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Zorluk: OrtaLinear and Exponential Growth

An agricultural scientist is tracking the biomass, in grams, of two different plant species under controlled conditions. Species A starts with a biomass of 250250 grams and increases by 1212 grams each day. Species B starts with a biomass of 180180 grams and its biomass increases exponentially by a constant daily percentage. On day 55, the biomass of Species A and Species B is equal. What is the daily percentage increase in the biomass of Species B, to the nearest tenth of a percent?

  1. 11.5%11.5\%Cevap
  2. B
    14.4%14.4\%
  3. C
    34.4%34.4\%
  4. D
    10.0%10.0\%

Cevap

The correct answer is 11.5%11.5\%.
The correct answer is 11.5%11.5\%. First, find the biomass of Species A on day 55 using the linear equation 250+12(5)=310250 + 12(5) = 310 grams. Because the biomass of both species is equal on day 55, set up the exponential equation for Species B as 180(1+r)5=310180(1 + r)^5 = 310, where rr is the daily rate of increase. Divide both sides by 180180 to get (1+r)51.7222(1 + r)^5 \approx 1.7222. Taking the fifth root of both sides yields 1+r1.11481 + r \approx 1.1148, which means r0.1148r \approx 0.1148, or 11.5%11.5\%.

Adım Adım Çözüm

1
Calculate the biomass of Species A on day 55.
The biomass of Species A on day 55 is 250+12(5)=310250 + 12(5) = 310 grams.
This establishes the common biomass value that both species reach on day 55 using the linear growth model.
2
Set up the exponential growth equation for Species B.
The equation is 180(1+r)5=310180(1 + r)^5 = 310, where rr is the daily growth rate.
Species B grows exponentially, starting at 180180 grams and reaching 310310 grams on day 55.
3
Solve for the daily growth rate, rr.
(1+r)5=3101801.7222(1 + r)^5 = \frac{310}{180} \approx 1.7222, which yields 1+r(1.7222)1/51.11481 + r \approx (1.7222)^{1/5} \approx 1.1148, so r0.1148r \approx 0.1148 or 11.5%11.5\%.
Isolating rr by taking the fifth root of both sides gives the daily percentage increase.

Anahtar Kavram

Linear vs. Exponential Growth Models
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