_{c})) ; β(_{c})) = β(_{c}), and this is just β as desired.

Exercise 3.58

We have a category C and a preorder P, considered as a category.
1. Suppose that F, G : C → P are functors and α, β : F G are natural transformations; we need to show that α = β.

It suffices to check that αc(_{c}) = β(_{c}) for each object c (in) Ob(C). But α(_{c}) and β(_{c}) are morphisms F(c) → G(c) in P, which is a preorder, and the definition of a preorder considered as a category is that it has at most one morphism between any two objects. Thus α(_{c}) = β(_{c}), as desired.

2. This is false. Let P := 1, let C := (stackrel{a}{ullet} frac{f_{1}}{f_{2}} ext { b }), let F(1) := a, let G(1) := b, let α(_{1}) := f(_{1}), and let β(_{1}) := g(_{2}).

Exercise 3.62

We need to write down the following

as a Gr-instance, as in Eq. (3.61). The answer is as follows:

Exercise 3.64

Let G, H be the following graphs:

and let’s believe the authors that there is a unique graph homomorphism α: G H for which α(_{Arrow})(a) = d.

  1. We have α(_{Arrow})(b) = e and α(_{Vertex})(1) = 4, α(_{Vertex})(2) = 5, and α(_{Vertex})(3) = 5.

  2. We roughly copy the tables and then draw the lines (shown in black; ignore the dashed lines for now):

3. It works! One example of the naturality is shown with the help of dashed blue lines above. See how both paths starting at a end at 5?

Exercise 3.67

We just need to write out the composite of the following functors

in the form of a database, and then draw the graph. The results are given below.

Exercise 3.73

We are interested in how the functors − × B and (−)(^{B}) should act on morphisms for a given set B.

We didn’t specify this in the text—we only specified − × B and (−)(^{B}) on objects so in some sense this exercise is open: you can make up anything you want, under the condition that it is functorial. However, the authors cannot think of any such answers except the one we give below.

1. Given an arbitrary function f : X Y, we need a function X × B Y × B. We suggest the function which might be denoted f × B; it sends (x, b) to ( f (x), b). This assignment is functorial: applied to id(_{X}) it returns id(_{X x B}) and it preserves composition.

2. Given a function f : X Y, we need a function X(^{B}) → Y(^{B}). The canonical function would be denoted f(^{B}); it sends a function g : B X to the composite (g ; f ) :B X Y. This is functorial: applied to id(_{X}) it sends g to g, i.e. f(^{B})(id(_{X})) = id(_{X})(^{B}), and applied to the composite (f(_{1}) ; f(_{2})): X Y Z, we have

(f(_{1}) ; f(_{2}))(^{B})(g) = g ; (f(_{1}) ; f(_{2})) = (g ; f(_{1})) ; f(_{2}) = (f(_{1})(^{B}) ; f(_{2})(^{B}))(g)

for any g (in) X(^{B}).

3. If p : (mathbb{N}) → (mathbb{N})(^{mathbb{N}}) is the result of currying +: (mathbb{N})×(mathbb{N}) → (mathbb{N}), then p(3) is an element of (mathbb{N})(^{mathbb{N}}), i.e. we have p(3): (mathbb{N}) → (mathbb{N}) what function is it? It is the function that adds three. That is p(3)(n) := n + 3.

Exercise 3.76

The functor !: C → 1 from Eq. (3.75) sends each object c (in) C to the unique object 1 (in) 1 and sends each morphism f : c d in C to the unique morphism id(_{1}) : 1 → 1 in 1.

Exercise 3.78

We want to draw the graph corresponding to the instance I : G → Set shown below:

Here it is, with names and emails shortened (e.g. B=Bob, 3=Em_3):

Exercise 3.81

An object z is terminal in some category C if, for every c (in) C there exists a unique morphism c z. When C is the category underlying a preorder, there is at most one morphism between any two objects, so the condition simplifies: an object z is terminal iff, for every c (in) C there exists a morphism c z. The morphisms in a preorder are written with ≤ signs, so z is terminal iff, for every c (in) P we have c z, and this is the definition of top element.

Exercise 3.82

The terminal object in Cat is 1 because by Exercise 3.76 there is a unique morphism (functor) C → 1 for any object (category) C (in) Cat.

Exercise 3.83

Consider the graph 2V := • • with two vertices and no arrows, and let C = Free(2V); it has two objects and two morphisms (the identities). This category does not have a terminal object because it does not have any morphisms from one object to the other.

Exercise 3.88

A product of x and y in P is an object z (in) P equipped with maps z x and z y such that for any other object z′ and maps z′ → x and z′ → y, there is a unique morphism z′ → z making the evident triangles commute. But in a preorder, the maps are denoted ≤, they are unique if they exist, and all diagrams commute. Thus the above becomes: a product of x and y in P is an object z with z x and z y such that for any other z′, if z′ ≤ x and z′ ≤ y then z′ ≤ z. This is exactly the definition of meet, z = x (land) y.

