Anonymous asked in Science & MathematicsMathematics · 2 weeks ago

Math Quesiton?

In 1990 a fatal disease evolves to which 40 percent of a population of 5 million trees is susceptible. The proportion of susceptible trees which survive for a period of t years beyond 1990 is 𝑒^−𝑡.

(b) How quickly is the disease killing off trees at the start of 1992?

270671  trees per year

(c) When will the population be reduced to 80 percent of the level in 1990?

I got B, but I keep getting C wrong can someone help?

1 Answer

  • 2 weeks ago
    Favorite Answer

    dN/dt = -No*e^(-t).

    When No = 0.40*5,000,000 = 2,000,000,

    then at t = 1 you have -dN/dt = 2000000/e, which is much more than 270671.  So I guess what is meant by "t years beyond 1990" is t years beyond the START of 1990.  Then the start of 1992 is t = 2, and

    -dN/dt = 2000000/e^2 = 270671 as you said.

    For (c), remember that 3000000 trees are not susceptible to the disease at all.  So you are looking for a time when the disease has killed half of the susceptible trees, i.e., e^(-t) = 1/2, or e^t = 2, so t = 0.693 years.  It already happens in early September of 1990.

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