Jocelyn invests $1200 in an account that earns 2.4% annual interest. marcus invests $400 in an account that earns 5.2% annual interest. find when the value of marcus's investment equals the value of jocelyn's investment and find the common value of the investments at that time.

Let the time that the two investments be n, then the value of the investment at time n is given by P(1 + r)^n where P is the invested amount, r is the rate.
Thus, 1200(1 + 2.4/100)^n = 400(1 + 5.2)^n
1200(1 + 0.024)^n = 400(1 + 0.052)^n
1200(1.024)^n = 400(1.052)^n
1200 / 400 = (1.052)^n / (1.024)^n
3 = (1.052 / 1.024)^n = (1.02734375)^n
log 3 = log (1.02734375)^n = n log (1.02734375)
n = log 3 / log (1.02734375) = 40.72 years.
Therefore, the two investments becomes equal after 40.72 years.

The common value of the investment after 40.72 years is 1200(1.024)^40.72 = 1200 x 2.627 = $3,152.42

Answer:

$4034.40

step-by-step explanation:

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40