Middle School

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------------------------------------------------ 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

Answer :

Answer:

The correct option is A.

Step-by-step explanation:

The given function is

[tex]F(t)=pe^{rt}[/tex]

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

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

[tex]F(3)=pe^{0.03(3)}[/tex]

[tex]F(3)=pe^{0.09}[/tex]

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

[tex]191.5=p(1.0942)[/tex]

Divide both sides by 1.0942.

[tex]\frac{191.5}{1.0942}=p[/tex]

[tex]175.0137=p[/tex]

[tex]p\approx 175[/tex]

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

Answer:

A. 175

Step-by-step explanation: