complete the table below giving the amount p that must be invested at interest rate 5.5% compounded…

Explanation:

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 (initial investment), $r$ is the annual interest rate (in decimal form), and $t$ is the number of years. We need to solve for $P$, so $P=\frac{A}{e^{rt}}$. Given $A = 110000$ and $r=0.055$.

Step2: Calculate $P$ for $t = 1$

Substitute $A = 110000$, $r = 0.055$, and $t = 1$ into the formula $P=\frac{A}{e^{rt}}$. So $P=\frac{110000}{e^{0.055\times1}}=\frac{110000}{e^{0.055}}\approx\frac{110000}{1.05654}\approx104113.47$.

Step3: Calculate $P$ for $t = 10$

Substitute $A = 110000$, $r = 0.055$, and $t = 10$ into the formula $P=\frac{A}{e^{rt}}$. So $P=\frac{110000}{e^{0.055\times10}}=\frac{110000}{e^{0.55}}\approx\frac{110000}{1.73325}\approx63452.81$.

Step4: Calculate $P$ for $t = 20$

Substitute $A = 110000$, $r = 0.055$, and $t = 20$ into the formula $P=\frac{A}{e^{rt}}$. So $P=\frac{110000}{e^{0.055\times20}}=\frac{110000}{e^{1.1}}\approx\frac{110000}{3.00417}\approx36615.03$.

Step5: Calculate $P$ for $t = 30$

Substitute $A = 110000$, $r = 0.055$, and $t = 30$ into the formula $P=\frac{A}{e^{rt}}$. So $P=\frac{110000}{e^{0.055\times30}}=\frac{110000}{e^{1.65}}\approx\frac{110000}{5.10792}\approx21535.94$.

Step6: Calculate $P$ for $t = 40$

Substitute $A = 110000$, $r = 0.055$, and $t = 40$ into the formula $P=\frac{A}{e^{rt}}$. So $P=\frac{110000}{e^{0.055\times40}}=\frac{110000}{e^{2.2}}\approx\frac{110000}{9.02501}\approx12188.38$.

Step7: Calculate $P$ for $t = 50$

Substitute $A = 110000$, $r = 0.055$, and $t = 50$ into the formula $P=\frac{A}{e^{rt}}$. So $P=\frac{110000}{e^{0.055\times50}}=\frac{110000}{e^{2.75}}\approx\frac{110000}{15.6426}\approx7031.90$.

Answer:

$t$	$P$
1	$104113.47$
10	$63452.81$
20	$36615.03$
30	$21535.94$
40	$12188.38$
50	$7031.90$