Suppose th $$y = \frac { 2 - \log ( 100 - x ) } { 0.39 }$$ can be used to calculate the number of years y for x percent of a population of 823 web-footed sparrows to die. Approximate the percentage (to the nearest whole per cent) of web-footed
Sparrows that died after 2 yr.
A) 83%
B) 85%
C) 77%
D) 81%

Question

Quizplus · Accepted Answer

To find the percentage of web-footed sparrows that died after 2 years, we substitute $y = 2$ into the given equation and solve for $x$. $$2 = \frac{2 - \log(100 - x)}{0.39}$$Multiplying both sides by 0.39 gives:$$0.78 = 2 - \log(100 - x)$$Subtracting 2 from both sides:$$-1.22 = - \log(100 - x)$$Taking the negative of both sides:$$1.22 = \log(100 - x)$$To find $100 - x$, we use the inverse of the logarithm, which is 10 raised to the power of both sides:$$10^{1.22} = 100 - x$$$$10^{1.22} \approx 16.6$$So,$$100 - x = 16.6$$Solving for $x$:$$x = 100 - 16.6 = 83.4$$Rounded to the nearest whole percent, $x = 83\%$.

Suppose Th y=2−log⁡(100−x)0.39y = \frac { 2 - \log ( 100 - x ) } { 0.39 }y=0.392−log(100−x)​

Suppose Th $y = \frac { 2 - \log ( 100 - x ) } { 0.39 }$