Question

The length of a parasite in experiment A is 2.5 × 10–3 inch. The length of a parasite in experiment B is 1.25 × 10–4 inch. How much greater is the length of the parasite in experiment A compared to the length of the parasite in experiment B? A

Answer 1

We have been given that :-

The length of a parasite in experiment A is
`2.5 * 10^(-3)`

The length of a parasite in experiment B is
`1.25 * 10^(-4)`

Let us write the the length of the parasite in experiment A in the exponent of -3.

`1.25 * 10^(-4)= 0.125 * 10^(-3)`

Clearly, the length of parasite in experiment A is greater than the length of parasite in experiment B.

The difference in the length is given by

`2.5 * 10^(-3) -0.125 * 10^(-4)`

`=2.375 * 10^(-3)`

Therefore, the length of the parasite in experiment A is
`=2.375 * 10^(-3)` inches greater than the length of the parasite in experiment B.

Answer 2

Answer: 20 times

Explanation:

Given : The length of a parasite in experiment A =
`2.5*10^(-3)\text{ inch}`

The length of a parasite in experiment B =
`1.25*10^(-4)\text{ inches}`

Then, the number of times the length of the parasite in experiment A is greater as compared to the length of the parasite in experiment B:-

`n=(2.5*10^(-3))/(1.25*10^(-4))\\\\=(2.5)/(1.25)*10^(-3-(-4))\\\\=2*10^(-3+4)=2*10^(1)=20`

Hence, the length of the parasite in experiment A is 20 times greater as compared to the length of the parasite in experiment B