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:
18
Step-by-step explanation:
53.83 rounds to 54
36.464 rounds to 36
54-46=18
Answer:
2.64575131106
Step-by-step explanation:
Answer:
15
measure of third side is = 25^2 - 20^2
= 625 - 400
= 225
= √225
= 15
hope it helps
