Answer:
At 6% $3,529.412 will be invested
At 11% $6,470.588 will be invested
Explanation:
Let x be the investment for 6% stock
And (10,000-x) is the investment it 11% stock
Let I be interest earned on both investments.
Using the formula
Principal(p)= Interest(I)*Rate(r)*Time(t)
p/RT= I
So considering both investments
x/(6%*1)= (10,000-x)/(11%*1)
x/0.06= (10,000-x)/0.11
Cross-multiply
0.11x= 0.06(10,000-x)
0.11x= 600- 0.06x
Rearranging
0.11x+ 0.06x= 600
0.17x= 600
x= 600/0.17= 3,529.412 amount invested at 6%
Amount invested at 11%= 10,000-3,529.412
= 6,470.588
Answer:
Value of x maximising profit : x = 5
Explanation:
Cost : C(x) = x^3 - 6x^2 + 13x + 15 ; Revenue: R(x) = 28x
Profit : Revenue - Cost = R(x) - C(x)
28x - [x^3 - 6x^2 + 13x + 15] = 28x - x^3 + 6x^2 - 13x - 15
= - x^3 + 6x^2 + 15x - 15
To find value of 'x' that maximises total profit , we differentiate total profit function with respect to x & find that x value.
dTP/dx = - 3x^2 + 12x + 15 = 0 ► 3x^2 - 12x - 15 = 0
3x^2 + 3x - 15x - 15 = 0 ► 3x (x +1) - 15 (x + 1) = 0 ► (x+1) (3x-15) = 0
x + 1 = 0 ∴ x = -1 [Rejected, production quantity cant be negative] ;
3x - 15 = 0 ∴ 3x = 15 ∴ x = 15/3 = 5
Double derivate : d^2TP/dx^2 = - 6x + 12
d^2TP/dx^2 i.e - 6x + 12 at x = 5 is -6(5) + 12 = - 30+ 12 = -8 which is negative. So profit function is maximum at x = 5
The one who will most likely have a higher BAC is the father because a person who is older will most likely have the higher BAC, as the father is already seventy five and much older to his son, he will be therefore have a higher BAC compared to his son.
Answer:
25.55 days
Explanation:
first we must calculate the accounts receivable turnover ratio = net sales / average accounts receivable
net sales = $1,000,000
average accounts receivable ($80,000 + $60,000) / 2 = $70,000
accounts receivable turnover ratio = $1,000,000 / $70,000 = 14.286
average collection period = 365 days / accounts receivable turnover ratio = 365 / 14.286 = 25.55 days
Answer and Explanation:
a. The computation of depreciation for each of the first two years by the straight-line method is shown below:-
Depreciation
= (Assets cost - Salvage value) ÷ Useful life
= ($171,000 - 0) ÷ 25
= $6,840
For First year = $6,840
For Second year = $6,840
It would be the same for the remaining useful life
b. The computation of depreciation for each of the first two years by the double-declining-balance method is shown below:-
First we have to determine the depreciation rate which is shown below:
= One ÷ useful life
= 1 ÷ 25
= 4%
Now the rate is double So, 8%
In year 1, the original cost is $171,000, so the depreciation is $13,680 after applying the 8% depreciation rate
And, in year 2, the ($171,000 - $13,680) × 8% = $12,585.60
