Answer:
a) $3480
b) $4036.8
Step-by-step explanation:
The compound interest formula is given by:
![A(t) = P(1 + \frac{r}{n})^{nt}](https://tex.z-dn.net/?f=A%28t%29%20%3D%20P%281%20%2B%20%5Cfrac%7Br%7D%7Bn%7D%29%5E%7Bnt%7D)
Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per year and t is the time in years for which the money is invested or borrowed.
Suppose that $3000 is placed in an account that pays 16% interest compounded each year.
This means, respectively, that ![P = 3000, r = 0.16, n = 1](https://tex.z-dn.net/?f=P%20%3D%203000%2C%20r%20%3D%200.16%2C%20n%20%3D%201)
So
![A(t) = P(1 + \frac{r}{n})^{nt}](https://tex.z-dn.net/?f=A%28t%29%20%3D%20P%281%20%2B%20%5Cfrac%7Br%7D%7Bn%7D%29%5E%7Bnt%7D)
![A(t) = 3000(1 + \frac{0.16}{1})^{t}](https://tex.z-dn.net/?f=A%28t%29%20%3D%203000%281%20%2B%20%5Cfrac%7B0.16%7D%7B1%7D%29%5E%7Bt%7D)
![A(t) = 3000(1.16)^{t}](https://tex.z-dn.net/?f=A%28t%29%20%3D%203000%281.16%29%5E%7Bt%7D)
(a) Find the amount in the account at the end of 1 year.
This is A(1).
![A(t) = 3000(1.16)^{t}](https://tex.z-dn.net/?f=A%28t%29%20%3D%203000%281.16%29%5E%7Bt%7D)
![A(1) = 3000(1.16)^{1} = 3480](https://tex.z-dn.net/?f=A%281%29%20%3D%203000%281.16%29%5E%7B1%7D%20%3D%203480)
(b) Find the amount in the account at the end of 2 years.
This is A(2).
![A(2) = 3000(1.16)^{2} = 4036.8](https://tex.z-dn.net/?f=A%282%29%20%3D%203000%281.16%29%5E%7B2%7D%20%3D%204036.8)
3p : 8s
1p : (8/3)s
So for every one 'p' there is (8/3) 's'
Ratio of p to s is 1 : 8/3
Answer:
log x^13.
Step-by-step explanation:
Using the laws of logarithms
a log b = log b^a and log a + log b = log ab:
3 log x^2 + 7 log x
= log (x^2)^3 + log x^7
= log x^6 + log x^7
= log (x^6*x^7)
= log x^13.
Answer:
8 in^2.
Step-by-step explanation:
F = P * A
22 = 2.75 * A
A = 22/ 2.75
A = 8 in^2.