I don't know if I'm right, but if each pencil is 10.5 grams then you multiply it by 5 ( number of pencils) and you get 52.5, which would not meet the standards of a pack weighing 60 to 95 grams.
Answer:
18:23
Step-by-step explanation:
18:23 (cannot be simplified further as there is no common factor)
Answer:
Do no reject null hypothesis.
Conclusion:
there is no sufficient statistical evidence at 0.025 level of significance to support the claim.
Step-by-step explanation:
Given that;
mean x" = 5.4
standard deviation σ = 0.7
n = 6
Null hypothesis H₀ : μ = 5.0
Alternative hypothesis H₁ : μ > 5.0
∝ = 0.025
now,
t = ( 5.4 - 5.0) / ( 0.7/√6) = 0.4 / 0.2857 = 1.4
degree of freedom df = n-1 = 6 - 1 = 5
T critical = 2.571
Therefore; t < T critical,
Do no reject null hypothesis.
Conclusion:
there is no sufficient statistical evidence at 0.025 level of significance to support the claim.
Answer:
There is enough evidence to support the claim that the population mean is greater than 100
Step-by-step explanation:
<u>Step 1</u>: We state the hypothesis and identify the claim
and
(claim)
<u>Step 2</u>: Calculate the test value.


<u>Step 3</u>: Find the P-value. The p-value obtained from a calculator is using d.f=39 and test-value 1.126 is 0.134
<u>Step 4</u>: We fail to reject the null hypothesis since P-value is greater that the alpha level. (0.134>0.05).
<u>Step 5</u>: There is enough evidence to support the claim that the population mean is greater than 100.
<u>Alternatively</u>: We could also calculate the critical value to obtain +1.685 for
and d.f=39 and compare to the test-value:
The critical value (1.685>1.126) falls in the non-rejection region. See attachment.
NB: The t- distribution must be used because the population standard deviation is not known.
B.) 85
They are alternate interior angles so they are the same degree.