solve the problems. you will need a scientific or graphing calculator.\n3.1) using the formula $a = p(1 +…

solve the problems. you will need a scientific or graphing calculator.\n3.1) using the formula $a = p(1 + r)^t$, calculate the value of an initial investment of $4,500 after 10 years at 4% interest.\n3.2) using the formula $a = p(1+\frac{r}{n})^{nt}$, calculate the value of an initial investment of $4,500 after 10 years at 4% interest, compounded quarterly (four times per year).\n3.3) a medical team has found that the blood concentration of a particular medicine has a decay rate of 40% in 24 hours. how much of an initial dose of 1,000 mg of the medicine will be detected after 48 hours? round to the nearest mg. use the formula $a = a_0e^{-kt}$. (the value of $t$ is $\frac{48}{24}=2$.)

solve the problems. you will need a scientific or graphing calculator.\n3.1) using the formula $a = p(1 + r)^t$, calculate the value of an initial investment of $4,500 after 10 years at 4% interest.\n3.2) using the formula $a = p(1+\frac{r}{n})^{nt}$, calculate the value of an initial investment of $4,500 after 10 years at 4% interest, compounded quarterly (four times per year).\n3.3) a medical team has found that the blood concentration of a particular medicine has a decay rate of 40% in 24 hours. how much of an initial dose of 1,000 mg of the medicine will be detected after 48 hours? round to the nearest mg. use the formula $a = a_0e^{-kt}$. (the value of $t$ is $\frac{48}{24}=2$.)

Answer

3.1

Explanation:

Step1: Identify values

$P = 4500$, $r=0.04$, $t = 10$

Step2: Substitute into formula

$A=4500\times(1 + 0.04)^{10}$

Step3: Calculate

$A=4500\times1.04^{10}\approx4500\times1.4802442849\approx6661.10$

Answer:

$$6661.10$

3.2

Explanation:

Step1: Identify values

$P = 4500$, $r = 0.04$, $n=4$, $t = 10$

Step2: Substitute into formula

$A=4500\times\left(1+\frac{0.04}{4}\right)^{4\times10}$

Step3: Simplify exponent and fraction

$A=4500\times(1 + 0.01)^{40}$

Step4: Calculate

$A=4500\times1.01^{40}\approx4500\times1.4888637336\approx6699.89$

Answer:

$$6699.89$

3.3

Explanation:

Step1: Find decay constant $k$

If the decay rate is $40%$ in 24 hours, then when $t = 1$ (in units of 24 - hour periods), $A=(1 - 0.4)A_0=0.6A_0$. Using $A = A_0e^{-kt}$, when $t = 1$, $0.6A_0=A_0e^{-k\times1}$, so $e^{-k}=0.6$, and $k=-\ln(0.6)\approx0.5108$.

Step2: Identify values for the problem

$A_0 = 1000$, $k\approx0.5108$, $t = 2$ (since 48 hours is 2 twenty - four hour periods)

Step3: Substitute into formula

$A = 1000\times e^{-0.5108\times2}$

Step4: Calculate

$A=1000\times e^{-1.0216}\approx1000\times0.3597\approx360$

Answer:

$360$ mg