usse the following formula to find the efficiency of energy transfer between the harbor seal (2,500 Kcal) and a polar bear (375 Kcal).

Question

10%
12%
15%
20%

Answer 1

To calculate the efficiency of energy transfer from the harbor seal to the polar bear, you can use the formula:

\[ \text{Efficiency} = \left( \frac{\text{Energy received by predator}}{\text{Energy available from prey}} \right) \times 100 \]

In this case, the energy received by the polar bear (predator) is 375 Kcal, and the energy available from the harbor seal (prey) is 2,500 Kcal.

Now, substituting the values:

\[ \text{Efficiency} = \left( \frac{375 , \text{Kcal}}{2500 , \text{Kcal}} \right) \times 100 \]

Calculating this gives:

\[ \text{Efficiency} = \left( 0.15 \right) \times 100 = 15% \]

Therefore, the efficiency of energy transfer between the harbor seal and the polar bear is 15%.

Answer 2

are you sure this is right

Answer 3

Yes, the calculation I provided is correct based on the information given.

To recap:

1. Energy from the harbor seal (prey) = 2,500 Kcal
2. Energy received by the polar bear (predator) = 375 Kcal

Using the formula for efficiency of energy transfer:

\[ \text{Efficiency} = \left( \frac{\text{Energy received by predator}}{\text{Energy available from prey}} \right) \times 100 \]

Substituting the values:

\[ \text{Efficiency} = \left( \frac{375 , \text{Kcal}}{2500 , \text{Kcal}} \right) \times 100 \]

Calculating this gives:

\[ \text{Efficiency} = \left( 0.15 \right) \times 100 = 15% \]

So, the efficiency of energy transfer from the harbor seal to the polar bear is indeed 15%. If there is any specific part you want to clarify or if you have additional context, please let me know!