A biologist uses the formula E=km^(2)/(4) to model the relationship between the mass of a bird's egg, E grams, and the mass of the bird, m grams. The number k is a constant that is close to 0.25 for healthy eggs. a) A ruby-throated hummingbird has a mass of 3.4 g. What is the mass of its egg? b) An ostrich egg has a mass of 1.4 kg. What is the mass of the mother bird? c) A 7.3 -kg whooping crane lays an egg whose mass is 208 g Does this bird appear to be healthy? Explain.

**a) Mass of hummingbird egg:**<br /><br />* Given: m = 3.4 g, k ≈ 0.25<br />* Formula: E = km^(2/4) = km^(1/2)<br />* Calculation: E = 0.25 * (3.4)^(1/2) <br />* E ≈ 0.25 * 1.84<br />* E ≈ 0.46 g<br /><br />Therefore, the mass of the hummingbird's egg is approximately 0.46 grams.<br /><br />**b) Mass of the mother ostrich:**<br /><br />* Given: E = 1.4 kg = 1400 g (since 1 kg = 1000 g), k ≈ 0.25<br />* Formula: E = km^(1/2)<br />* Rearranging the formula to solve for m: m = (E/k)^2<br />* Calculation: m = (1400/0.25)^2<br />* m = (5600)^2<br />* m = 31,360,000 g<br />* m = 31,360 kg (since 1000 g = 1 kg)<br /><br />Therefore, the mass of the mother ostrich is approximately 31,360 kg.<br /><br />**c) Health of the whooping crane:**<br /><br />* Given: m = 7.3 kg = 7300 g, E = 208 g, k ≈ 0.25<br />* We will calculate the expected egg mass for a healthy whooping crane with a mass of 7.3 kg using the given formula and compare it to the actual egg mass.<br />* Calculation: E = 0.25 * (7300)^(1/2)<br />* E ≈ 0.25 * 85.44<br />* E ≈ 21.36 g<br /><br />The expected mass of a healthy whooping crane's egg is approximately 21.36 g. The actual egg mass is 208 g, which is significantly larger (almost 10 times larger) than the expected mass.<br /><br />Therefore, based on this model, the whooping crane does *not* appear to be healthy. The egg is much larger than expected for a bird of that mass. There could be various reasons for this discrepancy, including measurement error, individual variation, or a health issue affecting the bird or the egg.<br />