Frank will need 4 gallons of juice.
35*12=420 fluid ounces
420/32=13.125 quarts
13.125/4=3.28 gallons
He will have some juice cups left over.
Answer:
-9*F
Step-by-step explanation:
Answer:
2miles
Step-by-step explanation:
56/4=14
28/2=14
Answer:
22.4
Step-by-step explanation:
you can add or mulitiply all the sides together
The missing part of the question is highlighted in bold format
The Wall Street Journal reported that the age at first startup for 90% of entrepreneurs was 29 years of age or less and the age at first startup for 10% of entrepreneurs was 30 years of age or more.
Suppose a sample of 200 entrepreneurs will be taken to learn about the most important qualities of entrepreneurs. Show the sampling distribution of p where p is the sample proportion of entrepreneurs whose first startup was at 29 years of age or less. If required, round your answers to four decimal places. np = n(1-p) = E(p) = σ(p) = (b) Suppose a sample of 200 entrepreneurs will be taken to learn about the most important qualities of entrepreneurs. Show the sampling distribution of p where p is now the sample proportion of entrepreneurs whose first startup was at 30 years of age or more. If required, round your answers to four decimal places.
Answer:
(a)
np = 180
n(1-p) = 20
E(p) = p = 0.9
σ(p) = 0.0212
(b)
np = 20
n(1 - p) = 180
E(p) = p = 0.1
σ(p) = 0.0212
Step-by-step explanation:
From the given information:
Let consider p to be the sample proportion of entrepreneurs whose first startup was at 29 years of age or less
So;
Given that :
p = 90% i.e p = 0.9
sample size n = 200
Then;
np = 200 × 0.9 = 180
n(1-p) = 200 ( 1 - 0.9)
= 200 (0.1)
= 20
Since np and n(1-p) are > 5 ; let assume that the data follows a normal distribution ;
Then:
The expected value of the sampling distribution of p = E(p) = p = 0.9
Variance 



The standard error of σ(p) = 

= 0.0212
(b)
Here ;
p is now the sample proportion of entrepreneurs whose first startup was at 30 years of age or more
p = 10% i.e p = 0.1
sample size n = 200
Then;
np = 200 × 0.1 = 20
n(1 - p) = 200 (1 - 0.1 ) = 180
Since np and n(1-p) are > 5 ; let assume that the data follows a normal distribution ;
Then:
The Expected value of the sampling distribution of p = E(p) = p = 0.1
Variance 



The standard error of σ(p) = 

= 0.0212