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Masja [62]
3 years ago
9

Consider the function ƒ(x) = (x + 1)2 – 1. Which of the following functions stretches ƒ(x) vertically by a factor of 4?

Mathematics
1 answer:
laila [671]3 years ago
5 0

Answer:

C f(x) = 4(x+1)2-1

Step-by-step explanation:

factor of 4 = 2^2

(x+1)2-1 = 4(x+1) 2-1 = with x

           =   4(+1) 2-1  = without x

           =  (4 - 4) 2 = individual products of -1

           =  (8 - 8 ) = individual products of 2

           =   8 - 8  = 2^2 -2^2

           =  2^2 - 2^2

(x+1)2-1 = 4(x+1)2-1 = with x

            = 2x^2 -2^2

          -x = 2^2 -2^2

            x = -2^2-2^2

            x = 4

which proves f(x) is a factor of 4

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Answer:

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We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

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Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample. In this problem, \sigma = 0.25

(a) The desired margin of error is $0.10.

This is n when M = 0.1. So

M = z*\frac{\sigma}{\sqrt{n}}

0.1 = 1.96*\frac{0.25}{\sqrt{n}}

0.1\sqrt{n} = 1.96*0.25

\sqrt{n} = \frac{19.6*0.25}{0.1}

(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.1})^{2}

n = 24.01

Rounding up to the nearest whole number, 25.

(b) The desired margin of error is $0.06.

This is n when M = 0.06. So

M = z*\frac{\sigma}{\sqrt{n}}

0.06 = 1.96*\frac{0.25}{\sqrt{n}}

0.06\sqrt{n} = 1.96*0.25

\sqrt{n} = \frac{19.6*0.25}{0.06}

(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.06})^{2}

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Rounding up, 67

(c) The desired margin of error is $0.05.

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M = z*\frac{\sigma}{\sqrt{n}}

0.05 = 1.96*\frac{0.25}{\sqrt{n}}

0.05\sqrt{n} = 1.96*0.25

\sqrt{n} = \frac{19.6*0.25}{0.05}

(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.05})^{2}

n = 96.04

Rounding up, 97

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