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Vilka [71]
4 years ago
11

Translate the following written expression into a mathematical expression. Then, find the expression's value when e = 65 and f =

37.
"The difference of e and f, all multiplied by two, increased by four"

A.
56
B.
52
C.
60
D.
97
Mathematics
2 answers:
mote1985 [20]4 years ago
6 0
First write out the problem with the given variables:

(E-F)2+4

Next plug in your given values:
(65-37)2+4

Solve:
(28)2+4
56+4
60
joja [24]4 years ago
3 0

Answer:

Try A. because it stands out to me the most and that's what my gut is telling me to go with. So I hope that helps.

Step-by-step explanation:

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Researchers believe that 24.5% of the adults in the United States are obese. A Gallup poll examined the rate of obesity in Ameri
Setler79 [48]

Answer:

The p-value of the test is 0 < 0.05(standard significance level), which means that this is enough evidence to state that more than 24.5% of all adults in the U.S. are obese

Step-by-step explanation:

Researchers believe that 24.5% of the adults in the United States are obese. Test if there is enough evidence to state that more than 24.5% of all adults in the U.S. are obese.

At the null hypothesis, we test if the proportion is of 24.5% or less, that is:

H_0: p \leq 0.245

At the alternative hypothesis, we test if this proportion is greater than 24.5%, that is:

H_1: p > 0.245

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.245 is tested at the null hypothesis:

This means that \mu = 0.245, \sigma = \sqrt{0.245*0.755}

Survey 86,664 randomly sampled U.S. adults. Of the adults surveyed, 23,053 said that they were obese.

This means that n = 86664, X = \frac{23053}{86664} = 0.266

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.266 - 0.245}{\frac{\sqrt{0.245*0.755}}{\sqrt{86664}}}

z = 14.37

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion of 0.266 or higher, which is 1 subtracted by the p-value of z = 14.37.

The p-value of z = 14.37 is 1.

1 - 1 = 0

The p-value of the test is 0 < 0.05(standard significance level), which means that this is enough evidence to state that more than 24.5% of all adults in the U.S. are obese

4 0
3 years ago
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