Answer:
Explanation:
The total equivalent units of direct materials and conversion costs for the month has been computed and attached.
Note that the conversion cost for the ending work in process was calculated as:
= $35,000 × 28%
= $35,000 × 0.28
= $9,800
Check the attachment for further analysis.
Answer: $34,696
Explanation: $1000,000/25=$40,000 as deposit to be made 25 times, but out of this amount 8% interest will be subtracted which gives 8/100×40,000=$3200.
Hence annual deposit will be $40000-$3200=$36800.
But annual raise of 3% should be subtracted as well making deposit to be 3/100×36800=$1104.
Substracting we have $36800-$1104=$35696.
Hence i must deposit $35696 first to meet this goal.
I believe that it is B. <span>but i could be wrong that seems to be the most logical answer
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Answer:
The mean withdraw has increased during weekend.
Explanation:
Assume that the withdraw amounts are normal distributed. To test whether the mean withdrawal has increased during weekends, we take a z-test. The z-test is possible because the observed sample (weekend transactions) is greater than 30.
The null hypothesis (
) is when the mean withdrawal is greater than 550. The alternative hypothesis (
) is when the mean withdrawal is equal to 550 or smaller. At an alpha of 0.05% is selected with a two-tailed test, , there is 0.025% of the samples in each tail, and the alpha has a critical value of 1.96 or -1.96. If the z-value is greater than 1.96 or less than -1.96, the null hypothesis is rejected.
z-value = (600-550) / 70 / 36^(1/2) = 0.1190
At α=0.05, the z-value < 1.96 and > -1.96, the null hypothesis is not rejected. Therefore, the mean withdraw has increased during weekend.
