Tami would like to withdraw $10,364.10 at the end of each year, for 10 years, from an account paying 2.3% compounded annually. determine the amount needed in the account for tami to do this. round to the nearest cent.
a.
$115,052.98
b.
$103,641.00
c.
$91,651.92
d.
$93,759.91

Option c -$91,651.92

Step-by-step explanation:

Given : Tami would like to withdraw $10,364.10 at the end of each year, for 10 years, from an account paying 2.3% compounded annually.

To find : Determine the amount needed in the account for Tami to do this.    

Solution :

We apply the present value formula to get the amount,

Where, PV is the present value ,the required amount

P is the principal P=$10,364.10

r is the rate r=2.3% annually, r=0.023

n is the time n=10 years

Substitute in the formula,

Therefore, The amount needed is $91651.92.

Hence, Option c is correct.

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2