Exercise 3.90

1. The identity morphism on the object (c, d) in the product category C × D is (id(_{c}), id(_{d})).

2. Suppose given three composable morphisms in C × D

((c(_{1}), d(_{1})) stackrel{(f(_{1}), g(_{1})})}{ ightarrow} (c(_{2}), d(_{2}) stackrel{(f(_{2}), g(_{2})})}{ ightarrow} (c(_{3}), d(_{3}) stackrel{(f(_{3}), g(_{3})})}{ ightarrow} (c(_{4}), d(_{4}))

We want to check that ((f(_{1}), g(_{1})) ; (f(_{2}), g(_{2})) ; (f(_{3}), g(_{3})) = (f(_{1}), g(_{1})) ; ((f(_{2}), g(_{2})) ; (f(_{3}), g(_{3})). But composition in a product category is given component-wise. That means the left-hand side is ((f(_{1}) ; f(_{2})) ; f(_{3}), (g(_{1}) ; g(_{2})) ; g(_{3})), whereas the right-hand side is (f(_{1}) ; (f(_{2}) ; f(_{3})), g(_{1}) ; (g(_{2}) ; g(_{3}))), and these are equal because both C and D individually have associative composition.

3. The product category 1 × 2 has two objects (1, 1) and (1, 2) and one non-identity morphism (1,1) → (1,2). It is not hard to see that it looks the same as 2. In fact, for any C there is an isomorphism of categories 1 × C (cong) C.

4. Let P and Q be preorders, let X = P × Q be their product preorder as defined in Example 1.56, and let P, Q, and X be the corresponding categories. Then X = P × Q.

Exercise 3.91

A product of X and Y is an object Z equipped with morphisms (X stackrel{px}{leftarrow} Z stackrel{py}{ ightarrow} Y) such that for any other object Z' equipped with morphisms (X stackrel{p'x}{leftarrow} Z stackrel{p'y}{ ightarrow} Y), there is a unique morphism f : Z Z making the triangles commute, f ; p(_{X}) = p′(_{X}) and f ; p(_{Y}) = p′(_{Y}). But “an object equipped with morphisms to X and Y” is exactly the definition of an object in Cone(X, Y), and a morphism f making the triangles commute is exactly the definition of a morphism in Cone(X, Y). So the definition above becomes: a product of X and Y is an object Z (in) Cone(X, Y) such that for any other object Z′ there is a unique morphism Z′ → Z in Cone(X, Y). This is exactly the definition of Z being terminal in Cone(X, Y).

Exercise 3.97

Suppose J is the graph (egin{array}{l} v1 ullet end{array})(egin{array}{l} v2 ullet end{array}) and D : J → Set is given by two sets, D(v(_{1})) = A and D(v(_{2})) = B for sets A, B. The product of these two sets is A × B. Let’s check that the limit formula in Theorem 3.95 gives the same answer. It says

(egin{aligned}
&lim _{g} D:=left{left(d_{1}, ldots, d_{n} ight) mid d_{i} in Dleft(v_{i} ight) ext { for all } 1 leq i leq n ight. ext { and }
& ext { for all } a: v_{i} ightarrow v_{j} in A, ext { we have } left.D(a)left(d_{i} ight)=d_{j} ight}
end{aligned})

But in our case n = 2, there are no arrows in the graph, and D(v(_{1})) = A and D(v(_{2})) = B. So the formula reduces to

lim(_{J})D := (d(_{1}), d(_{2})) | d(_{1}) (in) A and d(_{2}) (in) B.

which is exactly the definition of A × B.

Exercise 3.101

Given a functor F : C → D, we define its opposite F(^{op}) : C(^{op}) → D(^{op}) as follows.

For each object c (in) Ob(C(^{op})) = Ob(C), put F(^{op})(c) := F(c). For each morphism f : c(_{1}) → c(_{2}) in C(^{op}), we have a corresponding morphism f′: c(_{2}) → c(_{1}) in C and thus a morphism F(f′): F(c(_{2})) → F(c(_{1})) in D, and thus a morphism F(f ) : F(^{op})(c(_{1}))→F(^{op})(c(_{2})).

Hence we can define F(^{op})(f ) := F(f' )'. Note that the primes (−') are pretty meaningless, we only put them there to differentiate between things that are very closely related. It is easy to check that our definition of Fop is functorial: it sends identities to identities and composites to composites.


NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.3

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Interest

interest on capital is the interest paid to owners for providing a firm with the required capital to start a business. It is similar to obtaining a loan from any financial institution. The partners are paid interest on the capital that remains outstanding.

Simple interest

When money is borrowed on simple interest then the interest is calculated uniformly on the original principal throughout the loan period.
S. I. = P X R X T 100

To Find Compound Interest When Interest Is Compounded Annually

In such cases where interest is compounded yearly, the interest accrued during the first year is added to the principal and the amount so obtained becomes the principal for the second year. The amount at the end of the second year becomes the principal for the third year, and so on.

Find the compound interest on 25000 for 3 years at 10% per annum, compounded annually.

Principal for the first year = 25000.
S. I. = (P X R X T)/100
Interest for the first year = (25000 x 10 x 1)/100 = Rs. 2500
Amount at the end of the first year = (25000 + 2500) = Rs. 27500.
Principal for the second year = 27500.
Interest for the second year = (27500 x 10 x 1)/100 = Rs. 2750
Amount at the end of the second year = (27500 + 2750) = 30250.
Principal for the third year = 30250.
Interest for the third year = (30250 x 10 x 1)/100 = Rs. 3025
Amount at the end of the third year = (30250 + 3025) = 33275.
Compound interest = (33275 – 25000) = 8275.

