Answer:
Present Value = ![X [\frac{1}{(1 + 0.12)^{1} } + \frac{1}{(1 + 0.12)^{2} } + \frac{1}{(1 + 0.12)^{3} } + \frac{1}{(1 + 0.12)^{4} } + \frac{1}{(1 + 0.12)^{5} } ]](https://tex.z-dn.net/?f=X%20%5B%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B1%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B2%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B3%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B4%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B5%7D%20%7D%20%5D)
Step-by-step explanation:
To find - If discount rate is 12%, the present value of Rs X received at the end of each year for the next five years is equal to .... ?
Solution -
We know that, formula for finding the Present vale is given by
Present value = Future value / (1 + r)ⁿ
where r is the rate of interest
and n is Number of periods
Now,
Here in the question, we have
r = 12% = 12/100 = 0.12
n = 5
Also, Given that, we have received Rs X at the end of each year
So,
Present Value = 
= ![X [\frac{1}{(1 + 0.12)^{1} } + \frac{1}{(1 + 0.12)^{2} } + \frac{1}{(1 + 0.12)^{3} } + \frac{1}{(1 + 0.12)^{4} } + \frac{1}{(1 + 0.12)^{5} } ]](https://tex.z-dn.net/?f=X%20%5B%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B1%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B2%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B3%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B4%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B5%7D%20%7D%20%5D)
⇒Present Value = ![X [\frac{1}{(1 + 0.12)^{1} } + \frac{1}{(1 + 0.12)^{2} } + \frac{1}{(1 + 0.12)^{3} } + \frac{1}{(1 + 0.12)^{4} } + \frac{1}{(1 + 0.12)^{5} } ]](https://tex.z-dn.net/?f=X%20%5B%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B1%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B2%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B3%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B4%7D%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B%281%20%2B%200.12%29%5E%7B5%7D%20%7D%20%5D)
Answer:
Step-by-step explanation:
The amount of money that Jason earns is dependent on the amount of time, t in hours that he works
The function f(x) = 15t represents the money he earns for working t hours. Therefore, the dependent variable is f(x) and the independent variable is t
Jason works at least 5 hours but not more than 10. This means that The minimum amount that he can earn is 15×5 = $75
The maximum amount that he can earn is 15×10 = $150
The practical domain is all possible values for t. It becomes
5 lesser than or equal to t lesser than or equal to 10
The practical range is all possible values for f(x). It becomes
75 lesser than or equal to f(x) lesser than or equal to 150
Answer:
$58.20
Step-by-step explanation:
600 /200 = 3
3 * 19.40 = 58.20
Answer:
37×2×(-5+5+1/2) = inverse property of addition
37×2×(0+1/2) = commutative property
37×2×1/2 = identity property of addition
37×1 = inverse property of multiplication
37 = identity property of multiplication
I hope it helps
Step-by-step explanation:
Only -3 satisfies this equation. so i think it has only one solution! hope it will help you..