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finlep [7]
3 years ago
10

Juanita is 13 years older than her cousin. The sum of their ages is no less than 103 years. Enter an inequality that can be used

to represent this situation in the first box, where x represents Juanita's cousin's age.
Mathematics
1 answer:
Bond [772]3 years ago
6 0
Please see below solution:

y = x + 13<span>
y + x ≥ 103
x + 13 + x ≥ 103
Inequality is:
2 x + 13 ≥ 103
2 x ≥ 103 - 13
2 x ≥ 90
x ≥ 90 : 2
x ≥ 45
<span>The youngest age Juanita`s cousin can be is 45.</span></span>

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a) 25

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Step-by-step explanation:

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

\alpha = \frac{1-0.95}{2} = 0.025

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

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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}

n = 66.7

Rounding up, 67

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

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0.05\sqrt{n} = 1.96*0.25

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

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Rounding up, 97

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