Cricket Expression

Cricket Expression
N=(5.63×10^10)e^-(6290K)t …what is the temp in fahrenheit? please help DUE TOMORROW at 4?

the rate of chirping of the snowy tree cricket varies with temperature. a linear relationship provides a good match to the chirp rate, but an even more accurate relationship is the following:
N=(5.63 . 10^10)e^-(6290K)/T
In this expression, N is the number of chirps in 13.0s and T is the temperatrue in Kelvins. if a cricket is obswerved to chirp 185 times in 60.0s, what is the temperature in degrees Fahrenheit?

first youll have to find your N. N in this case would be 185 * (13/60) = 40.08333….. You divide N by the factor in front of the e value. 40.083/5.63*10^10 = 7.1196 * 10^-10 = e^(-6290/T).

Take the natural log (LN) of both sides to get:

-21.063 = -6290/T

solve for T
T = 298.6 degrees kelvin
convert to Celsius = T(kelvin) – 273.15 = 25.45
convert to Fahrenheit = T(cel) * (9/5) + 32 = 77.81.

I lost some accuracy with my rounding. doing the procedure straight through with my calculator gives me 77.86 deg F.

Hope this explains it clearly.