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Nitella [24]
3 years ago
10

A flashlight battery manufacturer makes a model of battery whose mean shelf life is three years and four months, with a standard

deviation of three months. The distribution is approximately normal. One production run of batteries in the factory was 25,000 batteries. How many of those batteries can be expected to last between three years and one month and three years and seven months?
Mathematics
1 answer:
telo118 [61]3 years ago
3 0

Answer:

17000 batteries

Step-by-step explanation:

Three years and one month is equivalent to the mean minus one standard deviation.

Three years and seven months is equivalent to the mean plus one standard deviation.

For a normal distribution, we know that 68% of population is between mean ± 1 sd, then can be expected that 25000*68% = 17000 of batteries last between three years and one month and three years and seven months

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Find the volume of the cylinder with a diameter of 12 inches and a height of 10
kati45 [8]

Answer:

360 pi in ^3

Step-by-step explanation:

The volume of a cylinder is given by

V = pi r^2 h

We know the diameter is 12 so the radius is 1/2  the diameter

r = d/2 = 12/2 = 6

V = pi (6)^2 * 10

V = pi (36)*10

V = 360 pi in ^3

We can approximate pi by 3.14

V =1130.4 in ^3

Or we can approximate pi by using the pi button

V =1130.973355  in ^3

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3 years ago
If you answer this you are a certified cool kid
Oksana_A [137]

Answer:

60

Step-by-step explanation:

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An amusement park would like to determine if they want to add in a new roller coaster. In order to decide whether or not they sh
Rzqust [24]

Using the z-distribution, it is found that the 90% confidence interval is given by: (0.6350, 0.6984).

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

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

In which:

  • \pi is the sample proportion.
  • z is the critical value.
  • n is the sample size.

In this problem, we have a 90% confidence level, hence\alpha = 0.9, z is the value of Z that has a p-value of \frac{1+0.9}{2} = 0.95, so the critical value is z = 1.645.

The sample size and the estimate are given by:

n = 600, \pi = \frac{400}{600} = 0.6667

Hence, the bounds of the interval are given by:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6667 - 1.645\sqrt{\frac{0.6667(0.3333)}{600}} = 0.6350

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6667 + 1.645\sqrt{\frac{0.6667(0.3333)}{600}} = 0.6984

The 90% confidence interval is given by: (0.6350, 0.6984).

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

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I need to know the answer my sister keeps bugging me and i do not know
solong [7]
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