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Rama09 [41]
3 years ago
15

Carol is one-sixth of an inch taller than dave.dave is fifty-eight and a half inches tall.how tall is carol

Mathematics
1 answer:
katrin2010 [14]3 years ago
5 0
C=1/6 of x
D=55/1/2
1/6 of 111/2=110.167
55.08
=55 inches
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Usually it means to solve it. Sometimes they give you values to substitute in for variables (letters) in the math sentence/equation/expression.

Does that make sense?

Step-by-step explanation:

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The National Center for Health Statistics reported that of every 883 deaths in recent years, 24 resulted from an automobile acci
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There is a 2.7% chance that  a particular death is due to an automobile accident?

Step-by-step explanation:

Automobile _____ 24 deaths

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3 years ago
Suppose a random sample of 100 observations from a binomial population gives a value of pˆ = .63 and you wish to test the null h
irakobra [83]

Answer:

We conclude that the population proportion is equal to 0.70.

Step-by-step explanation:

We are given that a random sample of 100 observations from a binomial population gives a value of pˆ = 0.63 and you wish to test the null hypothesis that the population parameter p is equal to 0.70 against the alternative hypothesis that p is less than 0.70.

Let p = <u><em>population proportion.</em></u>

(1) The intuition tells us that the population parameter p may be less than 0.70 as the sample proportion comes out to be less than 0.70 and also the sample is large enough.

(2) So, Null Hypothesis, H_0 : p = 0.70      {means that the population proportion is equal to 0.70}

Alternate Hypothesis, H_A : p < 0.70      {means that the population proportion is less than 0.70}

The test statistics that would be used here <u>One-sample z-test</u> for proportions;

                           T.S. =  \frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }  ~  N(0,1)

where, \hat p = sample proportion = 0.63

            n = sample of observations = 100

So, <u><em>the test statistics</em></u>  =  \frac{0.63-0.70}{\sqrt{\frac{0.70(1-0.70)}{100} } }

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The value of z-test statistics is -1.528.

<u>Now at 0.05 level of significance, the z table gives a critical value of -1.645 for the left-tailed test.</u>

Since our test statistics is more than the critical value of z as -1.528 > -1.645, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.

Therefore, we conclude that the population proportion is equal to 0.70.

(c) The observed level of significance in part B is 0.05 on the basis of which we find our critical value of z.

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