Question

1.

Answer 1

Answer:

Part 1) $r=3.5\%$  

Part 2) $t=6\ years$

Part 3) $21.9\ ft$

Step-by-step explanation:

Part 1) we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest in decimal  

t is Number of Time Periods  

e is the mathematical constant number

Let

x -----> amount of money to be invested

2x ---> Final Investment Value

we have  

$t=20\ years\\ P=\ x\\A=\ 2x\\r=?$  

substitute in the formula above  

$2x=x(e)^{20r}$  

$2=(e)^{20r}$  

Apply ln both sides

$ln(2)=ln[(e)^{20r}]$  

$ln(2)=(20r)ln(e)$  

$ln(2)=(20r)$  

$0.693=(20r)$  

$r=0.693/20$  

$r=0.035$  

$r=3.5\%$  

Part 2) we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest in decimal  

t is Number of Time Periods  

e is the mathematical constant number

Let

x -----> amount of money to be invested

2x ---> Final Investment Value

we have  

$t=?\ years\\ P=\ x\\A=\ 2x\\r=0.126$  

substitute in the formula above  

$2x=x(e)^{0.126t}$  

$2=(e)^{0.126t}$  

Apply ln both sides

$ln(2)=ln[(e)^{0.126t}]$  

$ln(2)=(0.126t)ln(e)$  

$ln(2)=(0.126t)$  

$0.693=(0.126t)$  

$t=0.693/0.126$  

$t=5.5=6\ years$

Part 3) we have

$f(x)=18.4+2.5ln(x+1)$

where

f(x) -----> is the long jump record in feet

x -----> the number of years since the school was opened

so

For x=3 years

substitute

$f(3)=18.4+2.5ln(3+1)$

$f(3)=18.4+2.5ln(4)$

$f(3)=18.4+2.5(1.386)$

$f(3)=21.9\ ft$