Question

jane is a new engineer at a factory, and part of her job is lowering the number of defective parts produced. in her first month, there were 1,280 defective parts. in her second month, there were 1,088 defective parts.
write an exponential equation in the form ( y = a(b)^x ) that can model the monthly number of defective parts, ( y ), ( x ) months after jane was hired.
use whole numbers, decimals, or simplified fractions for the values of ( a ) and ( b ).
( y = )
how many months after jane was hired will the monthly number of defective parts be less than 500?
(square) months

Explanation:

Step1: Identify initial value ($a$)

When $x=0$ (month Jane was hired), the initial number of defective parts is the first month's count: $a = 1280$.

Step2: Find growth/decay factor ($b$)

Divide the second month's defective parts by the first month's:
$$b = \frac{1088}{1280} = \frac{17}{20} = 0.85$$

Step3: Write the exponential equation

Substitute $a$ and $b$ into $y=a(b)^x$:
$$y = 1280(0.85)^x$$

Step4: Solve for $x$ when $y<500$

Set up the inequality:
$$1280(0.85)^x < 500$$
Divide both sides by 1280:
$$(0.85)^x < \frac{500}{1280} = \frac{25}{64} \approx 0.3906$$
Take natural log of both sides (logarithm preserves inequality for positive bases <1, so reverse the inequality sign):
$$x > \frac{\ln(0.3906)}{\ln(0.85)}$$
Calculate the values:
$\ln(0.3906) \approx -0.940$, $\ln(0.85) \approx -0.1625$
$$x > \frac{-0.940}{-0.1625} \approx 5.78$$
Since $x$ must be a whole number of months, round up to 6.

Answer:

Exponential equation: $y = 1280(0.85)^x$
Number of months: 6 months