Answer:
To Part A
Explanation:
Basic and non-basic are the two broad industry categories. Basic industries (primary) consist of small and large businesses that sell primarily to external customers. Non-basic industries (secondary) consist of primarily small businesses that sell to local customers, including basic and non-basic businesses.
Assuming that oil production will peak in about 2020 and
then decline at about 3% per year, when will production
be half of that in 2020? What could be the consequences?
Why? How could these consequences be avoided?
Answer:
Given that oil production will peak in about 2020 and then decline at about 3% per year, to determine when will production be half of that in 2020 the following calculation must be performed:
50 x (1 + 0.03) ^ X = 100
50 x 1.03 ^ X = 100
50 x 1.03 ^ 23.45 = 100
0.45 x 365 = 164.25
Thus, oil production will be cut in half in 23 years, 5 months and 10 days. After that, the communications and transportation system will be seriously affected, since they depend heavily on oil. These consequences could be avoided if energy alternatives were obtained that could replace oil.
Job A3B was ordered by a customer on September 25. During the month of September, Jaycee Corporation requisitioned $1,600 of direct materials and used $3,100 of direct labor. The job was not finished by the end of September, but needed an additional $2,100 of direct materials and additional direct labor of $5,600 to finish the job in October. The company applies overhead at the end of each month at a rate of 200% of the direct labor cost incurred. What is the total cost of the job when it is completed in October?
Multiple Choice:
$12,400
$18,900
$29,800
$22,000
$25,700
Answer: $29800
Explanation:
The total cost of the job when it is completed in October will be:
Direct materials = $1600 + $2100 = $3700
Direct labor = $3100 + $5600 = $8700
Factory overhead = 200% × $8700 = $17400
Total cost = $3700 + $8700 + $17400
= $29800
The distribution of average wait times in drive-through restaurant lines in one town was approximately normal with mean μ=242 seconds and standard deviation σ=13 seconds.
Amelia only likes to use the drive-through for restaurants where the average wait time is in the bottom 15% for that town.
What is the maximum average wait time for restaurants where Amelia likes to use the drive-through? Round to the nearest whole second.
You're looking for the 15th percentile of a normal distribution with µ = 242 and σ = 13, which is to say you want to find x * such that
P(X ≤ x *) = 0.15
Transform the wait-time random variable X to Z, which follows the standard normal distribution with mean 0 and s.d. 1 :
Z = (X - µ) / σ → X = µ + σ Z = 242 + 13 Z
Now,
P(X ≤ x *) = P(242 + 13 Z ≤ x *) = P(Z ≤ (x * - 242)/13) = 0.15
Use a calculator (something with an inverse CDF function) or Z-table to look up the z-score, z = (x * - 242)/13, associated with a probability of 0.15. You would find that
z = (x * - 242)/13 ≈ -1.03643
Solve for x * :
x * ≈ 242 + 13 (-1.03643) ≈ 228.526 ≈ 229