Name and locate points in all four quadrants of the coordinate plane.

If you inflate a beach ball with 14,130 cubic inches of air, what will its radius be? Use 3.14 for π. Round to the nearest whole

GREYUIT [131]

Answer: 15

Step-by-step explanation:

The answer is because if you divide 14,130 by 3.14. You'll get 4500 and then divide that by 300 and your finally is 15

8 0

3 years ago

Triangle ABC has been rotated 90° to create triangle DEF. Write the equation, in slope-intercept form, of the side of triangle A

Ket [755]

the equation in the slope-intercept of the side of triangle ABC that is perpendicular to segment EF is y = x + 1

<h3>How to determine the equation</h3>

From the figure given, we can deduce the coordinates of the sides

For A

A ( 4,2)

For B

B ( 4, 5)

C ( 1, 2)

D ( 2, -4 )

E ( 5, -4)

F ( 2, -1)

The slope for BC

Slope = $\frac{y2 - y1}{x2 - x1}$

Substitute the values for both B and C coordinates, we have

Slope = $\frac{2- 5}{1 - 4}$

Find the difference for both the numerator and denominator

Slope = $\frac{-3}{-3}$

Slope = 1

We have the rotation for both point ( 0, 1)

y - y1 = m ( x - x1)

The values for y1 and x1 are 1 and 0 respectively and the slope m is 1

Substitute the values

y - 1 = 1 ( x - 0)

y - 1 = x

Make 'y' the subject of formula

y = x + 1

Thus, the equation in the slope-intercept of the side of triangle ABC that is perpendicular to segment EF is y = x + 1

Learn more about linear graphs here:

brainly.com/question/4074386

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4 0

2 years ago

A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$