Answer:
We accept the null hypothesis that the breaking strength mean is less and equal to 1750 pounds and has not increased.
Step-by-step explanation:
The null and alternative hypotheses are stated as
H0: u ≥ 1750 i.e the mean is less and equal to 1750
against the claim
Ha: u > 1750 ( one tailed test) the mean is greater than 1750
Sample mean = x`= 1754
Population mean = u = 1750
Population deviation= σ = 65 pounds
Sample size= n = 100
Applying the Z test
z= x`- u / σ/ √n
z= 1754- 1750 / 65/ √100
z= 4/6.5
z= 0.6154
The significance level alpha = 0.1
The z - value at 0.1 for one tailed test is ± 1.28
The critical value is z > z∝.
so
0.6154 is < 1.28
We accept the null hypothesis that the breaking strength mean is less and equal to 1750 pounds and has not increased.
Answer:
yes
Step-by-step explanation:
because its arranged in well functional statement
Hi miss you did not mark you thank you for asking me thank
Answer:
C. 56.25π
Step-by-step explanation:
Circumference=15π=πd⇒d=15
Area=πr^2⇒r=d/2⇒r=7.5⇒π7.5^2=56.25π
