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vaieri [72.5K]
1 year ago
12

Dave has 25 3/4 feet of piping and needs to cut it in 1 1/2 foot measurements for a project. How many pieces of piping will he e

nd up with?
Mathematics
1 answer:
Ostrovityanka [42]1 year ago
5 0

Answer: 17 1/6 or  17-18

(u guess which is right if its an integer ^^)

Step-by-step explanation:

25 3/4 ÷ 1 1/2

First change the mixed fractions to improper fractions.

<u>25 3/4 </u>= <u>103/4</u>

<u>1 1/2 </u>= <u>3/2</u>

Now divide: 103/4 ÷ 3/2

103/4 * 2/3 (multiply by flipping the second fraction)

= 206/12

Now divide: 206 ÷ 12

= 17 2/12 (<u>simplified:</u> 17 1/6)

<u><em>The answer will be 17 1/6 </em></u> (OR 17-18)

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Hello!

There are two variables of interest:

X₁: number of college freshmen that carry a credit card balance.

n₁= 1000

p'₁= 0.37

X₂: number of college seniors that carry a credit card balance.

n₂= 1000

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a. You need to construct a 90% CI for the proportion of freshmen  who carry a credit card balance.

The formula for the interval is:

p'₁±Z_{1-\alpha /2}*\sqrt{\frac{p'_1(1-p'_1)}{n_1} }

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0.37±1.648*\sqrt{\frac{0.37*0.63}{1000} }

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[0.35;0.39]

With a confidence level of 90%, you'd expect that the interval [0.35;0.39] contains the proportion of college freshmen students that carry a credit card balance.

b. In this item, you have to estimate the proportion of senior students that carry a credit card balance. Since we work with the standard normal approximation and the same confidence level, the Z value is the same: 1.648

The formula for this interval is

p'₂±Z_{1-\alpha /2}*\sqrt{\frac{p'_2(1-p'_2)}{n_2} }

0.48±1.648* \sqrt{\frac{0.48*0.52}{1000} }

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[0.45;0.51]

With a confidence level of 90%, you'd expect that the interval [0.45;0.51] contains the proportion of college seniors that carry a credit card balance.

c. The difference between the width two 90% confidence intervals is given by the standard deviation of each sample.

Freshmen: \sqrt{\frac{p'_1(1-p'_1)}{n_1} } = \sqrt{\frac{0.37*0.63}{1000} } = 0.01527 = 0.015

Seniors: \sqrt{\frac{p'_2(1-p'_2)}{n_2} } = \sqrt{\frac{0.48*0.52}{1000} }= 0.01579 = 0.016

The interval corresponding to the senior students has a greater standard deviation than the interval corresponding to the freshmen students, that is why the amplitude of its interval is greater.

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