To Find Compound Interest When Interest is Compounded Half- Yearly

In such cases, if the rate of interest is R% per annum then it is clearly % per half-year. The amount alter the first half-year becomes the principal for the next half-year, and so on. The method for calculating the compound interest in such cases is shown in the example given below.

Find the compound interest on 5000 for 1 year at 8% per annum, compounded half-yearly.

Rate of interest = 8% per annum = 4% per half-year.
Time = 1 year = 2 half-years.
Original principal = 5000.
Interest for the first half-year = Rs. (5000 x 4 x 1)/100 = Rs. 200
Amount at the end of the first half-year = (5000 + 200) = 5200.
Principal for the second half-year = 5200.
Interest for the second half-year = Rs. (5200 x 4 x 1)/100 = Rs. 208.
Amount at the end of the second half-year = (5200 + 208) = 5408
Compound interest = (5408 – 5000) = 408.

To Find Compound Interest When Interest is Compounded Quarterly

If the rate of interest is R% per annum and the interest is compounded quarterly, then the rate is R/4% per quarter.

Find the compound interest on 8000 for 1 year at 20% per annum, interest being payable quarterly.

Principal = 8000
Rate of Interest = 20% per annum = 20/4% = 5% per quarter
Time = 1 year = 4 quarters
Interest for 1st quarter = Rs. (8000 x 5 x 1)/100 = Rs. 400
Amount at the end of 1st quarter = (8000 + 400) = 8400.
Principal for 2nd quarter = 8400.
Interest for 2nd quarter = Rs. (8400 x 4 x 1)/100 = 420.
Amount at the end of 2nd quarter = (8400 + 420) = 8820.
Principal for 3rd quarter = 8820
Interest for 3rd quarter = Rs. (8820 x 5 x 1)/100 = 441.
Amount at the end of 3rd quarter = (8820 + 441) = 9261.
Principal for 4th quarter = 9261
Interest for 4th quarter = Rs. (9261 x 5 x 1)/100 = 463.05.
Amount at the end of 4th quarter = (9261 + 463.05) = 9724.05.
So, Compound Interest = (9724.05 – 8000) = 1724.05

Calculating Compound Interest by Using Formulae

The method of calculating the compound interest discussed previously is quite long and cumbersome, especially when the time period is large.
We can derive general formulae for calculating compound interest in various cases, as given below. It is very easy to calculate compound interest by using these formulae.
When the interest is compounded annually
Let principal = P, rate = R% per annum and time = n years. Then, the amount A is given by the formula.
A = Rs. P (1 + r/100)ⁿ

Find the amount of 8000 for 3 years, compounded annually at 10% per annum. Also, find the compound interest.

Here, p = 8000, R = 10% per annum and n = 3 years
amount after 3 years = Rs. 8000 (1 + 10/100)3
= 8000 x 11/10 x 11/10 x 11/10 = 10648
Thus, amount after 3 years = 10648. And, compound interest = (10648 – 8000) = 2648.

Applications of Compound Interest Formula

In each of the following situations we use the CI formula.
(i) Increase (or decrease) in population
(ii) The growth of bacteria when the rate of growth is known
(iii) Depreciation in the values of machines, etc., at a given rate

Note: For decrease (or depreciated value), we use the formula:
Decreased value (or depreciated value) = P <1 – R/100>ⁿ

Interest Compounded Half-Yearly

Let principal = P, rate = R% per annum, time = n years.
Suppose that the interest is compounded half-yearly. Then,
Rate = R/2% per half-year, time = (2n) half-years, and amount = Rs. P x (1 + R/200)2n
Compound interest = (amount) – (principal)

Interest Compounded Quarterly

Let principal = P, rate = R% per annum, time = n years.
Suppose that the interest is compounded quarterly, Then,
Rate = R/4% per half-year, time = (4n) half-years, and amount = Rs. P x (1 + R/400)4n
Compound interest = (amount) – (principal)

Unitary Method

(i) Suppose A can finish a piece of work in 8 days.
Then, work done by A in 1 day = [by unitary method].
(ii) Suppose that the work done by A is 1 day is.
Then, time taken by A to finish the whole work = 6 days.

General Rules

(i) Suppose A can finish a work in n days.
Then, work done by A in 1 day = 1/n
(ii) Suppose that the done by A in 1 day is 1/n.
Then, time taken by A to finish the whole work = n days.

A alone can finish a piece of work in 12 days and B alone can do it in 15 days. If both of them work at it together, how much time will they take to finish it?

Time taken by A to finish the work = 12 days.
Time taken by B to finish the work = 15 days.
Work done by A in one day = 1/12
Work done B in 1 day = 1/15
Work done by (A + B) in 1 day = (1/12 + 1/15) = 9/60 = 3/20
Time taken by (A + B) to finish the work = 20/3 days
Hence, both can finish the work in 6(2/3) days.

