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3 years ago

Proportions, assist me please!

Mathematics

1 answer:

Anit [1.1K]3 years ago

6 0

Step-by-step explanation:

$w = \frac{1}{2} {x}^{2} y \\ (2)w = \frac{1}{2} {x}^{2} y(2) \\ 2w = {x}^{2} y \\ \frac{2w}{ {x}^{2} } = \frac{{x}^{2} y}{ {x}^{2} } \\ \frac{2w}{ {x}^{2} } = y$

<h3>Answer:</h3>

$y = \frac{2x}{ {x}^{2} } \\$

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3 years ago

i think of a number. if i divide the sum of 5 and the number by 3, the result is 2 less than one quarter of the number. find the

Marrrta [24]

Answer:

x = 4.75

Step-by-step explanation:

<u>(5 +X</u><u> )/3</u> = 2 - <u> X/4</u>

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3 years ago

Mrs. Ibarra wants to create a right triangle for a geometry test. She plans to use 15, 36, and 41 as side lengths.

barxatty [35]

Answer: A, C, D and E

Step-by-step explanation:

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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

Becky bought a bag of mixed chocolates that have either a toffee or nut center. Her and four friends pulled out the first eleven

denpristay [2]

Answer

Part A

Probability that the next chocolate will have a toffee center = (8/11)

Part B

Expected number of chocolates with nut centers = 18 nut centers

Part C

Probability of the picking three consecutive chocolates with nut centers

= (27/1331)

= 0.0203

Explanation

The first 11 chocolates had 8 toffee and 3 nut centers.

The result from these first 11 picks will be used to calculate the rest of the probability or make predictions.

So, for the first part of the question.

Probability that the next chocolate will have a toffee center

= (8/11)

Part B

If there are 66 chocolates left in the bag, how many do you expect to have a nut center?

Probability of picking a chocolate with nut center = p = (3/11)

Number of chocolates in total = n = 66

Expected number of chocolates with nut centers

= np

= (66) (3/11)

= 18 nut centers

Part C

What is the probability that the next 3 pulled do have a nut in the center?

Probability of picking a chocolate with nut center = (3/11)

Probability of the picking three consecutive chocolates with nut centers

= (3/11) × (3/11) × (3/11)

= (27/1331)

= 0.0203

Hope this Helps!!!

3 0

1 year ago

Proportions, assist me please! ​

Proportions, assist me please!