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weqwewe [10]
3 years ago
8

Given y = a(x - h)^2 + k, p = A. a B. 4a C. 1/4a

Mathematics
1 answer:
Flauer [41]3 years ago
8 0

9514 1404 393

Answer:

  C.  1/(4a)

Step-by-step explanation:

We assume you're comparing the vertex form ...

  y = a(x -h)^2 +k

to the form used to write the equation in terms of the focal distance p.

  y = 1/(4p)(x -h)^2 +k

That comparison tells you ...

  a = 1/(4p)

  p = 1/(4a) . . . . . . multiply by p/a; matches choice C

__

<em>Additional comment</em>

When using plain text to write a rational expression, parentheses are needed around any denominator that has is more than a single constant or variable. The order of operations requires 1/4a to be interpreted as (1/4)a. The value of p is 1/(4a).

When rational expressions are typeset, the fraction bar serves as a grouping symbol identifying the entire denominator:

  p=\dfrac{1}{4a}

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In a survey of 1005 adult Americans, 46.6% indicated that they were somewhat interested or very interested in having web access
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Answer:

No, the marketing manager was not correct in his claim.

Step-by-step explanation:

We are given that in a survey of 1005 adult Americans, 46.6% indicated that they were somewhat interested or very interested in having web access in their cars.

Suppose that the marketing manager of a car manufacturer claims that the 46.6% is based only on a sample and that 46.6% is close to half, so there is no reason to believe that the proportion of all adult Americans who want car web access is less than 0.50.

<em>Let p = population proportion of all adult Americans who want car web access</em>

SO, Null Hypothesis, H_0 : p \geq 50%   {means that the proportion of all adult Americans who want car web access is more than or equal to 0.50}

Alternate Hypothesis, H_a : p < 50%  {means that the proportion of all adult Americans who want car web access is less than 0.50}

The test statistics that will be used here is<u> One-sample z proportion statistics</u>;

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

where,  \hat p = sample proportion of Americans who indicated that they were somewhat interested or very interested in having web access in their cars =  46.6%

            n = sample of Americans = 1005

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<em>Since in the question we are not given the level of significance so we assume it to be 5%. Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z so we sufficient evidence to reject our null hypothesis as it will fall in the rejection region.</em>

Therefore, we conclude that the proportion of all adult Americans who want car web access is less than 0.50 which means the marketing manager was not correct in his claim.

3 0
3 years ago
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