Problems on Pipes and Cistern

A cistern or a water tank is connected with two types of pipes.
(i) Inlet: The pipe which fills the tank is called on inlet.
(ii) Outlet: The pipe which empties the tank is called an outlet.
Rule 1. Suppose an outlet fills a tank in n hours.
Then, part of the tank filled in 1 hour = 1/n,
i.e., work done by the inlet in 1 hour = 1/n.
Rule 2. Suppose an outlet empties a full tank in m hours.
Then, part of the tank emptied in 1 hour = 1/m
i.e., work done by the outlet in 1 hour = (-1/m)

A tap A can fill a cistern in 8 hours while tap B can fill it in 4 hours. In how much time will the cistern be filled if both A and B are opened together?

Time taken by tap A to fill the cistern = 8 hours.
Time taken by tap B to fill the cistern = 4 hours.
Work done tap A in 1 hour = 1/8
Work done by tap B in 1 hour = ¼
Work done by (A + B) in 1 hour = (1/8 + ¼) = 3/8
Time taken (A + B) to fill the cistern = 8/3 hours = 2 hours 40 min.

What is the difference between simple interest and compound interest Why do you end up with more money with compound interest?

Why do you end up with more money with compound interest? Simple interest is interest paid only on the original investment whereas compound interest paid both on the original investment and on all interest that has been added to the original investment.

The population of a town was 176400 in the years 2009. It increases at the rate of 5% per annum. (i) What would be its population in the year 2011? (ii) What was its population in the year 2007?

Population in the year 2009 = 176400.
Rate of increase = 5% per annum.
(i) population in the year 2011 = population after 2 years
= 176400 <1 + 5/100>2
= 176400 <21/20 x 21/20>
= 194481.
Hence, the population of the town in 2011 would be 194481.
(ii) Let the population in the year 2007 be x. Then,
(ii) Let the population in the year 2007 be x. Then,
population in 2009 = x (1 + 5/100)2
x (21/20 x 21/20) = (441/400) x
(441/400) x = 176400
So, x = (176400 x 400)/441 = 160000
Hence, the population of the town in 2007 was 160000.


8.3: Solutions for Chapter 3

New International Version
“In that day,” declares the Sovereign LORD, “the songs in the temple will turn to wailing. Many, many bodies—flung everywhere! Silence!”

New Living Translation
In that day the singing in the temple will turn to wailing. Dead bodies will be scattered everywhere. They will be carried out of the city in silence. I, the Sovereign LORD, have spoken!”

English Standard Version
The songs of the temple shall become wailings in that day,” declares the Lord GOD. “So many dead bodies!” “They are thrown everywhere!” “Silence!”

Berean Study Bible
“In that day,” declares the Lord GOD, “the songs of the temple will turn to wailing. Many will be the corpses, strewn in silence everywhere!”

King James Bible
And the songs of the temple shall be howlings in that day, saith the Lord GOD: there shall be many dead bodies in every place they shall cast them forth with silence.

New King James Version
And the songs of the temple Shall be wailing in that day,” Says the Lord GOD— “Many dead bodies everywhere, They shall be thrown out in silence.”

New American Standard Bible
The songs of the palace will turn to wailing on that day,” declares the Lord GOD. “The corpses will be many in every place they will throw them out. Hush!”

NASB 1977
“The songs of the palace will turn to wailing in that day,” declares the Lord GOD. “Many will be the corpses in every place they will cast them forth in silence.”

Amplified Bible
In that day, the songs of the palace shall turn to wailing,” says the Lord GOD. “There will be many dead bodies in [sacred] silence they will throw them everywhere.”

Christian Standard Bible
In that day the temple songs will become wailing” —this is the Lord GOD’s declaration. “Many dead bodies, thrown everywhere! Silence! ”

Holman Christian Standard Bible
In that day the temple songs will become wailing"—this is the Lord GOD’s declaration. “Many dead bodies, thrown everywhere! Silence!”

American Standard Version
And the songs of the temple shall be wailings in that day, saith the Lord Jehovah: the dead bodies shall be many: in every place shall they cast them forth with silence.

Aramaic Bible in Plain English
And the praises of the temple shall fall in that day in that day, says The Lord of Lords, corpses shall abound in every place and they shall be cast to destruction

Brenton Septuagint Translation
And the ceilings of the temple shall howl in that day, saith the Lord God: there shall be many a fallen one in every place I will bring silence upon them.

Contemporary English Version
Instead of singing in the temple, they will cry and weep. Dead bodies will be everywhere. So keep silent! I, the LORD, have spoken!"

Douay-Rheims Bible
And the hinges of the temple shall screak in that day, saith the Lord God: many shall die: silence shall be cast in every place.

English Revised Version
And the songs of the temple shall be howlings in that day, saith the Lord GOD: the dead bodies shall be many in every place shall they cast them forth with silence.

Good News Translation
On that day the songs in the palace will become cries of mourning. There will be dead bodies everywhere. They will be cast out in silence."

GOD'S WORD® Translation
On that day the songs of the temple will become loud cries," declares the Almighty LORD. "There will be dead bodies scattered everywhere. Hush!"

International Standard Version
At that time," declares the Lord GOD, "the temple songs will be wailing. Many bodies will accumulate everywhere.

