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3 years ago

YALL PLSSSSSSSS HELP IM SHAKING I NEED HELP PLSS

Mathematics

1 answer:

Vilka [71]3 years ago

8 0

Answer:

someone is adding dont worrie

Step-by-step explanation:

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At many golf clubs, a teaching professional provides a free 10-minute lesson to new customers. A golf magazine reports that go

Airida [17]

Answer:

Step-by-step explanation:

a) The objective of the study is test the claim that the average gain in the green fees , lessons or equipment expenditure for participating golf facilities is less than $2,100 under the claim the null and alternative hypothesis are,

H₀ : μ = $2,100

H₀ : μ < $2,100

B) Suppose you selects α = 0.01

The probability that the null hypothesis is rejected when the average gain is $2,100 is 0.01

C) For α = 0.01

specify the rejection region of a large sample test

At the given level of significance 0.01 and the test is left-tailed then rejection level of a large-sample = < - 1.28

7 0

3 years ago

PLEASE I NEED THE SLOPE OF THE RED LINE AND THE BLUE LIKE BY NOW!!!!

netineya [11]

Red line slope=3/4
Blue line slope=2/3

3 0

3 years ago

Read 2 more answers

Do you know how to use the Pythagorean identity?

goblinko [34]

The Pythagorean theorem or identity is given by

H²=P²+B²
P²=H²-B²
B²=H²-P²

Where

B is base of right angle traingle
H is Hypotenuse
P is perpendicular

8 0

2 years ago

Find the solution of the system of equations.<br><br> -7x+4y=-32<br> 9x-2y=38

zlopas [31]

Answer:

(4,-1)

x=4

y=-1

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an

Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

Normal probability distribution

Problems of normally distributed samples can be 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.

In this problem, we have that:

$\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12$