Question

find how many years it would take for an investment of 4400 dollars to grow to 7100 dollars at an annual interest rate of 4.5% compounded continuously. round your answer to 2 decimal places as needed. question help: video post to forum submit question question 5 0/5 pts 2 99 details find the time it takes for $5,900 to double when invested at an annual interest rate of 19%, compounded continuously. years give your answer accurate to the tenths place value. find the time it takes for $590,000 to double when invested at an annual interest rate of 19%, compounded continuously. years give your answer accurate to the tenths place value. question help: video post to forum submit question question 6 0/5 pts 2 99 details the number of bacteria in a culture is given by the function, $n(t)=950e^{0.25t}$, where $t$ is measured in hours. a) what is the relative rate of growth of this bacterium population? % b) what is the initial population of the culture? bacteria c) how many bacteria will the culture contain at time $t = 5$? round to the nearest whole bacteria.

Explanation:

Question 1:

Step1: Recall continuous - compounding formula

The formula for continuous - compounding is $A = Pe^{rt}$, where $A$ is the final amount, $P$ is the principal amount, $r$ is the annual interest rate (in decimal form), and $t$ is the time in years. Given $P = 4400$, $A = 7100$, and $r=0.045$. Substitute these values into the formula: $7100 = 4400e^{0.045t}$.

Step2: Isolate the exponential term

Divide both sides of the equation by 4400: $\frac{7100}{4400}=e^{0.045t}$, which simplifies to $\frac{71}{44}=e^{0.045t}$.

Step3: Take the natural logarithm of both sides

$\ln(\frac{71}{44})=\ln(e^{0.045t})$. Since $\ln(e^{x}) = x$, we have $\ln(\frac{71}{44}) = 0.045t$.

Step4: Solve for $t$

$t=\frac{\ln(\frac{71}{44})}{0.045}$. Calculate $\ln(\frac{71}{44})\approx\ln(1.61364)\approx0.478$ and $t=\frac{0.478}{0.045}\approx10.62$.

Step1: Use continuous - compounding formula for doubling

When an amount doubles, $A = 2P$. The continuous - compounding formula is $A = Pe^{rt}$. Substituting $A = 2P$ into it gives $2P=Pe^{0.19t}$. Since $P
eq0$, we can divide both sides by $P$ to get $2 = e^{0.19t}$.

Step2: Take the natural logarithm of both sides

$\ln(2)=\ln(e^{0.19t})$. Since $\ln(e^{x}) = x$, we have $\ln(2)=0.19t$.

Step3: Solve for $t$

$t=\frac{\ln(2)}{0.19}$. Calculate $\ln(2)\approx0.693$ and $t=\frac{0.693}{0.19}\approx3.6$.

Step1: Use the same logic as above

Since the time $t$ for an amount to double when compounded continuously does not depend on the principal amount $P$ (as shown in the previous part where $P$ cancels out), for $P = 590000$ and $r = 0.19$, when $A = 2P$, we have $2P=Pe^{0.19t}$, then $2 = e^{0.19t}$.

Step2: Take the natural logarithm of both sides

$\ln(2)=\ln(e^{0.19t})$, so $\ln(2)=0.19t$.

Step3: Solve for $t$

$t=\frac{\ln(2)}{0.19}\approx3.6$.

Answer:

$10.62$

Question 5 (first part):