<h2><u>
Answer with explanation</u>
:</h2>
Let
be the distance traveled by deluxe tire .
As per given , we have
Null hypothesis : 
Alternative hypothesis : 
Since
is left-tailed and population standard deviation is known, thus we should perform left-tailed z-test.
Test statistic : 
where, n= sample size
= sample mean
= Population mean
=sample standard deviation
For
, we have

By using z-value table,
P-value for left tailed test : P(z≤-2.23)=1-P(z<2.23) [∵P(Z≤-z)=1-P(Z≤z)]
=1-0.9871=0.0129
Decision : Since p value (0.0129) < significance level (0.05), so we reject the null hypothesis .
[We reject the null hypothesis when p-value is less than the significance level .]
Conclusion : We do not have enough evidence at 0.05 significance level to support the claim that t its deluxe tire averages at least 50,000 miles before it needs to be replaced.
Answer:
unlikely
Step-by-step explanation:
11% is closer to 0 than it is to 1/2. when something may occur, but significantly less than half of the time, we say it is unlikely
Answer:
sorry but this question does not make any sense try rewording it
Step-by-step explanation:
Answer:
-24.5
Step-by-step explanation:
-56+-8x=137
-8x=196
x=-24.5