<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:
2 possible coordinate pairs
N (8, 17) OR (-12, -3)
Step-by-step explanation:
The rule:
(x, y) => (x + 10, y + 10) OR (x - 10, y -10)
(-2 + 10, 7 + 10) OR (-2 - 10, 7 -10)
(8, 17) OR (-12, -3)
Hope this helps!
Answer step-by-step explanation:
We know that the critical t value for a 98% confidence level for a sample of size n = 18:
Finding degrees of freedom (n-1) = 17
-search the information in table t for 0.98 and find that t = 2.567
The best estimate to the nearest one percent of the fraction; 7/15 is; 47%.
<h3>What is the best estimate of the fraction to the nearest percent?</h3>
From the task content, it follows that the fraction given whose estimate is to be determined is; 7/15.
The fraction expressed as a percentage is;
(7/15) × 100 %
= 46.667%.
Hence, when rounded to the nearest one percent; = 47%.
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