ABCDEFGH = 22.
HEFG = 12
ABCD = 10
Answer:
2(n-12)
Step-by-step explanation:
Answer:
4th is the answer
Step-by-step explanation:
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.
the quickest way to do this would be to multiply 1,375 by 3 and then add 2.
1,375 * 3 = 4,125 + 2 = 4,127
correct!
