MATH SOLVE

4 months ago

Q:
# In the year 2006, a company made $7 million in profit. For each consecutive year after that, their profit increased by 9%. How much would the company's profit be in the year 2008, to the nearest tenth of a million dollars?

Accepted Solution

A:

Answer:The company's profit be in the year 2008 is $8.3 million.Step-by-step explanation:Given:In the year 2006, a company made $7 million in profit, their profit increased by 9%.So, we need to calculate the company's profit be in the year 2008.Now, by putting the formula to find the profit(P) after the two year:Difference between the year = 2008 - 2006 = 2 year.So, number of years(n) = 2 yearRate of profit increasing(r) = 9%Amount company made in 2006 (A) = $7 million.[tex]P=A(1+\frac{r}{100})^{n}[/tex][tex]P=7(1+\frac{9}{100})^{2}[/tex][tex]P=7(1+0.09)^{2}[/tex][tex]P=7(1.09)^{2}[/tex][tex]P=7\times 1.188[/tex][tex]P=8.316[/tex]Profit in the year 2008 would be 8.316 million, and nearest to the tenth of a million dollars is $8.3 as 3 is in the tenth place of the decimal and 1 in the hundredth so rounding will change $8.316 to $8.3.Therefore, the company's profit be in the year 2008 is $8.3 million.