Question

IVE .tex 1. The percentage of a certain brand of computer chips that will fail after t years of use is estimated to be P (1) = 100 (He) What percentage of this brand of computer chips are expected to fail after 3 years, and how will that percentage be changing? (a) Percent = 25.9%, Rate of change = 7.4 percent/year (b) Percent = 12.2%, Rate of change = 8.8 percent/year (c) Percent = 92.6%; Rate of change = 74.1 percent/year (d) Percent = 74.1%; Rate of change = 2.6 percent/year (e) None of the above.

Answer 1

$P(t)=100\left(1-e^{-0.1t}\right)$

take = 3 year

P(3) = 100 (1 - -0.13)

$P(3)= 100\left(1-e^{-0.3}\right)$

P(3) = 100 (1 - 0.74081)

$P(3)= 100\cdot \:0.2591$

${\color{Red} P(3) \approx 25.9 \%}$

.

$P(t)=100\left(1-e^{-0.1t}\right)$

to find the rate of change, take the derivative

$P'(t)=\frac{d}{dt}\left(100\left(1-e^{-0.1t}\right)\right)$

$P'(t)=100\frac{d}{dt}\left(1-e^{-0.1t}\right)$

$P'(t)=100\left(\frac{d}{dt}\left(1\right)-\frac{d}{dt}\left(e^{-0.1t}\right)\right)$

$P'(t)=100\left(0-\left(-0.1e^{-0.1t}\right)\right)$

$P'(t)=100\left(0.1e^{-0.1t}\right)$

$P'(t)=10e^{-0.1t}$

take = 3 year

$P'(3)= 10e^{-0.1\cdot 3}$

${\color{Red} P'(3) \approx 7.4 \%}$

.

final answer is

${\color{Red} percent=25.9\%,\:\:rate\:\:of\:\:change=7.4\%\:\:per/\:\:year}$

option A is correct