JPS Tanakh 1917
And the songs of the palace shall be wailings in that day, Saith the Lord GOD The dead bodies shall be many In every place silence shall be cast.

Literal Standard Version
And female singers have howled of a palace in that day,” A declaration of Lord YHWH, “Many [are] the carcasses, into any place throw—hush!”

NET Bible
The women singing in the temple will wail in that day." The sovereign LORD is speaking. "There will be many corpses littered everywhere! Be quiet!"

New Heart English Bible
The songs of the temple will be wailings in that day," says the LORD. "The dead bodies will be many. In every place they will throw them out with silence.

World English Bible
The songs of the temple will be wailings in that day," says the Lord Yahweh. "The dead bodies will be many. In every place they will throw them out with silence.

Young's Literal Translation
And howled have songstresses of a palace in that day, An affirmation of the Lord Jehovah, Many are the carcases, into any place throw -- hush!

Lamentations 2:10
The elders of the Daughter of Zion sit on the ground in silence. They have thrown dust on their heads and put on sackcloth. The young women of Jerusalem have bowed their heads to the ground.

Amos 5:16
Therefore this is what the LORD, the God of Hosts, the Lord, says: "There will be wailing in all the public squares and cries of 'Alas! Alas!' in all the streets. The farmer will be summoned to mourn, and the mourners to wail.

Amos 5:23
Take away from Me the noise of your songs! I will not listen to the music of your harps.

Amos 6:4
You lie on beds inlaid with ivory, and lounge upon your couches. You dine on lambs from the flock and calves from the stall.

Amos 6:5
You improvise songs on the harp like David and invent your own musical instruments.

Amos 6:8
The Lord GOD has sworn by Himself--the LORD, the God of Hosts, has declared: "I abhor Jacob's pride and detest his citadels, so I will deliver up the city and everything in it."

Amos 6:10
And when the relative who is to burn the bodies picks them up to remove them from the house, he will call to one inside, "Is anyone else with you?" "None," that person will answer. "Silence," the relative will retort, "for the name of the LORD must not be invoked."

And the songs of the temple shall be howlings in that day, said the Lord GOD: there shall be many dead bodies in every place they shall cast them forth with silence.

Amos 8:10 And I will turn your feasts into mourning, and all your songs into lamentation and I will bring up sackcloth upon all loins, and baldness upon every head and I will make it as the mourning of an only son, and the end thereof as a bitter day.

Amos 5:23 Take thou away from me the noise of thy songs for I will not hear the melody of thy viols.

Hosea 10:5,6 The inhabitants of Samaria shall fear because of the calves of Bethaven: for the people thereof shall mourn over it, and the priests thereof that rejoiced on it, for the glory thereof, because it is departed from it…

Amos 4:10 I have sent among you the pestilence after the manner of Egypt: your young men have I slain with the sword, and have taken away your horses and I have made the stink of your camps to come up unto your nostrils: yet have ye not returned unto me, saith the LORD.

Isaiah 37:36 Then the angel of the LORD went forth, and smote in the camp of the Assyrians a hundred and fourscore and five thousand: and when they arose early in the morning, behold, they were all dead corpses.

Jeremiah 9:21,22 For death is come up into our windows, and is entered into our palaces, to cut off the children from without, and the young men from the streets…

Amos 6:9,10 And it shall come to pass, if there remain ten men in one house, that they shall die…

Jeremiah 22:18 Therefore thus saith the LORD concerning Jehoiakim the son of Josiah king of Judah They shall not lament for him, saying, Ah my brother! or, Ah sister! they shall not lament for him, saying, Ah lord! or, Ah his glory!

Leviticus 10:3 Then Moses said unto Aaron, This is it that the LORD spake, saying, I will be sanctified in them that come nigh me, and before all the people I will be glorified. And Aaron held his peace.

Psalm 39:9 I was dumb, I opened not my mouth because thou didst it.


Ex 8.3 Class 7 Maths Question 1.
Tell what is the profit or loss in the following transactions. Also, find profit percent or loss percent in each case.
(a) Gardening shears bought for ₹ 250 and sold for ₹ 325.
(b) A refrigerator bought for ₹ 12,000 and sold at ₹ 13,500.
(c) A cupboard bought for ₹ 2,500 and sold at ₹ 3,000.
(d) A skirt bought for ₹ 250 and sold at ₹ 150.
Solution:


Ex 8.3 Class 7 Maths Question 2.
Convert each part of the ratio to percentage:
(a) 3 : 1
(b) 2 : 3 : 5
(c) 1 : 4
(d) 1 : 2 : 5
Solution:


Ex 8.3 Class 7 Maths Question 3.
The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Solution:
The original population of the city = 25000
The decreased population of the city = 24500
∴ Decrease in population = 25000 – 24500 = 500
∴ Percentage of decrease = ( × 100 ) % = 2%

Ex 8.3 Class 7 Maths Question 4.
Arun bought a car for ₹ 3,50,000. The next year, the price went up to ₹ 3,70,000. What was the percentage of the price increase?
Solution:
Original cost of the car = ₹ 3,50,000
Increased cost of the car = ₹ 3,70,000
∴ Increase in price = ₹ (370000 – 350000) = ₹ 20,000
∴ Percentage increase = ( × 100 )% = % = 5 %

Ex 8.3 Class 7 Maths Question 5.
I buy a T.V. for ₹ 10,000 and sell it at a profit of 20%. How much money do I get for it?
Solution:

Ex 8.3 Class 7 Maths Question 6.
Does Juhi sell a washing machine? 13,500. She loses 20% in the bargain. What was the price at which she bought it?
Solution:
We have, S.P. = ₹13,500 and loss = 20%.

