A person invests 4500 dollars in a bank. The bank pays 4.75% interest compounded semi-annually. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 5900 dollars?

It should take 5.8 years to reach his goal.

Step-by-step explanation:

In order to solve this question we must use the compounded interest's formula shown below:

M = C*(1 + r/n)^(t*n)

Where M is the final amount, C is the initial amount, r is the interest rate, n is the compound rate and t is the time elapsed in years. Applying the data from the problem we have:

5900 = 4500*(1 + 0.0475/2)^(2*t)

5900 = 4500*(1 + 0.02375)^(2*t)

5900 = 4500*(1.02375)^(2*t)

4500*(1.02375)^(2*t) = 5900

(1.02375)^(2*t) = 5900/4500

(1.02375)^(2*t) = 59/45

log[1.02375^(2*t)] = log(59/45)

2*t*log(1.02375) = log(59/45)

t = log(59/45)/[2*log(1.02375)] = 5.77008

It should take 5.8 years to reach his goal.

$2416.6

step-by-step explanation:

per hour = $ 25

total work time = 35

commission = 15%

total she has to make = $ 4,500

let she sold x dollar products

so her commission will be 15 % of x

now

total = weekly basic pay + commission

4500= (25*35) + 15 % of x

4500=875 + * x

4500-875 =   * x

3625= * x

multiplying wwith 100 on both sides gives

36250 = 15 x

now

x =  

x= $2416.6