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The atmospheric pressure in the lower stratosphere decreases exponentially from 473 lb \(/ \mathrm{ft}^{2}\) at \(36,152 \mathrm{ft}\) to \(51 \mathrm{lb} / \mathrm{ft}^{2}\) at \(82,345 \mathrm{ft}\) Data: grc.nasa.gov a) Find the exponential decay rate \(k,\) and write an equation for an exponential function that can be used to estimate the atmospheric pressure in the stratosphere \(h\) feet above \(36,152\) ft. b) Estimate the atmospheric pressure at \(50,000 \mathrm{ft}\) \((h=50,000-36,152)\) c) At what height is the atmospheric pressure \(100 \mathrm{lb} / \mathrm{ft}^{2} ?\) d) What change in altitude will result in atmospheric pressure being halved?

The decay rate of krypton- 85 is \(6.3 \%\) per year. What is its half-life?

The moose population in New York is growing exponentially. The number of moose in the state \(t\) years after 1997 can be approximated by $$ M(t)=50(1.25)^{t} $$ a) Estimate the number of moose in New York in \(1997,\) in \(2012,\) and \(\operatorname{in} 2020\) b) Graph the function. (IMAGE CANT COPY)

Atmospheric pressure \(P\) at an elevation \(a\) feet above sea level can be estimated by $$ P=P_{0} e^{-0.00004 a} $$ where \(P_{0}\) is the pressure at sea level, which is approximately 29.9 in. of mercury (Hg). Explain how a barometer, or some other device for measuring atmospheric pressure, can be used to find the height of a skyscraper.

Growth of Bacteria. The bacteria Escherichia coli (E. coli) are commonly found in the human bladder. Suppose that 3000 of the bacteria are present at time \(t=0 .\) Then \(t\) minutes later, the number of bacteria present is $$ N(t)=3000(2)^{t / 20} $$ If \(100,000,000\) bacteria accumulate, a bladder infection can occur. If, at 11: 00 A.M., a patient's bladder contains \(25,000 E\) coli bacteria, at what time can infection occur?

Simplify. \(\sqrt[3]{40 a^{5} b^{12}}\)

Solve $$ \log _{6} x=2 $$

Simplify. \(i^{43}\)

Solve Explain why a logarithm base must be positive.

Solve for \(x\). Give an approximation to four decimal places. $$ \frac{3.01}{\ln x}=\frac{28}{4.31} $$