Answer:
a) The data distribution consists of ( 7 )1's (denoting a foreign student) and ( 43 )0's (denoting a student from the U.S.).
b) The population distribution consists of the x-values of the population of 12,152 full-time undergraduate students at theuniversity, ( 6 )% of which are 1's (denoting a foreign student) and ( 94 )% of which are 0's (denoting a student from the U.S.).
c) The mean is ( 0.06 )
The standard deviation is ( 0.0336 )
The sampling distribution represents the probability distribution of the ( sample ) proportion of foreign students in a random sample of ( 50 ) students. In this case, the sampling distribution is approximately normal with a mean of ( 0.06 ) and a standard deviation of ( 0.0336 )
Step-by-step explanation:
Answer:
B. (1/2, 3)
Step-by-step explanation:
It is perhaps easiest to try the point values in the equations.
A — 4·2+1 = 9; -2·2 +4 ≠9 . . . . not the answer
B — 4·1/2 +1 = 3; -2·(1/2) +4 = 3 . . . . this is the answer
we need go no further since we have the answer
We have to compute first for the total cost of the loan by obtaining the monthly amortization:
<span> A= P/ [(1+ r/n)^n-1] /r(1+r/n)^n</span>
where
P= loan amount
P= $35125
r=interest rate of 7.44%
n = compounding frequency of 4
t= length of loan 10 years
A= $1252.7
Total cost = $1252.7x40=$50108
Computing the percentage of the finance charges
Finance charge = debt charge+service charge
Finance charge% =($50108-$32125+$5180.7)($50108+$5180.7)x 100
=36.47%
The answer is letter b.36.47%
