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DENIUS [597]
2 years ago
12

A company performed power tests on a set of batteries of the same type. The company determined that the equation y = 100 - 8.9x,

where x is the number of hours of use and y is the percent of battery power remaining, models the battery life. Based on the equation, what is the best prediction of the percent of remaining power for a battery after 11 hours of use? *
A.1.2%
B.2.1%
C.10%
D.97.9%
Mathematics
2 answers:
Fudgin [204]2 years ago
8 0

Answer: 2.1%

Step-by-step explanation:

y = 100 - 8.9 (11)

y = 100 - 97.9

y = 2.1

Paraphin [41]2 years ago
7 0

Answer:

B) 2.1

Step-by-step explanation:

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Answer:

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Here is a typical statement made by the media: "Based on a recent study, pennies weigh an average of 2.5 grams with a margin of
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The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 ci
Olenka [21]

Answer:

The correct answer is

(0.0128, 0.0532)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence interval 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

Z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}

For this problem, we have that:

In a random sample of 300 circuits, 10 are defective. This means that n = 300 and \pi = \frac{10}{300} = 0.033

Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool.

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{300}} = 0.033 - 1.96\sqrt{\frac{0.033*0.967}{300}} = 0.0128

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{300}} = 0.033 + 1.96\sqrt{\frac{0.033*0.967}{300}} = 0.0532

The correct answer is

(0.0128, 0.0532)

4 0
1 year ago
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