Find a 92.9% confidence interval for the difference p1−p2p1−p2 of the population proportions.
Hint: For a level of confidence (1−α)⋅100%(1−α)⋅100%, z∗z∗ is the value that leaves an area of (1−α)/2(1−α)/2 to its right in a standard normal distribution
I hope that helps
B 0.05 since the thousandth digit is higher than 5, it needs to be rounded up.
Answer:
Jerry is 90, Marty 45 and Phil is 50
Answer:
B. -5.
Step-by-step explanation:
fg(x) = (x - 1)(5x^2 + 19x - 3)
= 5x^3 + 14x^2 - 22x + 3
So fg(-4) = 5(-4)^3 + 14(-4)2 -22(-4) + 3
= -320 + 224 + 88 + 3
= -5.
In case you want the working out -
You will want to put all these fractions over the same denominator, the lowest common multiple is 180
-2/9
180/9 = 20
20 x -2 = -40/180
A -7/12
180/12 = 15
15 x -7 = -105/180
B -1/15
180/15 = 12
12 x -1 = -12/180
C -1/3
180/3 = 60
60 x -1 = -60/180
D -1/9
180/9 = 20
20 x -1 = -20/180
-105 < -60 < -40 < -20 < -12
A < C < x < D < B
B and D
