Answer:
0.6210
Step-by-step explanation:
Given that a Food Marketing Institute found that 39% of households spend more than $125 a week on groceries
Sample size n =87
Sample proportion will follow a normal distribution with p =0.39
and standard error = 
the probability that the sample proportion of households spending more than $125 a week is between 0.29 and 0.41
=
There is 0.6210 probability that the sample proportion of households spending more than $125 a week is between 0.29 and 0.41
To solve this problem, we simply use the equation of
volume for hollow cylinders:
V = π h (R^2 – r^2)
where V is volume, h is height = 10, R is outer radius =
8.4/2 = 4.2, r is inner radius = 6/2 = 3
V = π (10) (4.2^2 – 3^2)
<span>V = 86.4π = 271.43 mm^3</span>
P(x) = 2x² - 4xq(x) = x - 3
To find the answer, we plug q(x) into p(x):
p(q(x)) = 2(x - 3)² - 4(x - 3)p(q(x)) = 2(x² - 6x + 9) - 4x + 12p(q(x)) = 2x² - 12x + 18 - 4x + 12p(q(x)) = 2x² - 16x + 30
The third option is correct.