Answer:
$15 million
Explanation:
Data provided in the question:
Inventory turn ratio = 60
Annual sales = $50 million
Average inventory = $250,000
Now,
we know,
Inventory turn ratio = ( Cost of goods sold ) ÷ ( Average inventory )
thus,
60 = ( Cost of goods sold ) ÷ $250,000
or
Cost of goods sold = 60 × $250,000
or
Cost of goods sold = $15,000,000 or $15 million
Answer: 0.75
Explanation:
Mean(μ) = 300
Standard deviation(σ) = 20
Find the z-score of a person who scored 315 on the exam ;
Raw score (x) = 315
Z = (raw score - mean) ÷ standard deviation
Z = (x - μ) ÷ σ
Z = (315 - 300) ÷ 20
Z = 15 ÷ 20
Z = 0.75
Answer:
The most you would pay per share is $18.90 price per share today.
Explanation:
Note: See the attached file for the calculation of present values for year 1 to 3 dividends.
From the attached excel file, we have:
Previous year dividend in year 1 = Dividend just paid = $1.40
Total of PV of dividends from year 1 to year 3 = $4.50720663265306
Year 3 dividend = $1.55542091836735
Therefore, we have:
Year 4 dividend = Year 3 dividend * (100% + Dividend growth rate in year 4) = $1.55542091836735 * (100% + 4%) = $1.61763775510204
Price at year 3 = Year 4 dividend / (Rate of return - Perpetual dividend growth rate) = $1.61763775510204 / (12% - 4%) = $20.2204719387755
PV of price at year 3 = Price at year 3 / (100% + Required return)^Number of years = $20.2204719387755 / (100% + 12%)^3 = $14.3925325274857
Price per share today = Total of PV of dividends from year 1 to year 3 + PV of price at year 3 = $4.50720663265306 + $14.3925325274857 = $18.90
Therefore, the most you would pay per share is $18.90 price per share today.
Instructional<span> coordinators </span>can provide<span> training for </span>teachers<span> in curriculum</span>
The confidence interval is an interval estimate for a parameter value which gives an estimated range of values which is likely to include an unknown population parameter
At least 90% of (a large series of) 90% confidence intervals will include the unknown true values of the parameters.
<span>We use this interval to determine the probability that the confidence interval produced will contain the true parameter value</span>
