Answer:
The claim that the scores of UT students are less than the US average is wrong
Step-by-step explanation:
Given : Sample size = 64
Standard deviation = 112
Mean = 505
Average score = 477
To Find : Test the claim that the scores of UT students are less than the US average at the 0.05 level of significance.
Solution:
Sample size = 64
n > 30
So we will use z test

Formula : 


Refer the z table for p value
p value = 0.9772
α=0.05
p value > α
So, we accept the null hypothesis
Hence The claim that the scores of UT students are less than the US average is wrong
Answer:
167 243/386
Step-by-step explanation:
For steps, use this link:
https://mathsolver.microsoft.com/en/solve-problem/64705%20%60div%20%20386
Answer:
The length of river frontage for each lot are 96.55 ft. 98.85 ft, 101.15 ft and 103.45 ft.
Step-by-step explanation:
See the attached diagram.
The river frontage of 400 ft will be divided into 84 : 86 : 88 : 90 for each lot as AP, BQ, CR, DS and ET all are parallel.
Therefore, PQ : QR : RS : ST = 84 : 86 : 88 : 90 = 42 : 43 : 44 : 45
Let, PQ = 42x, QR = 43x, RS = 44x and ST = 45x
So, (42x + 43x + 44x + 45x) = 400
⇒ 175x = 400
⇒ x = 2.2988.
So, PQ = 42x = 96.55 ft.
QR = 43x = 98.85 ft.
RS = 44x = 101.15 ft and
ST = 45x = 103.45 ft
(Answer)
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