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jekas [21]
2 years ago
9

12 is what percent of 96

Mathematics
2 answers:
VikaD [51]2 years ago
7 0
Let us arrange an equation #

12 = x % of 96.

12 = (x/100) * 96.

12 = 96x/100              Cross Multiply.

12 * 100 = 96x

96x = 12 *100

x = (12*100)/96.    Cancel out or use your calculator, or 12 cancels itself is 1,
                           and into 96 is 8, so you have (100/8).
x = 100/8.            4 into 100 is 25 and into 8 is 2.

x = 25/2.
                          
x = 12.5

Therefore 12 is 12.5% of 96.  That's it.
Soloha48 [4]2 years ago
5 0
The answer is 12 is 12.5% of 96
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A, B, and C are points of tangency. CQ = 5, PQ = 10, and PR = 14. What is the perimeter
igomit [66]

*see attachment for diagram

Answer:

Perimeter = 38

Step-by-step explanation:

Recall: when two tangents are drawn to meet at a point outside a circle, the segments of the two tangents are congruent.

Given,

CQ = 5

PQ = 10

PR = 14

Perimeter of ∆PQR = RC + CQ + QB + BP + PA + AR

CQ = QB = 5 (tangents drawn from an external point)

BP = PQ - QB

BP = 10 - 5 = 5

BP = PA = 5 (tangents drawn from an external point)

AR = PR - PA

AR = 14 - 5 = 9

AR = RC = 9 (tangents drawn from an external point)

✔️Perimeter of ∆PQR = RC + CQ + QB + BP + PA + AR

= 9 + 5 + 5 + 5 + 5 + 9

Perimeter = 38

7 0
3 years ago
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

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