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Vitek1552 [10]
3 years ago
14

You remove four fuses of 10, 20, 30, and 30 amperes each, but you do not mark the corresponding circuits. If you insert the fuse

s so that each sequence is equally likely, what is the probability that the appropriate amperage fuse is assigned to all the circuits
Mathematics
1 answer:
iren2701 [21]3 years ago
4 0

Answer:

Standard amperage sizes of the Code are 15, 20, 25, 30, 35, 40, 45, 50, 60, ... Additional standard ampere ratings for fuses are 1, 3, 6, 10 and 601

Step-by-step explanation:

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What is the distance of a car traveling with a velocity of 20 m/s north in 5 seconds
blsea [12.9K]

distance=velocity×

(time)

20(5)

=100m/S

3 0
3 years ago
The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g
Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about \\ P(z>1.56) = 0.0594.

Step-by-step explanation:

This is a problem of the <em>distribution of sample means</em>. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves <em>normally</em> for samples sizes equal or greater than 30 \\ n \geq 30. Mathematically

\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}}) [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a <em>normal standard distribution</em>, i.e., a normal distribution that has a population mean \\ \mu = 0 and a population standard deviation \\ \sigma = 1.

\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}} [2]

From the question, we know that

  • The population mean is \\ \mu = 88 WPM
  • The population standard deviation is \\ \sigma = 14 WPM

We also know the size of the sample for this case: \\ n = 137 sixth graders.

We need to estimate the probability that a sample mean being greater than \\ \overline{X} = 89.87 WPM in the <em>distribution of sample means</em>. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}

\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}

\\ Z = 1.5634 \approx 1.56

This is a <em>standardized value </em> and it tells us that the sample with mean 89.87 is 1.56<em> standard deviations</em> <em>above</em> the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any <em>cumulative</em> <em>standard normal table</em> available in Statistics books or on the Internet. Of course, this probability is the same that \\ P(\overline{X} < 89.87). Then

\\ P(z

However, we are looking for P(z>1.56), which is the <em>complement probability</em> of the previous probability. Therefore

\\ P(z>1.56) = 1 - P(z

\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594

Thus, "The probability that the sample mean would be greater than 89.87 WPM" is about \\ P(z>1.56) = 0.0594.

5 0
3 years ago
Find the greatest common factor of 15x^2y^3 and -18x^3yz <br> The Answer has to be like _x_y_
Daniel [21]

Answer:

The greatest common factor of this would be 3x^2y

Step-by-step explanation:

In order to find this, first find the greatest common factor of the coefficients. Since 3 goes in evenly to both 15 and -18, then we know that it is a common factor.

From there we need to find the number of x's. Since the first term only has 2 x's and the second has 3, we take the lowest number. (x^2)

And since the first term has 3 y's and the second has just 1, we take the lowest number (y).

8 0
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4 b a 1. Joseph plans to give a bag of candy to each of the 22 students in his class. Each bag of candy will cost $2.87 after ta
zavuch27 [327]

Answer:

A

Step-by-step explanation:

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How do you find vertical asymptotes?
Keith_Richards [23]
You just have to set the denominator to zero and solve for x
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