Scientific Notation Online Practice

To determine how many years it will take for the bank deposit to yield $6.42×10^4 in interest given that it yields $802.50 in interest every year, we can set up the following equation:

\[
\text{Years} = \frac{\text{Total Interest Desired}}{\text{Annual Interest}}
\]

Substituting the values we have:

\[
\text{Years} = \frac{6.42 \times 10^4}{802.50}
\]

First, we can convert $802.50 into scientific notation:

\[
802.50 \approx 8.025 \times 10^2
\]

Now, we can perform the division:

\[
\text{Years} = \frac{6.42 \times 10^4}{8.025 \times 10^2} = \frac{6.42}{8.025} \times \frac{10^4}{10^2}
\]
\[
= 0.7995 \times 10^{4-2} = 0.7995 \times 10^2
\]

Now we want to express 0.7995 in standard scientific notation, which is between 1 and 10. To do that, we can write 0.7995 as \(7.995 \times 10^{-1}\).

Thus:

\[
0.7995 \times 10^2 = 7.995 \times 10^{2-1} = 7.995 \times 10^1
\]

Now rounding 7.995 to two significant figures, we have 8.0.

This leads us to the final answer:

\[
8.0 \times 10^1 \approx 8 \times 10^1
\]

So the answer is:

8×10^1