Ex 8.3 Class 7 Maths Question 7.
(i) Chalk contains calcium, carbon, and oxygen in the ratio 10:3:12. Find the percentage of carbon in chalk.
(ii) If in a stick of chalk, the carbon in 3g, what is the weight of the chalk stick?
Solution:

Ex 8.3 Class 7 Maths Question 8.
Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?
Solution:

Ex 8.3 Class 7 Maths Question 9.
Find the amount to be paid at the end of 3 years in each case :
(a) Principal = ₹ 1,200 at 12% p.a.
(b) Principal = ₹ 7,500 at 5% p.a.
Solution:

Ex 8.3 Class 7 Maths Question 10.
What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in 2 years?
Solution:

Ex 8.3 Class 7 Maths Question 11.
If Meena gives an interest of ₹ 45 for one year at a 9% rate p.a. What is the sum she has borrowed?
Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities, drop a comment below and we will get back to you at the earliest.


MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

Question 1.
Calculate the amount and compound interest on
(a) ₹ 10,800 for 3 years at 12(frac<1><2>)% per annum compounded annually.
(b) ₹ 18,000 for 2(frac<1><2>) years at 10% per annum compounded annually.
(c) ₹ 62,500 for 1(frac<1><2>) years at 8% per annum compounded half yearly.
(d) ₹ 8,000 for 1 year at 9% per annum compounded half yearly.
(e) ₹ 10,000 for 1 year at 8% per annum compounded half yearly.
Solution:
(a) We have,

(b) We have,
P = ₹ 18000
R = 10 % per annum
n = 2(frac<1><2>) years or 2.5 years

Now, we calculate S.I. on this amount for (frac<1><2>) year at 10 % per annum.
∴ Amount after 2.5 years

∴ Interest = A – P = 22869 -18000 = ₹ 4869

(c) We have,
P = ₹ 62500
R = 8 % per annum = 4 % per half year
n = 1(frac<1><2>) years = 3 half years

∴ Interest = A – P = 70304 – 62500 = ₹ 7804

(d) We have,
P = ₹ 8000
R = 9 % per annum
= 4 % per half year
n = 1 year = 2 half years

∴ Interest = A – P = 8736.20 – 8000 = ₹ 736.20

(e) We have,
P = ₹ 10000
R = 8% per annum = 4 % per half year
n = 1 year = 2 half years

∴ Interest = A – P = 10816 -10000 = ₹ 816

Question 2.
Kamla borrowed ₹ 26,400 from a bank to buy a scooter at a rate of 15% per annum compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?
(Hint : Find 4 for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for (frac<4><12>) years).
Solution:
We have,
P = ₹ 26400
R = 15 % per annum
n = 2 years 4 months
At the end of 2 years,

Now, P = ₹ 34914
R = 15 % per annum
n = 4 months = (frac<1><3>) years
At the end of 2 years and 4 months,

Question 3.
Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?
Solution:
For Fabina, P = ₹ 12500
R = 12 % per annum
n = 3 years
For simple interest,

Interest = 17000 – 12500 = ₹ 4500
For Radha, P = ₹ 12500
R = 10 % per annum
n = 3 years
As this is compound interest

Interest = 16637.50 -12500 = ₹ 4137.50
Hence, Fabina pays more interest by 4500 – 4137.50 = ₹ 362.50

Question 4.
I borrowed ₹ 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?
Solution:
We have, P = ₹ 12000
R = 6 % per annum
n = 2 years

So, the extra amount he would have to pay = 1483.20 -1440 = ₹ 43.20

Question 5.
Vasudevan invested X 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get
(i) after 6 months?
(ii) after 1 year?
Solution:
We have,
P = ₹ 60000
R = 12 % per annum = 6 % per half year
(i) n = 6 months = 1 half year

Question 6.
Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after 1(frac<1><2>) years if the interest is
(i) compounded annually.
(ii) compounded half yearly.
Solution:
We have,
P = ₹ 80000
R = 10 % per annum = 5 % per half year
(i) If the interest is compounded annually

(ii) If interest is compounded half yearly

∴ Difference in amounts = 92610 – 92400
= ₹ 210

Question 7.
Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find
(i) The amount credited against her name at the end of the second year.
(ii) The interest for the 3rd year.
Solution:
We have,
P = ₹ 8000
R = 5 % per annum
(i) n = 2 years

(ii) n = 3 years

Hence, interest for 3 rd year = 9261 – 8820 = ₹ 441

Question 8.
Find the amount and the compound interest on ₹ 10,000 for 1(frac<1><2>) years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually? Solution:
We have,
P = ₹ 10000
R = 10 % per annum = 5 % per half year
n = 1(frac<1><2>) years = 3 half years
(i) If interest is compounded half yearly

Interest = 11576.25 – 10000 = ₹ 1576.25

(ii) If interest is compounded annually Amount after 1 year,

Thus, more interest would be generated if interest is calculated half yearly.

