Answer:
Overhead application rate
= <u>Budgeted overhead</u>
Budgeted machine hours
= <u>$900,000</u>
30,000 hours
= $30 per machine hour
Overhead cost assigned to the product
= Overhead application rate x Actual machine hours
= $30 x 12,000 hours
= $360,000
Explanation:
In this case, there is need to determine the overhead application rate, which is the ratio of budgeted overhead to budgeted machine hours.
Then, we will obtain the overhead cost assigned to the product by multiplying the overhead application rate by actual machine hours.
Most likely a two year college, so a local community college nearby is a good option. It's also a lot cheaper than a traditional university.
EAR = (1 + periodic interest rate)^N - 1
<u>9.25 % Quarterly %</u>
EAR =
= 0.09575 or 9.58%
<u>16.75 Monthly %
</u>
EAR =
= 0.1809766 or 18.10%
<u>15.25 Daily %
</u>
EAR =
= 0.1647053 or 16.47%
<u>11.25 Semiannually %</u>
EAR =
= 0.115664 or 11.57%
Answer:
The probability that 2 or 3 customers will arrive in a 15-minute period is 0.4703
Explanation:
Firstly, we have to determine the segment unit, since the mean is 10 per hour, the segment unit is 1 hour.
The mean(m) = 10
since the period is 15 minutes = 0.25 hour, t= 0.25 hour / 1 hour. Therefore mt= 2*10 = 2.5
The poisson distribution formula P(x) = 
Therefore the probability that 2 or 3 customers will arrive in a 15-minute period
P(x=2) or P(x=3) = P(x=2) + P(x=3) =
= 0.2565 + 0.2138 = 0.4703
Therefore P(x=2) or P(x=3) = 0.4703
The probability that 2 or 3 customers will arrive in a 15-minute period is 0.4703
Answer:
a. $11,760.
Explanation:
Straight line depreciation expense = (Cost of asset - Salvage value) / useful life
Cost of asset = $60,000 + $8,000 + $2,800 = $70,800
($78,800 - $12,000) / 5 = $11,760.
I hope my answer helps you