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Alex17521 [72]
3 years ago
7

A music competition on television had five elimination rounds. After each elimination, only half of the contestants were sent to

the next round. The table below shows the number of contestants in each round of the competition: x 1 2 3 4 5 f(x) 64 32 16 8 4 Compute the average rate of change of f(x) from x = 3 to x = 5 and describe what it represents. −6 contestants per round, and it represents the number of contestants who will reach the final round −28 contestants per round, and it represents the number of contestants who will reach the final round −6 contestants per round, and it represents the average rate at which the number of contestants decreased from the third round to the fifth round −28 contestants per round, and it represents the average rate at which the number of contestants decreased from the third round to the fifth round
Mathematics
2 answers:
Luden [163]3 years ago
8 0
M = (Y-y)/(X-x)
m = (4-16)/(5-3)
m = -12/2
m = -6
<span>it represents the average rate at which the number of contestants decreased from the third round to the fifth round </span>
mafiozo [28]3 years ago
3 0
So if you guys did not catch that that would be C
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Answer:

Step-by-step explanation:

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

p'₂= 0.48

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} }

Z_{1-\alpha /2}= Z_{0.95}= 1.648

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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

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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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