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3 years ago

8

Each day, a computer company produces 20 more laptops computers than desktop computers. On a certain day, 5% of the laptop compu

ters and 7% of the desktop computers were found to be defective. If the total number of desktop and laptop computers that were found to be defective that day was 16, how many laptop computers did the computer company make that day?

2 answers:

Nezavi [6.7K]3 years ago

7 0

Let x be the no. of<span> laptop computers make that day
</span><span>20 more laptops computers than desktop computers
so no. of desktop computers = x-20

</span><span>5% of the laptop computers and 7% of the desktop computers
</span>= 0.05x + 0.07(x-20)
= 0.05x + 0.07x - 1.4
= 0.12x - 1.4
which equal to 16
so 0.12x-1.4 = 16
0.12x = 17.4
x=145
so the company builds 145 laptop computers that day

stich3 [128]3 years ago

7 0

l: laptop, d: desktop

d = l-20

(5l + 7d)/100 = 16

5l + 7l - 140 = 1600

12l = 1740

l = 145

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$m = 120$

For his brother:

Rate of change is calculated using:

$m = \frac{g(b) - g(a)}{b - a}$

$m = \frac{g(40) - g(32)}{40 - 32}$

$m = \frac{g(40) - g(32)}{8}$

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$m = \frac{g(4) - g(0)}{4}$

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