Problem: you want to start a savings account with $240. you want to save up to buy a large screen tv for $399.00. you find an investment opportunity that pays 6% apr interest compounded monthly. how long will it take to turn your initial investment into $399? answer the following questions showing all your work: a. write the equation for future value (amount) of a savings account with compound interest compounded multiple times per year. write it with vertical fractions or use parentheses correctly if you type it horizontally. use the variables a, p, r, n, and t, and explain what each of the variables represent. b. write the equation again substituting in the values from the problem above. make sure that the value that you use for "r" is correct. c. what is the value of the expression inside the parentheses? give your result to at least 7 to 9 decimal places if needed, unless it comes out exact to fewer places. d. solve this equation for "t" showing all your work. state the solution as a decimal value of t to 4 decimal places. (this is the number of years). e. convert this number to years and months. state the number of years and months it will take to save what you need. round to the nearest whole month. remember: 12 months per year!

A.  Equation for future value: A = P ( 1 + r / n )^(nt)

B.  Equation: $399.00 = $240.00 ( 1 + 0.06 / 12 )^(12t)

C.  The value  of the expression inside the parentheses: 1.005

D.  The number of years is t=8.4932

E.  It will take to save what you need 8 years and 6 months.

Step-by-step explanation:

A. Write the equation for future value (amount) of a savings account with compound interest compounded multiple times per year. Write it with vertical fractions or use parentheses correctly if you type it horizontally. Use the variables A, P, r, n, and t, and explain what each of the variables represent.

Future Value: A=$399.00

Present Value: P=$240.00

APR=Annual Percentage rate=Annual interest rate: r=6%=6/100→r=0.06

Number of compounding periods per year: n=12 (compounded monthly: 12 periods per year)

Time in years: t=?

A = P ( 1 + r / n )^(nt)

B. Write the equation again substituting in the values from the problem above. Make sure that the value that you use for "r" is correct.

Substituting the given values in the formula of part A:

$399.00 = $240.00 ( 1 + 0.06 / 12 )^(12t)

C. What is the value of the expression inside the parentheses? Give your result to at least 7 to 9 decimal places if needed, unless it comes out exact to fewer places.

Expression inside the parentheses:

1 + 0.06 / 12 = 1 + 0.005 →

1 + 0.06 / 12 = 1.005

D. Solve this equation for "t" showing all your work. State the solution as a decimal value of t to 4 decimal places. (This is the number of years).

  $399.00 = $240.00 ( 1.005 )^(12t)

Dividing both sides of the equation by $240.00

   $399.00 / $240.00 = $240.00 ( 1.005 )^(12t) / $240.00

1.6625 = ( 1.005 )^(12t)

Applying log both sides of the equation:

log (1.6625) = log ( 1.005 )^(12t)

Using that log a^b = b log a, with a=1.005 and b=12t:

log (1.6625) = 12t log (1.005)

Dividing both sides of the equation by 12 log (1.005):

log (1.6625) / [12 log (1.005)] = 12t log (1.005) / [12 log (1.005)]

log (1.6625) / [12 log (1.005)] = t

t =  log (1.6625) / [12 log (1.005)]

log (1.6625) = 0.220761654

log (1.005) = 0.002166062

t = 0.220761654 / [12 (0.002166062)]

t = 0.220761654 / 0.025992744

t = 8.493203103

Rounding to 4 decimal places:

t = 8.4932 years

E. Convert this number to years and months. State the number of years and months it will take to save what you need. Round to the nearest whole month. REMEMBER: 12 months per year!

t = 8.4932 years

t = 8 years + 0.4932 years

t = 8 years + 0.4932 years * (12 months / year)

t = 8 years + 5.9184 months

t = 8 years 5.9184 months

Rounding to the nearest whole month:

t = 8 years 6 months

It will take to save what you need 8 years and 6 months.

answer

the correct answer is

explanation

the given function is

the excluded values are obtained by solving:

we factor to obtain:

the excluded values are

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation: