Xavier is going to invest $3,900 and leave it in an account for 14 years. Assuming the interest is compounded daily, what interest rate, to the nearest hundredth of a percent, would be required in order for Xavier to end up with $9,200

the interest rate is 6%

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

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

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = $3900

A = $9200

t = 14 years

n = 365 because it was compounded 365 times in a year.

Therefore,

9200 = 3900(1 + r/365)^365 × 14

9200/3900 = (1 + 0.0027274r)^5110

2.359 = (1.0027274r)^5110

Taking log of both sides of the equation,it becomes

Log 2.359 = 5110 log (1 + 0.0027274r)

0.373 = 5110 log (1 + 0.0027274r)

0.373/5110 = log (1 + 0.0027274r)

0.000073 = log (1 + 0.0027274r)

Taking inverse log of both sides of the equation, it becomes

10^0.000073 = 10^log (1 + 0.0027274r)

1.000168 = 1 + 0.0027274r

0.0027274r = 1.000168 - 1

0.0027274r = 0.000168

r = 0.000168/0.0027274

r = 0.06

r = 0.06 × 100 = 6%

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step-by-step explanation:

i took the test: )

its 50 miles away from the starting point