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kiruha [24]
3 years ago
7

Cindy works two jobs to make ends meet. Both jobs pay her an hourly wage. At the first job, Cindy earns $6.50 per hour. At the s

econd job, she earns $10.25 per hour. Cindy works twice as many hours at the first job as she does at the second. Cindy earned $372 last week. Which of the following systems of equations correctly models this situation?
Mathematics
1 answer:
notsponge [240]3 years ago
3 0
The answer is B

6.5x + 1025y= 372
x= 2y
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99% confidence interval for the population proportion of employed individuals is [0.104 , 0.224].

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So, 99% confidence interval for the population proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5%

                                             level of significance are -2.5758 & 2.5758}  

P(-2.5758 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 2.5758) = 0.99

P( -2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

P( \hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

<em>99% confidence interval for p</em> = [\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

= [ 0.164-2.5758 \times {\sqrt{\frac{0.164(1-0.164)}{250} } } , 0.164+2.5758 \times {\sqrt{\frac{0.164(1-0.164)}{250} } } ]

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Therefore, 99% confidence interval for the population proportion of employed individuals who work at home at least once per week is [0.104 , 0.224].

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