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maks197457 [2]
2 years ago
12

Tell whether each graph is a function and justify your answer. Which graph is not a good representation of a​ real-world situati

on? Explain.

Mathematics
1 answer:
mina [271]2 years ago
5 0

Using the function concept, it is found that:

  • Graph B is a function because each value of x corresponds to exactly one y-value.

In a function, <u>one value of the input can be related to only one value of the output</u>.

  • In a graph, it means that for each value of x(horizontal axis), there can be only one respective value of y(vertical axis).

In this problem, at Graph A, when x = 5, for example a vertical line crosses the function 3 times, hence there are 3 respective values of y for x = 5, and the same is valid for other values of x, hence it is not a function.

At Graph B, <u>for each value of x, there is only one value of y</u>, hence it is a function.

Hence:

Graph B is a function because each value of x corresponds to exactly one y-value.

To learn more about the function concept, you can take a look at brainly.com/question/12463448

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The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college f
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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

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

0.37±1.648*0.015

[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} }

0.48±1.648*0.016

[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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