Question

question 9
an unknown radioactive element decays into non - radioactive substances. in 560 days the radioactivity of a sample decreases by 62 percent.
(a) find the decay constant k. (round your answer to 5 decimal places.)
k≈
(b) what is the half - life of the element? (round your answer to two decimal places)
half - life: (days)
(c) how long will it take for a sample of 100 mg to decay to 95 mg? (round your answer to two decimal places)
time needed: (days)
question help: video post to forum
submit question
question 10
a bacteria culture starts with 340 bacteria and grows at a rate proportional to its size. after 4 hours there will be 136 bacteria.
(a) express the population p after t hours as a function of t.
round the growth rate to five decimal places
p(t)=
(b) what will be the population after 7 hours?
bacteria
(c) how long will it take for the population to reach 1830? round your answer to two decimal places.
hours

Explanation:

Question 9

Step1: Find decay - constant $k$

The formula for radioactive decay is $A = A_0e^{-kt}$, where $A$ is the amount of the substance at time $t$, $A_0$ is the initial amount. If the radioactivity decreases by 62 percent in $t = 560$ days, then the remaining amount $A=(1 - 0.62)A_0=0.38A_0$. Substituting into the decay formula: $0.38A_0=A_0e^{-560k}$. Divide both sides by $A_0$ (since $A_0
eq0$) to get $0.38 = e^{-560k}$. Take the natural - logarithm of both sides: $\ln(0.38)=\ln(e^{-560k})$. Using the property $\ln(e^x)=x$, we have $\ln(0.38)=-560k$. Then $k=-\frac{\ln(0.38)}{560}\approx0.00173$.

Step2: Find the half - life $T$

The half - life formula is $T=\frac{\ln(2)}{k}$. Substituting $k\approx0.00173$ into the formula, we get $T=\frac{\ln(2)}{0.00173}\approx400.92$ days.

Step3: Find the time $t$ for decay from 100 mg to 95 mg

Using the decay formula $A = A_0e^{-kt}$, with $A_0 = 100$, $A = 95$, and $k\approx0.00173$. So $95 = 100e^{-0.00173t}$. Divide both sides by 100: $0.95 = e^{-0.00173t}$. Take the natural - logarithm of both sides: $\ln(0.95)=\ln(e^{-0.00173t})$. Since $\ln(e^x)=x$, we have $\ln(0.95)=-0.00173t$. Then $t=-\frac{\ln(0.95)}{0.00173}\approx29.14$ days.

Step1: Find the growth function $P(t)$

The population growth formula is $P(t)=P_0e^{rt}$, where $P_0$ is the initial population, $r$ is the growth rate, and $t$ is the time. Given $P_0 = 340$, and when $t = 4$, $P(4)=1360$. Substitute into the formula: $1360 = 340e^{4r}$. Divide both sides by 340: $4 = e^{4r}$. Take the natural - logarithm of both sides: $\ln(4)=\ln(e^{4r})$. Since $\ln(e^x)=x$, we have $\ln(4)=4r$. Then $r=\frac{\ln(4)}{4}\approx0.34657$. So $P(t)=340e^{0.34657t}$.

Step2: Find the population after 7 hours

Substitute $t = 7$ into $P(t)=340e^{0.34657t}$: $P(7)=340e^{0.34657\times7}=340e^{2.42599}\approx340\times11.29827\approx3841.41$.

Step3: Find the time $t$ for the population to reach 1830

Set $P(t)=1830$ in $P(t)=340e^{0.34657t}$. So $1830 = 340e^{0.34657t}$. Divide both sides by 340: $\frac{1830}{340}=e^{0.34657t}$, or $5.38235 = e^{0.34657t}$. Take the natural - logarithm of both sides: $\ln(5.38235)=\ln(e^{0.34657t})$. Since $\ln(e^x)=x$, we have $\ln(5.38235)=0.34657t$. Then $t=\frac{\ln(5.38235)}{0.34657}\approx4.99$.

Answer:

(a) $0.00173$
(b) $400.92$
(c) $29.14$

Question 10