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Dmitry [639]
3 years ago
12

A bank sampled its customers to determine the proportion of customers who use their debit card at least once each month. A sampl

e of 50 customers found that only 12 use their debit card monthly. Find a 99% confidence interval for the proportion of customers who use their debit card monthly.
Mathematics
1 answer:
Zinaida [17]3 years ago
3 0

Answer:

(0.084,0.396)

Step-by-step explanation:

The 99% confidence interval for the proportion of customers who use debit card monthly can be constructed as

p-z_{\frac{\alpha }{2}} \sqrt{\frac{pq}{n} }

p=\frac{x}{n}

p=\frac{12}{50}

p=0.24

q=1-p=1-0.24=0.76

\frac{\alpha }{2} =\frac{\0.01 }{2}=0.005

p-z_{\frac{\alpha }{2}} \sqrt{\frac{pq}{n} }

0.24-z_{0.005} \sqrt{\frac{0.24*0.76}{50} }

0.24-2.58(0.0604)

0.24-0.155832

By rounding to three decimal places we get,

0.084

The 99% confidence interval for the proportion of customers who use debit card monthly is (0.084,0.396).

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Answer:

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Step-by-step explanation:

The question is poorly formatted. The original question is:

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7

We have:

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7

Open bracket

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(8^ \frac{2}{3} *x^{3* \frac{2}{3}}) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}

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Express 8 as 2^3

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Express 2^3 as 8

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Expand each exponent

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Split

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{25x^2} *\sqrt{3x} + 8\sqrt{3x} * \sqrt{x^6}

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Factorize

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