Answer:
$0.013
0.010724
Explanation:
Given that :
Mean, m = 36500
Standard deviation, s = 5000
Refund of $1 per 100 mile short of 30,000 miles
A.) Expected cost of the promotion :
P(X < 30,000)
Using the Zscore relation :
Zscore = (x - m) / s
Zscore = (30000 - 36500) / 5000
= - 6500 / 5000
= - 1.3
100 miles = $1
1.3 / 100 = $0.013
b. What is the probability that Grear will refund more than $50 for a tire?
100 miles = $1
$50 = (100 * 50) = 5000 miles
Hence, more than $50 means x < (30000 - 5000) = x < 25000 miles
P(x < 25000) :
(25000 - 36500) / 5000
-11500 / 5000
= - 2.3
P(z < - 2.3) = 0.010724 (Z probability calculator)
Answer:
He will have $102,979 in his retirement account in 10 years.
Explanation:
Annual Payment = $2,000
Number of Year = n = 10
Interest rate = i = 5%
Compounded Quarterly
Future value after 10 years
FV = A [ ( ( 1 + ( r / m )^mt ) - 1 / ( r / m )
FV = $2,000 [ ( ( 1 + ( 0.05 / 4 )^40 ) - 1 / ( 0.05 / 4 )
Future value = $102,979
So, Ira Schwab will have $102,979 in his retirement account in 10 years.
Answer:
14%
Explanation:
required rate of return = risk free rate of return + ( risk premium x beta)
5% + 1.5 x 6% = 14%
Answer:
hi im just getting my points
Explanation:
and im good btw