Answer:
x must be greater than -53/14
(im not 100% sure but i think this is correct)
I believe it is C because I counted the dots in the shaded region.
Hope it is correct :)
Answer:
Average range(R-bar) = Sum of the sample range for each sample/number of samples = (9+8+1+8+7) /5 = 33/5 = 6.6
Sample size = 8
For a sample size of 8 the factors for control limit for range (D3) = 0.136 (obtained from table of constraints for x-bar and R chart
LCL = R-bar × D3 = 6.6 × 0.136 = 0.90
Step-by-step explanation:
Answer: 18
Step-by-step explanation:
Add them all up
18+21+20+14+17=90
Divide by 5 since that is how many numbers there are
90/5=18
Answer:
Step-by-step explanation:
Given that A be the event that a randomly selected voter has a favorable view of a certain party’s senatorial candidate, and let B be the corresponding event for that party’s gubernatorial candidate.
Suppose that
P(A′) = .44, P(B′) = .57, and P(A ⋃ B) = .68
From the above we can find out
P(A) = ![1-0.44 = 0.56](https://tex.z-dn.net/?f=1-0.44%20%3D%200.56)
P(B) = ![1-0.57 = 0.43](https://tex.z-dn.net/?f=1-0.57%20%3D%200.43)
P(AUB) = 0.68 =
![0.56+0.43-P(A\bigcap B)\\P(A\bigcap B)=0.30](https://tex.z-dn.net/?f=0.56%2B0.43-P%28A%5Cbigcap%20B%29%5C%5CP%28A%5Cbigcap%20B%29%3D0.30)
a) the probability that a randomly selected voter has a favorable view of both candidates=P(AB) = 0.30
b) the probability that a randomly selected voter has a favorable view of exactly one of these candidates
= P(A)-P(AB)+P(B)-P(AB)
![=0.99-0.30-0.30\\=0.39](https://tex.z-dn.net/?f=%3D0.99-0.30-0.30%5C%5C%3D0.39)
c) the probability that a randomly selected voter has an unfavorable view of at least one of these candidates
=P(A'UB') = P(AB)'
=![1-0.30\\=0.70](https://tex.z-dn.net/?f=1-0.30%5C%5C%3D0.70)