In a mountain range of California, the percent of moisture that falls as snow rather than rain can be approximated by the function $p(h) = 78 \ln(h) - 618$, where $h$ is the altitude in feet and $p(h)$ is the percent of an annual snow fall at the altitude $h$. Use the function to approximate the amount of snow at the altitudes 4000 feet and 8000 feet. The percent of annual precipitation that falls as snow at 4000 feet is approximately $oxed{}$ % (Round to the nearest integer.)
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The question mentions a function that represents the percent of annual precipitation that falls as snow at a certain elevation. The function is denoted as "func 78" and it is not explicitly given. Show more…
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In a mountain range of California, the percent of moisture that falls as snow rather than rain can be approximated by the function p(h) = 79 * ln(h) - 621, where h is the altitude in feet and p(h) is the percent of an annual snowfall at the altitude h. Use the function to approximate the amount of snow at the altitudes 3000 feet and 5000 feet. The percent of annual precipitation that falls as snow at 3000 feet is approximately __% (round to the nearest integer).
William S.
In the central Sierra Nevada of California, the percent of moisture that falls as snow rather than rain is approximated by $$f(x)=86.3 \ln x-680$$ where $x$ is the altitude in feet. (a) What percent of the moisture at $5000 \mathrm{ft}$ falls as snow? (b) What percent at $7500 \mathrm{ft}$ falls as snow?
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Common and Natural Logarithms
In the central Sierra Nevada of California, the percent of moisture that falls as snow rather than rain is approximated reasonably well by $$f(x)=86.3 \ln x-680$$ where $x$ is the altitude in feet. (a) What percent of the moisture at 5000 ft falls as snow? (b) What percent at 7500 ft falls as snow?
Donna D.
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