Hailey invested $2,000 in an account paying an interest rate of 2\tfrac{3}{8}2 8 3 ​ % compounded continuously. Sophie invested $2,000 in an account paying an interest rate of 2\tfrac{7}{8}2 8 7 ​ % compounded daily. To the nearest hundredth of a year, how much longer would it take for Hailey's money to double than for Sophie's money to double?

5.08 years

Step-by-step explanation:

We are to solve for time using compound interest formulas

For Hailey

Principal = $200

Amount = Double of Principal = $4000

Rate = 2 3/8% = 2.375.

Time = ?

Hence,

First, convert R as a percent to r as a decimal

r = R/100

r = 2.375/100

r = 0.02375 per year,

Then, solve the equation for t

t = ln(A/P) / r

t = ln(4,000.00/2,000.00) / 0.02375

t = 29.185 years

≈ 29.19 years

For Sophie

Principal = $200

Amount = Double of Principal = $4000

Rate = 2 7/8% = 2.875

Time = ?

First, convert R as a percent to r as a decimal

r = R/100

r = 2.875/100

r = 0.02875 per year,

Then, solve the equation for t

t = ln(A/P) / n[ln(1 + r/n)]

t = ln(4,000.00/2,000.00) / ( 365 × [ln(1 + 0.02875/365)] )

t = ln(4,000.00/2,000.00) / ( 365 × [ln(1 + 7.8767E-5)] )

t = 24.11 years

How much longer would it take for Hailey's money to double than for Sophie's money to double?

For Hailey it takes 29.19 years

For Sophie it takes 24.11 years

We find the difference

= 29.19 - 24.11

= 5.08 years

Therefore, it would take 5.08 years for Hailey's money to double than Sophie

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

eventually after a few more steps, your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x