Question
A scientist looks at a bacterium and a virus in a lab. The bacterium has a diameter of 10 Superscript negative 6 meters. The virus has a diameter of 10 Superscript negative 7 meters. Which statement accurately compares the sizes of the specimens?
The diameter of the bacterium is 10 times greater than that of the virus.
The diameter of the bacterium is StartFraction 1 Over 100 EndFraction times as great as that of the virus.
The diameter of the virus is 10 times greater than that of the bacterium.
The diameter of the virus is StartFraction 1 Over 100 EndFraction times as great as that of the bacterium.
The diameter of the bacterium is 10 times greater than that of the virus.
The diameter of the bacterium is StartFraction 1 Over 100 EndFraction times as great as that of the virus.
The diameter of the virus is 10 times greater than that of the bacterium.
The diameter of the virus is StartFraction 1 Over 100 EndFraction times as great as that of the bacterium.
Answers
GPT-4o mini
To compare the sizes of the bacterium and the virus, we start with their given diameters:
- Diameter of the bacterium = \(10^{-6}\) meters
- Diameter of the virus = \(10^{-7}\) meters
Now, we can compare their sizes:
1. To find out how many times larger the diameter of the bacterium is compared to the diameter of the virus, we can divide the diameter of the bacterium by the diameter of the virus:
\[
\text{Size Ratio} = \frac{\text{Diameter of Bacterium}}{\text{Diameter of Virus}} = \frac{10^{-6}}{10^{-7}}
\]
Calculating this gives:
\[
\text{Size Ratio} = \frac{10^{-6}}{10^{-7}} = 10^{(-6) - (-7)} = 10^{1} = 10
\]
This means that the diameter of the bacterium is 10 times greater than that of the virus.
Thus, the correct statement is:
**The diameter of the bacterium is 10 times greater than that of the virus.**
- Diameter of the bacterium = \(10^{-6}\) meters
- Diameter of the virus = \(10^{-7}\) meters
Now, we can compare their sizes:
1. To find out how many times larger the diameter of the bacterium is compared to the diameter of the virus, we can divide the diameter of the bacterium by the diameter of the virus:
\[
\text{Size Ratio} = \frac{\text{Diameter of Bacterium}}{\text{Diameter of Virus}} = \frac{10^{-6}}{10^{-7}}
\]
Calculating this gives:
\[
\text{Size Ratio} = \frac{10^{-6}}{10^{-7}} = 10^{(-6) - (-7)} = 10^{1} = 10
\]
This means that the diameter of the bacterium is 10 times greater than that of the virus.
Thus, the correct statement is:
**The diameter of the bacterium is 10 times greater than that of the virus.**