First we need to write the null and alternate hypothesis for this case.
Let x be the average number of text message sent. Then
Null hypothesis: x = 100
Alternate hypothesis: x > 100
The p value is 0.0853
If p value > significance level, then the null hypothesis is not rejected. If p value < significance level, then the null hypothesis is rejected.
If significance level is 10%(0.10), the p value will be less than 0.10 and we reject the null hypothesis and CAN conclude that:
The mean number of text messages sent yesterday was greater than 100.
If significance level is 5%(0.05), the p value will be greater than 0.05 and we cannot reject the null hypothesis and CANNOT conclude that:
The mean number of text messages sent yesterday was greater than 100.
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)
B is the answer I chose but there are the steps so you can see how to solve it
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Answer:
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Step-by-step explanation:
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