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dybincka [34]
2 years ago
15

These triangles are scaled copies of each other.

Mathematics
1 answer:
Mariana [72]2 years ago
3 0

Answer:

a. 4

b. 1/4

c. 16

d. 1/9

e. 4/9

f. 9/16

Step-by-step explanation:

The ratio of the areas is the square of the ratio of the lengths of the sides.

a. Triangles G and F

Select a side in triangle G and the corresponding side in triangle F:

side in F: 10

corresponding side in G: 5

ratio of lengths of F to G = 10/5 = 2

ratio of areas of G to F: (2)^2 = 4

b. Triangles G and B

Select a side in triangle G and the corresponding side in triangle B:

side in G: 5

corresponding side in B: 5/2

ratio of lengths of B to G = (5/2)/5 = 1/2

ratio of areas of B to G: (1/2)^2 = 1/4

c. Triangles B and F

Select a side in triangle B and the corresponding side in triangle F:

side in B: 5/2

corresponding side in F: 10

ratio of lengths of F to B = 10/(5/2) = 4

ratio of areas of F to B: (4)^2 = 16

Do the same for the other 3 pairs of triangles.

The answers are:

d. 1/9

e. 4/9

f. 9/16

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Canadians who visit the United States often buy liquor and cigarettes, which are much cheaper in the United States. However, the
fenix001 [56]

Answer:

(a): Marginal pmf of x

P(0) = 0.72

P(1) = 0.28

(b): Marginal pmf of y

P(0) = 0.81

P(1) = 0.19

(c): Mean and Variance of x

E(x) = 0.28

Var(x) = 0.2016

(d): Mean and Variance of y

E(y) = 0.19

Var(y) = 0.1539

(e): The covariance and the coefficient of correlation

Cov(x,y) = 0.0468

r \approx 0.2657

Step-by-step explanation:

Given

<em>x = bottles</em>

<em>y = carton</em>

<em>See attachment for complete question</em>

<em />

Solving (a): Marginal pmf of x

This is calculated as:

P(x) = \sum\limits^{}_y\ P(x,y)

So:

P(0) = P(0,0) + P(0,1)

P(0) = 0.63 + 0.09

P(0) = 0.72

P(1) = P(1,0) + P(1,1)

P(1) = 0.18 + 0.10

P(1) = 0.28

Solving (b): Marginal pmf of y

This is calculated as:

P(y) = \sum\limits^{}_x\ P(x,y)

So:

P(0) = P(0,0) + P(1,0)

P(0) = 0.63 + 0.18

P(0) = 0.81

P(1) = P(0,1) + P(1,1)

P(1) = 0.09 + 0.10

P(1) = 0.19

Solving (c): Mean and Variance of x

Mean is calculated as:

E(x) = \sum( x * P(x))

So, we have:

E(x) = 0 * P(0)  + 1 * P(1)

E(x) = 0 * 0.72  + 1 * 0.28

E(x) = 0   + 0.28

E(x) = 0.28

Variance is calculated as:

Var(x) = E(x^2) - (E(x))^2

Calculate E(x^2)

E(x^2) = \sum( x^2 * P(x))

E(x^2) = 0^2 * 0.72 + 1^2 * 0.28

E(x^2) = 0 + 0.28

E(x^2) = 0.28

So:

Var(x) = E(x^2) - (E(x))^2

Var(x) = 0.28 - 0.28^2

Var(x) = 0.28 - 0.0784

Var(x) = 0.2016

Solving (d): Mean and Variance of y

Mean is calculated as:

E(y) = \sum(y * P(y))

So, we have:

E(y) = 0 * P(0)  + 1 * P(1)

E(y) = 0 * 0.81  + 1 * 0.19

E(y) = 0+0.19

E(y) = 0.19

Variance is calculated as:

Var(y) = E(y^2) - (E(y))^2

Calculate E(y^2)

E(y^2) = \sum(y^2 * P(y))

E(y^2) = 0^2 * 0.81 + 1^2 * 0.19

E(y^2) = 0 + 0.19

E(y^2) = 0.19

So:

Var(y) = E(y^2) - (E(y))^2

Var(y) = 0.19 - 0.19^2

Var(y) = 0.19 - 0.0361

Var(y) = 0.1539

Solving (e): The covariance and the coefficient of correlation

Covariance is calculated as:

COV(x,y) = E(xy) - E(x) * E(y)

Calculate E(xy)

E(xy) = \sum (xy * P(xy))

This gives:

E(xy) = x_0y_0 * P(0,0) + x_1y_0 * P(1,0) +x_0y_1 * P(0,1) + x_1y_1 * P(1,1)

E(xy) = 0*0 * 0.63 + 1*0 * 0.18 +0*1 * 0.09 + 1*1 * 0.1

E(xy) = 0+0+0 + 0.1

E(xy) = 0.1

So:

COV(x,y) = E(xy) - E(x) * E(y)

Cov(x,y) = 0.1 - 0.28 * 0.19

Cov(x,y) = 0.1 - 0.0532

Cov(x,y) = 0.0468

The coefficient of correlation is then calculated as:

r = \frac{Cov(x,y)}{\sqrt{Var(x) * Var(y)}}

r = \frac{0.0468}{\sqrt{0.2016 * 0.1539}}

r = \frac{0.0468}{\sqrt{0.03102624}}

r = \frac{0.0468}{0.17614266944}

r = 0.26569371378

r \approx 0.2657 --- approximated

8 0
2 years ago
The next three terms of the sequence -2, 4, -8, 16 is -32, 60, -128<br> True or False
jekas [21]
False
common difference of this arithmetic sequence is -2
so next 3 terms are : -32, 64, -128
8 0
3 years ago
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