Question

If f(3)=191.5 when r =0.03 for the function F(t)=pe^rt, then what is the approximate value of P?

Answer 1

Answer:

A. 175

Step-by-step explanation:

Answer 2

Answer:

The correct option is A.

Step-by-step explanation:

The given function is

$F(t)=pe^{rt}$

It is given that f(3)=191.5 and r =0.03.

Substitute t=3 and r=0.03 in the given function.

$F(3)=pe^{0.03(3)}$

$F(3)=pe^{0.09}$

The value of F(3)=191.5. Put F(3)=191.5 in the above equation.

$191.5=p(1.0942)$

Divide both sides by 1.0942.

$\frac{191.5}{1.0942}=p$

$175.0137=p$

$p\approx 175$

The approximate value of p is 175. Therefore the correct option is A.