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:
Total Cost of Job X4A: $
Direct material cost ($9,000 x 500 units) 4,500,000
Direct labour cost (300 hrs x $15 x 500 units ) 2,250,000
Overhead applied (100 hrs x $22.50 x 500 units) 1,125,000
Total cost 7,875,000
Explanation:
The total cost of Job X4A is the aggregate of direct material cost, direct labour cost and overhead applied. Overhead is absorbed on the basis of machine hours. Thus, we will multiply the overhead rate by machine hours and number of units produced.
Answer:
E. The demand for loanable funds increases.
