Question

the approximate diameters of four objects are shown in the table. which statements correctly compare the approximate diameters of the objects? choose all that apply. object approximate diameter (millimeters) pet dander 5×10^(-4) dust mite 1.25×10^(-1) spider silk 4×10^(-3) human hair 8×10^(-2) a. the diameter of spider silk is about 8 times as large as the diameter of pet dander. b. the diameter of human hair is about 64 times as large as the diameter of a dust mite. c. the diameter of a dust mite is about 250 times as large as the diameter of pet dander. d. the diameter of spider silk is about 320 times as large as the diameter of a dust mite. e. the diameter of pet dander is about 4000 times as large as the diameter of a dust mite.

Explanation:

Step1: Calculate ratio of spider - silk to pet - dander diameter

To find how many times larger the diameter of spider - silk is than the diameter of pet - dander, divide the diameter of spider - silk by the diameter of pet - dander. Diameter of spider - silk is $4\times10^{-3}$ mm and diameter of pet - dander is $5\times10^{-4}$ mm. So, $\frac{4\times 10^{-3}}{5\times 10^{-4}}=\frac{4}{5}\times10^{-3 + 4}=0.8\times10^{1}=8$.

Step2: Calculate ratio of human - hair to dust - mite diameter

Diameter of human - hair is $8\times10^{-2}$ mm and diameter of dust - mite is $1.25\times10^{-1}$ mm. So, $\frac{8\times 10^{-2}}{1.25\times 10^{-1}}=\frac{8}{1.25}\times10^{-2+1}=6.4\times10^{-1}=0.64
eq64$.

Step3: Calculate ratio of spider - silk to dust - mite diameter

$\frac{4\times 10^{-3}}{1.25\times 10^{-1}}=\frac{4}{1.25}\times10^{-3 + 1}=3.2\times10^{-2}
eq320$.

Step4: Calculate ratio of pet - dander to dust - mite diameter

$\frac{5\times 10^{-4}}{1.25\times 10^{-1}}=\frac{5}{1.25}\times10^{-4 + 1}=4\times10^{-3}
eq4000$.

Answer:

A. The diameter of spider silk is about 8 times as large as the diameter of pet dander.