18x- 15
hope it helps
(11x-8)+7(x-1)
___________
(11x-8)+(7x-7)
___________
(11x+7x)+(-8-7)
___________
18x-15
Answer:
CI =[- 4.588; 15.4]
Since the calculated value of t= 4.244 falls in the critical region so we reject H0 and conclude that we are not confident that the mean difference in lifetime between front and rear brake pads is 99.5%
Step-by-step explanation:
The given data is
Car Rear Front d= rear- front
1 40.6 32.5 8.1
2 35.4 27.1 8.3
3 46.6 35.6 11
4 47.3 35.6 11.7
5 38.7 30.3 8.4
6 51.2 41.1 10.1
7 50.6 40.7 9.9
8 45.9 34.3 11.6
<u>9 47.2 36.5 10.7</u>
∑ = 89.8
Let the hypotheses be
H0: ud= 0 against the claim Ha: ud ≠0
The degrees of freedom = n-1= 9-1= 8
The significance level is 0.05
The test statistic is
t= d`/sd/√n
The critical region is ║t║≤ t (0.025,8) = ±2.306
Sd= 7.05434553
d`=∑d/n= 89.8/9= 9.978
Therefore
t= d`/ sd/√n
t= 9.978/ 7.054/√9
t= 4.244
1) Since the calculated value of t= 4.244 falls in the critical region so we reject H0 and conclude that we are not confident that the mean difference in lifetime between front and rear brake pads is 99.5%
2) the confidence interval for the difference of two samples can be calculated by
d ` ± td sd/√n
9.978 ±2.306* 7.054/√9
- 4.588; 15.4
In the table it shows that 1 hour corresponds to $25. Or if you want to check it, simply take any value and divide it by its corresponding hour ( ex. 50/2 or 75/3) Hope this helps!
The answer is C. m∠1 + m∠4 = 180°
B. 4
3n - (2 + n)
3(4) - (2 + 4)
12-6
=6
