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77julia77 [94]
3 years ago
10

Ajar contains 100 coins. If it is very unlikely that you will randomly choose a quarter out of the jar, how many quarters

Mathematics
2 answers:
astraxan [27]3 years ago
7 0
<h2>✧・゚: *✧・゚:*  Answer:  *:・゚✧*:・゚✧ </h2><h3> </h3><h3>✅ The answer is 10. </h3><h3> </h3><h2>✧・゚: *✧・゚:*  Explanation:  *:・゚✧*:・゚✧ </h2><h3> </h3><h3>❗ The answer is 10 because the probability is very unlikely and there are 100 coins so if there are 10 quarters it makes it very unlikely </h3><h3> </h3><h3> </h3><h3>~ ₕₒₚₑ ₜₕᵢₛ ₕₑₗₚₛ! :₎ ♡ </h3><h3> </h3><h3> </h3><h3>~ </h3>
marishachu [46]3 years ago
3 0
I would say 10. :)))))
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Campbell Middle school has 912 students.How many computers are in this schools computer lab?
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You can't really determine the amount of computers from this question, they're obviously not going to have one computer for everyone so half third, or even a fourth of the school population
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3 years ago
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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
Benny bought nine new baseball trading cards to add to his collection. The next day his dog ate half of his collection. There ar
hjlf

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stepan [7]

Answer:

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6+6= 12 cm

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To efficiently double the volume of a square pyramid, you need to double the height of the pyramid. 

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