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3 years ago

Answer the question.

Mathematics

1 answer:

Anettt [7]3 years ago

4 0

Answer:

wat give the question first so tat we can answer

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90x-71 should be correct

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3 years ago

Find the slope of the line through each pair of points (-10,17) , (-8,-1)

svet-max [94.6K]

Slope = (y2 - y1)/(x2 - x1) = (-1 - 17)/(-8 - (-10)) = -18/(-8 + 10) = -18/2 = -9

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3 years ago

A recent CBS News survey reported that 64% of adults felt the U.S. Treasury should continue making pennies. Suppose we select a

enyata [817]

Complete Question

A recent CBS News survey reported that 64% of adults felt the U.S. Treasury should continue making pennies. Suppose we select a sample of 18 adults.

a-1. How many of the 18 would we expect to indicate that the Treasury should continue making pennies

a-2) What is the standard deviation?

a-3) What is the likelihood that exactly 3 adults would indicate the Treasury should continue making pennies?

Answer:

a-1 $\= x = 11 .52$

a-2 $\sigma = 2.036$

a-3 $P(3) = 4.7*10^{-5}$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 18$

The proportion of adult that felt the U.S. Treasury should continue making pennies is p = 0.64

The proportion of adult that feel otherwise is

$q = 1- p = 1-0.64 = 0.36$

The mean is mathematically evaluated as

$\= x = n * p$

substituting values

$\= x = 18 * 0.64$

$\= x = 11 .52$

The standard deviation is mathematically represented as

$\sigma = \sqrt{ npq}$

substituting values

$\sigma = \sqrt{18 * 0.64 * 0.36}$

$\sigma = 2.036$

The likelihood that 3 adult would indicate the Treasury should continue making pennies is mathematically evaluated as

$P(3) = \left n} \atop \right. C_3 (p)^{3} * (q)^{n-3}$

Now

$\left n} \atop \right. C_3 = \frac{n! }{[n-3] ! 3!}$

substituting values

$\left n} \atop \right. C_3 = \frac{18! }{[15] ! 3!}$

$\left n} \atop \right. C_3 = \frac{18 * 17 * 16 * 15! }{[15] ! (3 *2 *1 )}$

$\left n} \atop \right. C_3 = \frac{18 * 17 * 16 }{ (3 *2 *1 )}$

$\left n} \atop \right. C_3 = 816$

$P(3) = 816 * (0.64 )^3 * (0.36 )^{18-3}$

$P(3) = 4.7*10^{-5}$

8 0

3 years ago

The average number of calories in a 1.5-ounce chocolate bar is 225. Suppose that the distribution of calories is approximately n

s344n2d4d5 [400]

Answer:

The probability that a randomly selected chocolate bar will have a. Between 200 and 220 calories will be 0.3023

Step-by-step explanation:

Given:

True mean=225

S.D=10

X1=200 and X2=220

To Find:

P(X1<x<X2)

Solution:

BY using Z-table for probability and Z-score we can proceed.

For 200 calories Z will be ,

Z=(sample mean -true mean)/S.D

=200-225/10

=-25/10

=-2.5

For 220 calories Z will be ,

Z=220-225/10

=-5/10

=-0.5

So required probability will be

$P(-2.5$

$=P(-2.5$

$=P(0>z>2.5)-P(0>z>0.5)$

$=0.4938-0.1915$

=0.3023

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3 years ago

Out of five men and five women, we form a committee consisting of four different people. Assuming that each committee of size fo

tekilochka [14]

An44swer:

Step-by-step explanation:

7 0

3 years ago