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2 years ago

Please help meee I need this by today please help me hellpppp

Mathematics

2 answers:

AfilCa [17]2 years ago

5 0

Answer:

A. 3x + 1.50y the variables are x and y

B. If x = 73 wristbands and 47 buttons then:

3(47) + 1.50(47)

141 + 70.5 = 211.5

Step-by-step explanation:

Igoryamba2 years ago

4 0

4.
A. 3x + 1.50y the variables are x and y
B. If x = 73 wristbands and 47 buttons then:
3(47) + 1.50(47)
141 + 70.5 = 211.5
There you go!!

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2 years ago

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The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).

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Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence interval $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

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In the study 340 babies were born, and 289 of them were girls. This means that $n = 340, \pi = \frac{289}{340} = 0.85$

Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born.

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{340}} = 0.85 - 2.575\sqrt{\frac{0.85*0.15}{400}} = 0.80$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{340}} = 0.85 + 2.575\sqrt{\frac{0.85*0.15}{400}} = 0.90$

The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).

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b)Yes, the proportion of girls is significantly different from 0.5.

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2 years ago