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AveGali [126]
2 years ago
10

A manufacturer knows that 2%2\% 2% 2, percent of its microchips are produced with a certain defect. They decide to change their

process to make it more efficient, and they want to test if the new process has the same defect rate or not.
Mathematics
1 answer:
kirza4 [7]2 years ago
3 0

Answer: 0.0776

Step-by-step explanation:

Since the manufacturer knows that 2% of its microchips are produced with a certain defect. Then

P( defective) = 2/100 = 0.02

P( non defective) = 1 - 0.02 = 0.98

Probability of all 4 not defective will be

P(all 4 not defective) = 0.98^4 = 0.9224

Probability that at least one of the selected chip is defective will be

P(at one defective) = 1 - 0.9224

= 0.0776

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charle [14.2K]

Answer:

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3 years ago
Which is correct regarding the statement: "If x is an odd integer, then the median of x, x + 2, x + 6, and x + 10 is an
nadya68 [22]

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Step-by-step explanation:

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2 years ago
Solving Equations by Adding or Subtracting Fractions<br> Show all working<br> p- 538 = -1124
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8 0
2 years ago
Read 2 more answers
The formula T= 2pi sqrt(L/32) relates the time, T, in seconds for a pendulum with the length, L, in feet, to make one full swing
tester [92]

The length of pendulum is 2.485 feet

<h3><u><em>Solution:</em></u></h3>

Given that,

The formula T= 2pi sqrt(L/32) relates the time, T, in seconds for a pendulum with the length, L, in feet, to make one full swing back and forth

<u><em>Therefore, the given formula is:</em></u>

T=2\pi \sqrt{\frac{L}{32} }

We have to find the length of pendulum that makes one full swing in 1.75 seconds

So the modify the given equation to find "L"

T=2\pi \sqrt{\frac{L}{32} }\\\\ \sqrt{\frac{L}{32} }=\frac{T}{2 \pi}\\\\\text{Taking square root on both sides }\\\\\frac{L}{32} = \frac{T^2}{4 \pi^2}\\\\L = \frac{T^2}{4 \pi^2} \times 32\\\\L = \frac{T^2}{\pi^2 } \times 8

Substitute T = 1.75 seconds and \pi = 3.14

L = \frac{1.75^2}{3.14 \times 3.14} \times 8\\\\L = \frac{3.0625}{9.8596} \times 8\\\\L = 2.485

Thus length of pendulum is 2.485 feet approximately

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