Answer:
34°
Step-by-step explanation:
If m∠ADE is with 34° smaller than m∠CAB, then denote
m∠ADE=x°,
m∠CAB=(x+34)°.
Since DE ║ AB, then
m∠ADE=m∠DAB=x°.
AD is angle A bisector, then
m∠EAD=m∠DAB=x°.
Thus,
m∠CAB=m∠CAD+m∠DAB=(x+x)°=2x°.
On the other hand,
m∠CAB=(x+34)°,
then
2x°=(x+34)°,
m∠ADE=x°=34°.
Answer:
No, there is not enough evidence to support the claim that true average task time with armor is less than 2 seconds.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that true average task time with armor is less than 2 seconds.
Then, the null and alternative hypothesis are:
The significance level is 0.01.
The sample has a size n=52.
The sample mean is M=1.95.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.2.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size are:
This test is a left-tailed test, with 51 degrees of freedom and t=-1.803, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.039) is bigger than the significance level (0.01), the effect is not significant.
The null hypothesis failed to be rejected.
There is not enough evidence to support the claim that true average task time with armor is less than 2 seconds.
This question is incomplete, the complete question is;
Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and preview ads before the movie starts. Many complain that the time devoted to previews is too long (The Wall Street Journal, October 12, 2012). A preliminary sample conducted by The Wall Street Journal showed that the standard deviation of the amount of time devoted to previews was 4 minutes. Use that as a planning value for the standard deviation in answering the following questions. Round your answer to next whole number,
a) If we want to estimate the population mean time for previews at movie theaters with a margin of error of 75 seconds, what sample size should be used? Assume 95% confidence.
b) If we want to estimate the population mean time for previews at movie theaters with a margin of error of 1 minute, what sample size should be used? Assume 95% confidence.
Answer:
a) the required Sample Size is 40
b) the required Sample Size is 62
Step-by-step explanation:
Given the data in the question;
standard deviation σ = 4 minutes
a)
margin of error E = 75 seconds = ( 75 / 60 )minutes = 1.25 minutes
And 95% confidence.
Now, Critical Value of z for 0.95 confidence interval;
∝ = 1 - 0.95 = 0.05
= 1.96
so, sample size n will be;
n = [ × σ/E ]²
we substitute;
n = [ 1.96 × (4/1.25) ]²
n = [ 1.96 × 3.2 ]²
n = [ 6.272 ]²
n = 39.338
Since we referring to a number of sample, its approximately becomes 40
Therefore, the required Sample Size is 40
b)
margin of error E = 1 minutes
And 95% confidence.
Now, Critical Value of z for 0.95 confidence interval;
∝ = 1 - 0.95 = 0.05
= 1.96
so, sample size n will be;
n = [ × σ/E ]²
we substitute;
n = [ 1.96 × (4/1) ]²
n = [ 1.96 × 4 ]²
n = [ 7.84 ]²
n = 61.466
Since we referring to a number of sample, its approximately becomes 62
Therefore, the required Sample Size is 62