Question

Allometric relations often can be modeled by ​f(x)= x^b​, where a and b are constants. One study showed that for a male fiddler crab weighing over 0.75​ gram, the weight of its claws can be estimated by ​f(x)= 0.448 x^1.21. The input x is the weight of the crab in​ grams, and the output​ f(x) is the weight of the claws in grams.
(a) Predict the weight of the claws of a 2-gram crab

(b) Approximate the weight of a crab that has 0.5-gram claws

Answer 1

Answer:

a) $f(2)= 0.448 (2)^{1.21}= 1.036 grams$

b) $0.5 = 0.448 x^{1.21}$

Dividing both sides by 0.448 we got:

$\frac{0.5}{0.448} = x^{1.21}$

We can appy the exponent $\frac{1}{1.21}$ in both sides of the equation and we got:

$(\frac{0.5}{0.448})^{\frac{1}{1.21}} = x= 1.095grams$

Step-by-step explanation:

For this case we know the following function:

$f(x) = 0.448 x^{1.21}$

The notation is: x is the weight of the crab in​ grams, and the output​ f(x) is the weight of the claws in grams.

Part a

For this case we just need to replace x = 2 gram in the function and we got:

$f(2)= 0.448 (2)^{1.21}= 1.036 grams$

Part b

For this case we know tha value for $f(x) =0.5$ and we want to find the value of x who satisfy this condition:

$0.5 = 0.448 x^{1.21}$

Dividing both sides by 0.448 we got:

$\frac{0.5}{0.448} = x^{1.21}$

We can appy the exponent $\frac{1}{1.21}$ in both sides of the equation and we got:

$(\frac{0.5}{0.448})^{\frac{1}{1.21}} = x= 1.095grams$