We use the z statistic:
z = (x – u) / s
But first let us calculate the standard deviation s,
sample x and mean u.
s = sqrt (n p q)
s = sqrt (112 * 0.77 * (1 – 0.77))
s = 4.45
x = 0.81 * 112 = 90.72
u = 0.77 * 112 = 86.24
So the z score is:
z = (90.72 – 86.24) / 4.45
z = 1.00
From the standard tables, the P value at z = 1.00 using
right tailed test is:
<span>P = 0.1587 = 15.87%</span>
the expected value for the 3 point shot = (3 * 0.30) + (0 * 0.70) = 0.90
the expected value for the 2 point shot = (2 * 0.48) + (0 * 0.52) = 0.96
the expected value for the 2 point shot is higher than the 3 point shot so he should pass the ball
Answer:
a) Narrower
b) Narrower
c) Wider
Step-by-step explanation:
We are given the following in the question:
Proportion of coworker who received flu vaccine = 32%
98% confidence interval: (0.231, 0.409)
Confidence interval:

a) Sample size had been 600 instead of 150
If we increase the sample size, thus the standard error of the interval decreases.
Since the standard error decreases, the confidence interval become narrower.
b) Confidence level had been 90% instead of 98%
As the confidence level increases, the confidence interval becomes narrower. This is due to a smaller value of z-statistic at 90% confidence level.
c) Confidence level had been 99% instead of 98%
As the confidence level increases, the width of the confidence interval increases and the confidence interval become wider. This is because of a larger value of z-statistic at 99% confidence interval.
Answer:
v(m) = 8 + 48m+ 180m² +216m³
Step-by-step explanation:
Let's first of all represent the edge of the the cube as a function of minutes.
Initially the egde= 2feet
As times elapsed , it increases at the rate of 6 feet per min, that is, for every minute ,there is a 6 feet increase.
Let the the egde be x
X = 2 + 6(m)
Where m represent the minutes elapsed.
So we Al know that the volume of an edge = edge³
but egde = x
V(m) = x³
but x= 2+6(m)
V(m) = (2+6m)³
v(m) = 8 + 48m+ 180m² +216m³