Answer:
Division
Step-by-step explanation:
Certain Outcomes/Total Outcomes = Probability
The operation used is division
The amount for the investment of $6000 will be a.$6369 b. $6090 and c.$6030.
<h3>What is compound interest?</h3>
Compound interest is the interest levied on the interest. The formula for the calculation of compound interest is given as:-
![A=P[1+\dfrac{r}{n}]^{nt}](https://tex.z-dn.net/?f=A%3DP%5B1%2B%5Cdfrac%7Br%7D%7Bn%7D%5D%5E%7Bnt%7D)
a) The amount in the bank after 6 years if interest is compounded annually.
![A=P[1+\dfrac{r}{1}]^{t}\\\\\\A=6000[1+\dfrac{0.01}{1}]^{ 6}](https://tex.z-dn.net/?f=A%3DP%5B1%2B%5Cdfrac%7Br%7D%7B1%7D%5D%5E%7Bt%7D%5C%5C%5C%5C%5C%5CA%3D6000%5B1%2B%5Cdfrac%7B0.01%7D%7B1%7D%5D%5E%7B%20%206%7D)
A= $6369
b) The amount in the bank after 6 years if interest is compounded quarterly.
![A=P[1+\dfrac{r}{4}]^{4t}\\\\\\A=6000[1+\dfrac{0.01}{4}]^{4\times 6}](https://tex.z-dn.net/?f=A%3DP%5B1%2B%5Cdfrac%7Br%7D%7B4%7D%5D%5E%7B4t%7D%5C%5C%5C%5C%5C%5CA%3D6000%5B1%2B%5Cdfrac%7B0.01%7D%7B4%7D%5D%5E%7B4%5Ctimes%206%7D)
A= $6090
c ) The amount in the bank after 6 years if interest is compounded monthly.
![A=P[1+\dfrac{r}{12}]^{4t}\\\\\\A=6000[1+\dfrac{0.01}{12}]^{12\times 6}](https://tex.z-dn.net/?f=A%3DP%5B1%2B%5Cdfrac%7Br%7D%7B12%7D%5D%5E%7B4t%7D%5C%5C%5C%5C%5C%5CA%3D6000%5B1%2B%5Cdfrac%7B0.01%7D%7B12%7D%5D%5E%7B12%5Ctimes%206%7D)
A=$6030
Hence the amount for the investment of $6000 will be a.$6369 b. $6090 and c.$6030.
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Answer:
f = 48/5
Step-by-step explanation:
5f = 12 * 4
5f = 48
f = 48/5
The interest is compounded quarterly because the formula is
A=p (1+r/k)^kt
A it's 18810.67 but I will assume it's unknown ?
P present value 7350
R interest rate 0.045
K compounded quarterly 4
Now solve the formula as the interest is compoundedquarterly
A=7,350×(1+0.045÷4)^(4×21)
A=18,810.67
So the answer is quarterly
Answer:

Step-by-step explanation:
The volume of the solid revolution is expressed as;

Given y = 2x²
y² = (2x²)²
y² = 4x⁴
Substitute into the formula
![V = \int\limits^2_0 {4\pi x^4} \, dx\\V =4\pi \int\limits^2_0 { x^4} \, dx\\V = 4 \pi [\frac{x^5}{5} ]\\](https://tex.z-dn.net/?f=V%20%3D%20%5Cint%5Climits%5E2_0%20%7B4%5Cpi%20x%5E4%7D%20%5C%2C%20dx%5C%5CV%20%3D4%5Cpi%20%5Cint%5Climits%5E2_0%20%7B%20x%5E4%7D%20%5C%2C%20dx%5C%5CV%20%3D%204%20%5Cpi%20%5B%5Cfrac%7Bx%5E5%7D%7B5%7D%20%5D%5C%5C)
Substituting the limits
![V = 4 \pi ([\frac{2^5}{5}] - [\frac{0^5}{5}])\\V = 4 \pi ([\frac{32}{5}] - 0)\\V = 128 \pi/5 units^3](https://tex.z-dn.net/?f=V%20%3D%204%20%5Cpi%20%28%5B%5Cfrac%7B2%5E5%7D%7B5%7D%5D%20-%20%5B%5Cfrac%7B0%5E5%7D%7B5%7D%5D%29%5C%5CV%20%3D%204%20%5Cpi%20%28%5B%5Cfrac%7B32%7D%7B5%7D%5D%20-%200%29%5C%5CV%20%3D%20128%20%5Cpi%2F5%20units%5E3)
Hence the volume of the solid is 