Question

Which statement is correct?
StartFraction 3.56 times 10 Superscript 2 Baseline Over 1.09 times 10 Superscript 4 Baseline EndFraction less-than-or-equal-to (4.08 times 10 Superscript 2 Baseline) (1.95 times 10 Superscript negative 6 baseline)
StartFraction 3.56 times 10 Superscript 2 Baseline Over 1.09 times 10 Superscript 4 Baseline EndFraction less-than (4.08 times 10 Superscript 2 Baseline) (1.95 times 10 Superscript negative 6 baseline)
StartFraction 3.56 times 10 Superscript 2 Baseline Over 1.09 times 10 Superscript 4 Baseline EndFraction greater-than (4.08 times 10 Superscript 2 Baseline) (1.95 times 10 Superscript negative 6 baseline)
StartFraction 3.56 times 10 Superscript 2 Baseline Over 1.09 times 10 Superscript 4 Baseline EndFraction = (4.08 times 10 Superscript 2 Baseline) (1.95 times 10 Superscript negative 6 baseline)

Answer 1

To determine which statement is correct, we need to evaluate both sides of the expressions.

Let's calculate the left side:

\[
\text{Left Side} = \frac{3.56 \times 10^2}{1.09 \times 10^4}
\]

Calculating this:

\[
= \frac{3.56}{1.09} \times 10^{2 - 4}
= \frac{3.56}{1.09} \times 10^{-2}
\]

Using a calculator (or computation), we find:

\[
\frac{3.56}{1.09} \approx 3.26
\]

So,

\[
\text{Left Side} \approx 3.26 \times 10^{-2} = 0.0326
\]

Now, let's calculate the right side:

\[
\text{Right Side} = (4.08 \times 10^2) \times (1.95 \times 10^{-6})
\]

Calculating this:

\[
= 4.08 \times 1.95 \times 10^{2 - 6}
= 4.08 \times 1.95 \times 10^{-4}
\]

Now, calculating \(4.08 \times 1.95\):

\[
4.08 \times 1.95 \approx 7.956
\]

So,

\[
\text{Right Side} \approx 7.956 \times 10^{-4} = 0.0007956
\]

Now we compare the two sides:

- Left Side: \(0.0326\)
- Right Side: \(0.0007956\)

Now we can determine the relationship:

\[
0.0326 > 0.0007956
\]

Thus, the correct statement is:

**StartFraction 3.56 times 10 Superscript 2 Baseline Over 1.09 times 10 Superscript 4 Baseline EndFraction greater-than (4.08 times 10 Superscript 2 Baseline) (1.95 times 10 Superscript negative 6 baseline)**.