Given :
A small pot holds 25% as much dirt as a large pot.
To Find :
If the small pot holds 8 cups of dirt, How many cups of dirt does the large pot hold.
Solution :
It is given that small pot holds 25% as much as dirt.
It means that large pot can hold 4 times as small pot.
Let, large pot holds x cups of dirt.
Therefore, large pot holds 32 cups.
- <u>Jan </u><u>purchased </u><u>1</u><u>4</u><u>0</u><u> </u><u>shares </u><u>of </u><u>stock </u><u>in </u><u>ABC </u><u>company </u><u>at </u><u>a </u><u>price </u><u>of </u><u>$</u><u>1</u><u>8</u><u>.</u><u>7</u><u>5</u><u> </u><u>per </u><u>share </u>
- <u>During </u><u>the </u><u>next </u><u>3</u><u> </u><u>days</u><u>, </u><u> </u><u>the </u><u>value </u><u>of </u><u>share </u><u>declined </u><u>by </u><u>$</u><u>1</u><u>.</u><u>0</u><u>0</u><u> </u><u>,</u><u> </u><u>$</u><u>1</u><u>.</u><u>7</u><u>5</u><u> </u><u>and </u><u>$</u><u>1</u><u>.</u><u>5</u><u>0</u>
- <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>value </u><u>of </u><u>a </u><u>share </u><u>of </u><u>ABC </u><u>stock </u><u>at </u><u>the </u><u>end </u><u>of </u><u>3</u><u> </u><u>days </u>
<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>
Cost of 1 share of ABC company = $18.75
<u>Therefore</u><u>, </u>
Cost of 140 shares purchased by Jan in the ABC company
<u>Now</u><u>, </u>
For next 3 days, the value of share declined
<u>Therefore</u><u>, </u>
The value of shares after 3 days will be
Hence, The value of share after 3 days will be $18.75 .
Answer:
Explanation:
<u>1. Model:</u>
i) Profit = Revenue - Cost
ii) Revenue = price × number of items
iv) Cost = variable cost + fixed expense
<u>2. Solution</u>
i) Revenue = $ 57 / item × 419 items
ii) Cost = $ 23 / item × 419 items + $5,984
iii) Profit = $ 57×419 - ($23×419 + $5,985)
Profit = $ 23,883 - ($ 9637 + $ 5,984) = $ 23,883 - $ 15,621
Profit = $ 8,262 ← answer
Answer:
7 / 10
Step-by-step explanation:
70%
= 70 / 100
= 7 / 10
Answer:
for(j = n; j > 0; j--)
System.out.print(\"*\");
Step-by-step explanation:
-A for loop is a repetition control structure.
-It allows the efficiency to write a loop that would otherwise be written a couple of times.
-The output is a single line comprising n asterisks,
-The number of asterisks printed will be equivalent to the n-value declared.