To find a reasonable estimate of the quotient of the highest and lowest audible frequencies of a porpoise, we need to perform the following steps:
1. Identify the highest frequency a porpoise can hear: [tex]\(15 \times 10^3\)[/tex] cycles per second.
2. Identify the lowest frequency a porpoise can hear: [tex]\(75 \times 10^1\)[/tex] cycles per second.
3. Calculate the quotient [tex]\( \frac{\text{highest frequency}}{\text{lowest frequency}} \)[/tex].
Let's break this down:
1. The highest frequency is [tex]\(15 \times 10^3 = 15,000\)[/tex] cycles per second.
2. The lowest frequency is [tex]\(75 \times 10^1 = 750\)[/tex] cycles per second.
To find the quotient:
[tex]\[ \text{Quotient} = \frac{15,000}{750} \][/tex]
When we divide 15,000 by 750, we get:
[tex]\[ \frac{15,000}{750} = 20 \][/tex]
Hence, the quotient of the highest and lowest audible frequencies of a porpoise is 20.
Now, we check which range the quotient falls into:
- Between 10,000 and 100,000
- Between 100,000 and 1,000,000
- Between 1,000 and 10,000
- Between 10,000,000 and 100,000,000
Since 20 does not fall into any of these given ranges, none of the ranges listed are reasonable estimates. This suggests that the quotient 20 is simpler and doesn't fit the ranges provided in the question.