Question 9.
Find the amount which Ram will get on ₹ 4096, if he gave it for 18 months at 12(frac<1><2>) % per annum, interest being compounded half yearly.
Solution:
We have,
P = ₹ 4096
R = 12.5 % per annum = 6.25 % per half year
n = 18 months = 3 half years

Question 10.
The population of a place increased to 54,000 in 2003 at a rate of 5% per annum
(i) find the population in 2001.
(ii) what would be its population in 2005?
Solution:
We have, population in 2003 = 54000
Rate = 5% per annum
(i) Let population in 2001 be x.
The population in 2003

∴ Population = 48980 in 2001

(ii) For population in 2005
n = 2 years

Question 11.
In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.
Solution:
Initial count of bacteria = 506000
Rate = 2.5 % per hour
n = 2 hours
∴ Count after 2 hours

Question 12.
A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Solution:
Initial price = ₹ 42000
Rate of depreciation = 8% per annum
n = 1 year


Chapter 8, Problem Problem_8-3 32

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NCERT Solutions of class 12 Maths exercise 8.3 -Application of Integrals

These NCERT Solutions of class 12 Maths exercise 8.3 of chapter 8-Application of Integrals are created by an expert CBSE teacher. The NCERT solutions written here are the solutions of class 12 NCERT maths exercise 8.3 of chapter 8-Application of Integrals. We are hundred percent sure that you will like and understand the concept of maths used in exercise 8.3-Application of Integrals because each solution is created by a step by step way,however, if you face any difficulty in understanding the method used here,you can write and suggest us in the comment box.

In this exercise you will study the questions based on the enclosed areas by the curves and the lines.

Q1.Find the area enclosed by the curve whose equations are : y = 2x², x =2 and x-axis

The equation of the curve y = 2x², shows that it is passing through the origin and symmetric about y axis.

Q2. Find the area enclosed by the curve whose equations are :y = 5x 4 ,x=3.x=7, and x-axis

The equation of curve(prabola) y = 5x 4 shows that the curve is symmetric about y-axis and it is passing through the origin.

Q3.Find the area enclosed between the curve y²= 3x and line y = 6x.

Ans. We are given the curve y²= 3x and line y = 6x.

The equation of parabola shows that its vertex is at (0,0) and it is symmetric about the x-axis located at the right side of the y-axis

Solving both equations, we get point of intersections of line and parabola

(6x)² = 3x⇒ 36x² = 3x⇒ 12x = 1⇒ x = 1/12

So, as the equation of line shows that it passes through the origin and we have got the point of intersection of the line and the curve is (1/12, 1/2)

Area covered by the curve and the line = Area enclosed by the curve (y²= 3x) OAB – area of the triangle OAB

The area enclosed by the curve (y²= 3x) OAB

Area covered by the curve and the line

Q4. Find the area enclosed by the curve y = 2x² and the lines y = 1 and y = 3 and the y-axis.

Ans. We are given the curve y = 2x² shows that its vertex is at (0,0) and is symmetric about the y-axis and the given lines are y =1 and y = 3

Solving the equation of curve for x

The area enclosed by the curve and the lines y = 1 and y=3 is ar FABDE

It is symmetric about y-axis, so

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Solutions for Chapter 8.3: Estimating a Population Mean

Solutions for Chapter 8.3: Estimating a Population Mean

  • 8.3.55: Critical values What critical value t* from Table B would you use f.
  • 8.3.56: Critical values What critical value t* from Table B should be used .
  • 8.3.57: Pulling wood apart How heavy a load (pounds) is needed to pull apar.
  • 8.3.58: Weeds among the corn Velvetleaf is a particularly annoying weed in .
  • 8.3.59: Should we use t? Determine whether we can safely use a t* critical .
  • 8.3.60: Should we use t? Determine whether we can safely use a t* critical .
  • 8.3.61: Blood pressure A medical study finds that x = 114.9 and sx = 9.3 fo.
  • 8.3.62: Travel time to work A study of commuting times reports the travel t.
  • 8.3.63: Willows in Yellowstone Writers in some fields summarize data by giv.
  • 8.3.64: Blink When two lights close together blink alternately, we see one .
  • 8.3.65: Bone loss by nursing mothers Breast-feeding mothers secrete calcium.
  • 8.3.66: Reading scores in Atlanta The Trial Urban District Assessment (TUDA.
  • 8.3.67: Men and muscle Ask young men to estimate their own degree of body m.
  • 8.3.68: A big-toe problem A bunion on the big toe is fairly uncommon in you.
  • 8.3.69: Give it some gas! Computers in some vehicles calculate various quan.
  • 8.3.70: Vitamin C content Several years ago, the U.S. Agency for Internatio.
  • 8.3.71: aired tires Researchers were interested in comparing two methods fo.
  • 8.3.72: Water Trace metals found in wells affect the taste of drinking wate.
  • 8.3.73: Estimating BMI The body mass index (BMI) of all American young wome.
  • 8.3.74: The SAT again High school students who take the SAT Math exam a sec.
  • 8.3.75: Multiple choice: Select the best answer for Exercises 75 to 78. One.
  • 8.3.76: Multiple choice: Select the best answer for Exercises 75 to 78. You.
  • 8.3.77: Multiple choice: Select the best answer for Exercises 75 to 78. A q.
  • 8.3.78: Multiple choice: Select the best answer for Exercises 75 to 78. Sci.
  • 8.3.79: Watching TV (6.1, 7.3) Choose a young person (aged 19 to 25) at ran.
  • 8.3.80: Price cuts (4.2) Stores advertise price reductions to attract custo.
Textbook: The Practice of Statistics
Edition: 5
Author: Daren S. Starnes, Josh Tabor
ISBN: 9781464108730

