(-2 1/2)×(-3 1/2) = (2 1/2)×(3 1/2) . . . . deal with the signs up front. The product of two negatives is a positive.
... = 2×(3 1/2) + (1/2)×(3 1/2) . . . . . use the distributive property
... = 2×3 + 2×(1/2) + (1/2)×3 + (1/2)×(1/2) . . . . and again
... = 6 + 1 + 3/2 + 1/4
... = 7 + (1 1/2) + 1/4
... = 8 + 2/4 + 1/4
... = 8 3/4
_____
On your calculator, -2.5 × -3.5 = 8.75.
X+3y=-13
x=-3y-13
5(-3y-13) +7y = -9
-15y -65 +7y = -9
-8y-65=-9
65 -
-----
-74
/-8
------
y= 9.25
x+3(9.25) = -13
x+27.75= -13
27.75-
-------
x= -40.75
<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.
Your answer is 140
so yeah i have to keep typing because its too short and im done