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Alex17521 [72]
3 years ago
15

K anyone help please

Mathematics
1 answer:
tigry1 [53]3 years ago
7 0

Answer:

Step-by-step explanation:

Given equation is,

a). y = \frac{6}{x}

x        0.5                1            2            3            4              5             6

y       \frac{6}{0.5}=12      \frac{6}{1}=6      \frac{6}{2}=3      \frac{6}{3}=2     \frac{6}{4}=1.5     \frac{6}{5}=1.2     \frac{6}{6}=1

b). By plotting the input-output values on the graph we get the curve as shown in option C.

Therefore, Option C will be the answer.

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g The average midterm score of students in a certain course is 70 points. From the past experience it is known that the midterm
spayn [35]

Answer:

0.98 = 98% probability that the average midterm score of these students is at most 75 points.

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, 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.

The average midterm score of students in a certain course is 70 points.

This means that \mu = 70

29 students are randomly selected and the standard deviation of their scores is found to be 13.15 points.

This means that \sigma = 13.15, n = 29, s = \frac{13.15}{\sqrt{29}} = 2.44

Find the probability that the average midterm score of these students is at most 75 points.

This is the pvalue of Z when X = 75. So

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

By the Central Limit Theorem

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

Z = \frac{75 - 70}{2.44}

Z = 2.05

Z = 2.05 has a pvalue of 0.98.

0.98 = 98% probability that the average midterm score of these students is at most 75 points.

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Answer: B

Step-by-step explanation:

2(—2) + 10 = 6.

2(0) + 6 = 6.

2(2) + 2 = 6.

2(4) + (—2) = 6

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