<h2><u>
Answer with explanation</u>
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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:
I think its C you can hate how much you want if I'm wrong im more of a no tester when I get an answer i check but not online
Step-by-step explanation:
Answer:
21:A 22:D
Step-by-step explanation:
hope that helps
M=0-5/1-0
m=-5/1=-5
y-y1=m*(x-x1)
y-5=-5(x-0)
y-5=-5x
y=-5x+5
Answer:
x | y | (x,y)
0 | 35 | (0,35)
3 | 50 | (3,50)
-7 | 0 | (-7,0)
Step-by-step explanation:
Given the equation: 
, to complete the table, plug in the value of x given in the table in each row, to find y.
When x = 0,
(0,35)
When x = 3,
(3,50)
When x = -7,
(-7,0)
