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Which statement is correct?

A. [tex]\(\left(2.06 \times 10^{-2}\right)\left(1.88 \times 10^{-1}\right)\ \textless \ \frac{7.69 \times 10^{-2}}{2.3 \times 10^{-5}}\)[/tex]

B. [tex]\(\left(2.06 \times 10^{-2}\right)\left(1.88 \times 10^{-1}\right) \geq \frac{7.69 \times 10^{-2}}{2.3 \times 10^{-5}}\)[/tex]

C. [tex]\(\left(2.06 \times 10^{-2}\right)\left(1.88 \times 10^{-1}\right)\ \textgreater \ \frac{7.69 \times 10^{-2}}{2.3 \times 10^{-5}}\)[/tex]

D. [tex]\(\left(2.06 \times 10^{-2}\right)\left(1.88 \times 10^{-1}\right)=\frac{7.69 \times 10^{-2}}{2.3 \times 10^{-5}}\)[/tex]


Sagot :

To determine which statement is correct, we need to evaluate both sides of each inequality or equality, step-by-step.

First, let's evaluate the left side:

[tex]\[ \left(2.06 \times 10^{-2}\right) \times \left(1.88 \times 10^{-1}\right) \][/tex]

Multiplying the constants:

[tex]\[ 2.06 \times 1.88 = 3.8728 \][/tex]

Next, we combine the exponents of 10:

[tex]\[ 10^{-2} \times 10^{-1} = 10^{-3} \][/tex]

Therefore:

[tex]\[ \left(2.06 \times 10^{-2}\right) \times \left(1.88 \times 10^{-1}\right) = 3.8728 \times 10^{-3} = 0.0038728 \][/tex]

Next, let's evaluate the right side:

[tex]\[ \frac{7.69 \times 10^{-2}}{2.3 \times 10^{-5}} \][/tex]

We first divide the constants:

[tex]\[ \frac{7.69}{2.3} = 3.344782608695652 \][/tex]

Next, we divide the powers of 10:

[tex]\[ 10^{-2} / 10^{-5} = 10^{-2 - (-5)} = 10^{3} \][/tex]

Therefore:

[tex]\[ \frac{7.69 \times 10^{-2}}{2.3 \times 10^{-5}} = 3.344782608695652 \times 10^{3} = 3343.4782608695655 \][/tex]

Now, we compare the left side and the right side:

[tex]\[ 0.0038728 \quad \text{and} \quad 3343.4782608695655 \][/tex]

Clearly:

[tex]\[ 0.0038728 < 3343.4782608695655 \][/tex]

Thus:

[tex]\[ \left(2.06 \times 10^{-2}\right)\left(1.88 \times 10^{-1}\right) < \frac{7.69 \times 10^{-2}}{2.3 \times 10^{-5}} \][/tex]

The correct statement is:

[tex]\[ \left(2.06 \times 10^{-2}\right)\left(1.88 \times 10^{-1}\right) < \frac{7.69 \times 10^{-2}}{2.3 \times 10^{-5}} \][/tex]

So, the first statement is correct.