Mathematics: If f(3)=191.5 when r =0.03 for the function f(t)=pe^rt, then what is the approximate value of p? a. 175 b. 78 c. 210 d. 471

If f(3)=191.5 when r =0.03 for the function f(t)=pe^rt, - id31915003439699, ayshajabeenmeox5jrh

JeroMii

11.03.2022

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.

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prince96

21.10.2021

-3/4

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

Step 2: Subtract 1/5x from both sides.

Step 3: Subtract 3/5 from both sides.

Step 4: Multiply both sides by 5/4.

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