Question

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?

Answer 1

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 year

Rate of profit increasing(r) = 9%

Amount company made in 2006 (A) = $7 million.

$P=A(1+\frac{r}{100})^{n}$

$P=7(1+\frac{9}{100})^{2}$

$P=7(1+0.09)^{2}$

$P=7(1.09)^{2}$

$P=7\times 1.188$

$P=8.316$

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.