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Gwar [14]
3 years ago
7

What is the answer?please help me

Mathematics
1 answer:
Maslowich3 years ago
6 0

Answer:

y = 2x - 1 Graph B

y = 2x + 2 Graph A

y = 2x - 5 Graph C

Step-by-step explanation:

y = mx + c

c is where the line intersect the Y axis.

<h2>Substitute x for 0:</h2>

y = 2x - 1

2 * 0 - 1 = -1.

This is graph B because it intersects at 0,-1.

y = 2x + 2

2 * 0 + 2 = 2

This is graph A because it intersects at 0,2.

y = 2x - 5

2 * 0 - 5 = -5

This is graph C because it intersects at 0,-5.

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True/False: All 3 angles of an equilateral triangle measure 60 degrees.<br> True<br> False
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true

Step-by-step explanation:

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The average score of all golfers for a particular course has a mean of 75 and a standard deviation of 4. Suppose 64 golfers play
Vilka [71]

Answer:

0.0228 = 2.28% probability that the average score of the 64 golfers exceeded 76.

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 problem, we have that:

\mu = 75, \sigma = 4, n = 64, s = \frac{4}{\sqrt{64}} = 0.5

Find the probability that the average score of the 64 golfers exceeded 76.

This is 1 subtracted by the pvalue of Z when X = 64.

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

By the Central Limit Theorem

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

Z = \frac{76 - 75}{0.5}

Z = 2

Z = 2 has a pvalue of 0.9772

1 - 0.9772 = 0.0228

0.0228 = 2.28% probability that the average score of the 64 golfers exceeded 76.

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3 years ago
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