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:
113.1 cubic inches
Step-by-step explanation:
Plug in the numbers into the formula and solve.
Answer:
6
Step-by-step explanation:
Answer:
he gets 4.33 dollars back
Step-by-step explanation:
add 2.75+1.20+1.35+0.37= 5.67
10-5.67=4.33 $
2x^2y(y^3-5)+14(y^3-5)
(2x^2y)+14(y^3-5)
2(x^2y+7)(y^3-5)
x^2y+7 Is a factor
