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2 years ago

Please help which of the following statements could be a function?

Mathematics

1 answer:

iragen [17]2 years ago

7 0

Answer:

The first one height vs shoe size.

Step-by-step explanation:

Let's evaluate each choice

1) Generally, it is possible that the taller a person is, that his/her shoe size would generally get bigger also. Hence we can say that it is possible that there is a function correlation between height and shoe size.

2) hair length can be cut down short or left long regardless of the height of a person, so we cannot say for sure that hair length correlates with height. Hence this is probably not a function)

3) hair color has no bearing on a persons age. hence this is probably not a function.

4) student ID number doesn't depend on the name of the student. Hence this is probably not a function.

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Step-by-step explanation:

take 3000 and minus that by 30 percent, then after that, add 7 percent to get your total.

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3 years ago

Which equation represents a relationship that will be a line graphed <br>

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Answer:

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3 years ago

M is the midpoint of line segment AB

SVEN [57.7K]

Answer:

The coordinates of A would be (-1, 2)

Step-by-step explanation:

In order to find this, use the mid-point formula.

(xA + xB)/2 = xM

In this, the xA is the x value of point A, xB is the x value of point B, and xM is the x value of M. Now we plug in the known information and solve for xA.

(xA + 5)/2 = 2

xA + 5 = 4

xA = -1

Now we can do the same using the midpoint formula and the y values.

(yA + yB)/2 = yM

(yA + 10)/2 = 6

yA + 10 = 12

yA = 2

This gives us the midpoint of (-1, 2)

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3 years ago

The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation

fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 197.5, \sigma = 8.3$

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{210.9 - 197.5}{8.3}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now $n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14$

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{196 - 197.5}{2.14}$

$Z = -0.7$

$Z = -0.7$ has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

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3 years ago

Jose earned 15 points in a video game. He lost 40 points, earned 87 points, then lost 30 more points. Write and evaluate an expr

KIM [24]

(15-40)+(87-30)

-25+57

32 total points

7 0

3 years ago