13. eric deposits $4,700 at 1.03% interest, compounded continuously for five years.\na. what is his ending…
13. eric deposits $4,700 at 1.03% interest, compounded continuously for five years.\na. what is his ending balance?\nb. how much interest did the account earn?\n14. write the verbal sentence that is the translation of $lim_{x \to +infty} f(x)=3.66$.\n15. write the verbal sentence given below symbolically using limit notation.\nthe limit of g(x), as x approaches zero, is fifteen.\n16. given the function $f(x)=\frac{2x - 17}{x}$, use a table to find $lim_{x \to +infty} f(x)$.\n17. find the balance for each compounding period on $50,000 for $2\frac{1}{2}$ years at a rate of 1.3%.\na. annually = 51640.88\nb. semiannually = 51,646.26\nc. quarterly = 51,642.97\nd. monthly = 51,650.74\ne. daily\nf. hourly = 51,651.69\ng. continuously\n18. a private university has an endowment fund that currently has 49 million dollars in it. if it is invested in a one - year cd that pays 2% interest compounded continuously, how much interest will it earn?\n19. use a table of increasing values of x to find each of the following limits.\na. $lim_{x \to +infty} f(x)$ if $f(x)=\frac{5x - 2}{x + 3}=5$\nb. $lim_{x \to +infty} g(x)$ if $g(x)=\frac{12x + 5}{4x+3}=3$\nc. $lim_{x \to +infty} f(x)$ if $f(x)=\frac{5x^{3}-100}{x^{2}}=+infty$\nd. $lim_{x \to +infty} f(x)$ if $f(x)=\frac{7x^{2}-1}{x^{3}+2}=0$\n20. find the interest earned on a $14,000 balance for nine months at 1.1% interest compounded continuously.\n21. assume you had p dollars to invest in an account that paid 5% interest compounded continuously. how long would it take your money to double? (hint: try substituting different numbers of years into the continuous compounding formula). round to the nearest year.\n34 financial algebra workbook 2 - 6
Answer
13.
a.
Explanation:
Step1: Recall continuous - compounding formula
The formula for continuous compounding is $A = Pe^{rt}$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal form), $t$ is the time in years, and $A$ is the ending balance. Given $P=$4700$, $r = 0.0103$ (since $1.03%=0.0103$), and $t = 5$.
Step2: Substitute values into the formula
$A=4700\times e^{0.0103\times5}=4700\times e^{0.0515}$. Using a calculator, $e^{0.0515}\approx1.0528$. So $A = 4700\times1.0528=$4958.16$.
Answer:
$$4958.16$
b.
Explanation:
Step1: Recall the interest - earning formula
The interest $I$ earned is $I=A - P$. We know $A = 4958.16$ and $P = 4700$ from part (a).
Step2: Calculate the interest
$I=4958.16−4700=$258.16$.
Answer:
$$258.16$
14.
Brief Explanations:
The limit notation $\lim_{x\rightarrow+\infty}f(x)=3.66$ can be translated as "The limit of the function $f(x)$ as $x$ approaches positive infinity is $3.66$".
Answer:
The limit of the function $f(x)$ as $x$ approaches positive infinity is $3.66$.
15.
Answer:
$\lim_{x\rightarrow0}g(x)=15$
16.
Explanation:
Step1: Rewrite the function
$f(x)=\frac{2x - 17}{x}=2-\frac{17}{x}$.
Step2: Analyze the limit as $x\rightarrow+\infty$
As $x\rightarrow+\infty$, $\frac{17}{x}\rightarrow0$. So $\lim_{x\rightarrow+\infty}f(x)=\lim_{x\rightarrow+\infty}(2-\frac{17}{x})=2$.
Answer:
$2$
17. g.
Explanation:
Step1: Recall continuous - compounding formula
The formula for continuous compounding is $A = Pe^{rt}$, where $P = 50000$, $r=0.013$ (since $1.3% = 0.013$), and $t = 2.5$.
Step2: Substitute values into the formula
$A = 50000\times e^{0.013\times2.5}=50000\times e^{0.0325}$. Using a calculator, $e^{0.0325}\approx1.0330$. So $A=50000\times1.0330=$51650$.
Answer:
$$51650$
18.
Explanation:
Step1: Recall continuous - compounding formula
The formula for continuous compounding is $A = Pe^{rt}$, where $P = 49000000$, $r = 0.02$ (since $2%=0.02$), and $t = 1$.
Step2: Calculate the ending amount
$A=49000000\times e^{0.02\times1}=49000000\times e^{0.02}$. Using a calculator, $e^{0.02}\approx1.0202$. So $A = 49000000\times1.0202=$49989800$.
Step3: Calculate the interest
$I=A - P=49989800 - 49000000=$989800$.
Answer:
$$989800$
20.
Explanation:
Step1: Convert time to years
Nine months is $t=\frac{9}{12}=0.75$ years. $P = 14000$ and $r = 0.011$ (since $1.1%=0.011$).
Step2: Use the continuous - compounding formula for interest
First, find $A$ using $A = Pe^{rt}=14000\times e^{0.011\times0.75}=14000\times e^{0.00825}$. $e^{0.00825}\approx1.0083$. $A = 14000\times1.0083=$14116.2$. Then $I=A - P=14116.2-14000=$116.2$.
Answer:
$$116.2$
21.
Explanation:
Step1: Set up the continuous - compounding equation
The continuous - compounding formula is $A = Pe^{rt}$. We want $A = 2P$, so $2P=Pe^{0.05t}$. Since $P\neq0$ (we have an amount of money to invest), we can divide both sides by $P$ to get $2=e^{0.05t}$.
Step2: Take the natural logarithm of both sides
$\ln(2)=\ln(e^{0.05t})$. Since $\ln(e^{x})=x$, we have $\ln(2)=0.05t$.
Step3: Solve for $t$
$t=\frac{\ln(2)}{0.05}$. $\ln(2)\approx0.6931$, so $t=\frac{0.6931}{0.05}=13.862\approx14$ years.
Answer:
$14$ years