The Practice of Statistics was written by and is associated to the ISBN: 9781464108730. Chapter 8.3: Estimating a Population Mean includes 26 full step-by-step solutions. Since 26 problems in chapter 8.3: Estimating a Population Mean have been answered, more than 107655 students have viewed full step-by-step solutions from this chapter. This expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: The Practice of Statistics, edition: 5.

A method of decomposing the total variability in a set of observations, as measured by the sum of the squares of these observations from their average, into component sums of squares that are associated with speciic deined sources of variation

The arithmetic mean of a set of numbers x1 , x2 ,…, xn is their sum divided by the number of observations, or ( / )1 1 n xi t n ? = . The arithmetic mean is usually denoted by x , and is often called the average

A qualitative characteristic of an item or unit, usually arising in quality control. For example, classifying production units as defective or nondefective results in attributes data.

An effect that systematically distorts a statistical result or estimate, preventing it from representing the true quantity of interest.

In experimental design, a group of experimental units or material that is relatively homogeneous. The purpose of dividing experimental units into blocks is to produce an experimental design wherein variability within blocks is smaller than variability between blocks. This allows the factors of interest to be compared in an environment that has less variability than in an unblocked experiment.

The portion of the variability in a set of observations that is due to only random forces and which cannot be traced to speciic sources, such as operators, materials, or equipment. Also called a common cause.

A subset selected without replacement from a set used to determine the number of outcomes in events and sample spaces.

A probability distribution for a continuous random variable.

A two-dimensional graphic used for a bivariate probability density function that displays curves for which the probability density function is constant.

A square matrix that contains the correlations among a set of random variables, say, XX X 1 2 k , ,…, . The main diagonal elements of the matrix are unity and the off-diagonal elements rij are the correlations between Xi and Xj .

A measure of association between two random variables obtained as the expected value of the product of the two random variables around their means that is, Cov(X Y, ) [( )( )] =? ? E X Y ? ? X Y .

For a random variable X, the function of X deined as PX x ( ) ? that is used to specify the probability distribution.

A quality tool that graphically shows the location of defects on a part or in a process.

The number of independent comparisons that can be made among the elements of a sample. The term is analogous to the number of degrees of freedom for an object in a dynamic system, which is the number of independent coordinates required to determine the motion of the object.

The error sum of squares divided by its number of degrees of freedom.

The numerical value of a point estimator.

A property of a collection of events that indicates that their union equals the sample space.

A series of tests in which changes are made to the system under study

In general, the agreement of a set of observed values and a set of theoretical values that depend on some hypothesis. The term is often used in itting a theoretical distribution to a set of observations.


27.5.2.1. Running testing client

You can run the testing client using the TestClient command-line parameter for client application startup. You can also specify the number of the TCP port to be used for testing manager – testing client communication. To do this, use the TPort option of the TestClient parameter. If no value is entered for this option, 1538 is used. This option must be specified if multiple testing clients run on a single computer.

So startup command line for the client application that is to be used as the testing client looks as shown below:

1cv8c ENTERPRISE /IBName "Test Application" /TestClient -TPort 1843

In this example, the thin client ( 1cv8c ) is used as the testing client for the Test Application infobase, and TCP-port 1843 is used for communicating with the testing manager.

Therefore, if you want to connect to the testing client represented by the thin or thick client, you should know two parameters: the IP-address (or computer name) where the testing client is running, and number of the TCP-port to be used for communicating.

Running a web client is a little more complicated. In addition to the TestClient option, you should specify another startup command-line option: the TestClientID<ID> that contains a unique identifier for the tested client application. If the web client is used, unique identification of the testing client requires three parameters:

„ Address of the web server where the tested application is deployed. You do not have to know the infobase address.

„ The number of the TCP-port to be used by the web server to transfer data between the testing manager and the testing client.

„ The identifier of a specific instance of the tested application running on the web client.

In other words, a single web server can work with multiple infobases and with multiple connections to a single infobase (also for testing purposes), but for the web server you have to specify which TCP-ports are to be used for communicating with the testing manager. You can do this in the testcfg.xml file stored in the web server extension settings directory. For details about this file, see "1C:Enterprise 8.3. Administrator Manual".


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Watch the video: Compound Interest: NCERT Exercise Question 4 - Comparing Quantities. Class 8 Maths (December 2021).