Answer:
a= 5
b= -3
c= -1
d= 1
Step-by-step explanation:
Edge answer
Now it would be
-2.3+(-5.9)
now just add those two together
and u get -8.2
So the answer is A
Step-by-step explanation:
A.
Null hypothesis
H0: σ1² ≤ σ2²
Alternative hypothesis
H1: σ1² > σ2²
We have level of significance = 0.01
Test statistic
F = S1²/S2²
= 170²/100²
= 28900/10000
F = 2.89
Df = degrees of freedom
Df1 = n1 - 1
= 26-1
= 25
Df2 = n2 - 1
= 25-1
= 24
B. We get the p value using f distribution
Fdist(2.89,25,24)
P value = 0.0057
The p-value is less than 0.01 so we reject null hypothesis.
So we conclude that automobiles that are of 4 years have their variances to be larger In annual repair costs than those that are of 2 years.
Reasonableness: we expect this since 4byears old automobiles are more likely to have more expenses during repair leading to greater variances
15 yards from the pass.
13 yards from the run.
The total is 28 yards. hope this helps you
Answer:
The answer is B
It's the same because it's still negative