Oliver invested $970 in an account paying an interest rate of 7 1/2 % compounded continuously. Carson invested $970 in an account paying an interest rate of 7 3/8% compounded annually. To the nearest of a hundredth of a year, how much longer would it take for Carson's money to double than for Oliver's money to double?

0.50 or about half a year longer.

Step-by-step explanation:

We can write an equation to model bot investments.

Oliver invested $970 in an account paying an interest rate of 7.5% compounded continuously.

Recall that continuous compound is given by the equation:

Where A is the amount afterwards, P is the principal amount, r is the rate, and t is the time in years.

Since the initial investment is $970 at a rate of 7.5%:

Carson invested $970 in an account paying an interest rate of 7.375% compounded annually.

Recall that compound interest is given by the equation:

Where A is the amount afterwards, P is the principal amount, r is the rate, n is the number of times compounded per year, and t is the time in years.

Since the initial investment is $970 at a rate of 7.375% compounded annually:

When Oliver's money doubles, he will have $1,940 afterwards. Hence:

Solve for t:

Take the natural log of both sides:

Simplify:

When Carson's money doubles, he will have $1,940 afterwards. Hence:

Solve for t:

Take the natural log of both sides:

Simplify:

Hence:

Then it will take Carson's money:

About 0.50 or half a year longer to double than Oliver's money.

y-y1=4(x-x1)

step-by-step explanation:

this is point-slope form since you only have 4 as the slope it goes in place of m

hope this : )