shinomiya
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PLEASE HELP ME!!!

A stock analyst attempts to express the price p of a share of XYZ stock as an exponentially increasing function of the time since XYZ's initial public offering (IPO):
p=10e^km, where m is the number of months since the IPO. The price of a share was $10.00 at the time of the IPO and $14.60 four months after the IPO. What is the approximate value of k? Round to the nearest thousandth.

ANSWER CHOICES:
A) 1.295 per month
B) 0.670 per month
C) 1.570 per month
D) 0.095 per month
E) 3.650 per month

Respuesta :

Answer: (d)

Step-by-step explanation:

Given

Price  of the stock is given by the function

[tex]P=10e^{km}[/tex]

after 4 months, price increases to $14.60

Insert the value in the function

[tex]\Rightarrow 14.60=10e^{4k}\\\\\Rightarrow 1.46=e^{4k}\\\text{Taking natural log both sides}\\\Rightarrow \ln 1.46=\ln e^{4k}\\\Rightarrow 0.378=4k\\\Rightarrow k=0.0946\approx 0.095\ \text{per month}[/tex]

hence, option (d) is correct.

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