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3 years ago

What is another way to group the factors? (3×2)×5

Mathematics

2 answers:

JulsSmile [24]3 years ago

7 0

Well you. an do (2×3)×5 or (3×5)×2 or (2×5)×3

Rina8888 [55]3 years ago

3 0

Well you do
(2x3) x5 or (3x5) x2 or (2x5)x3 is yours answer

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Which of the following is not a way of generating random numbers? A. random number tables B. using phone numbers selected at ran

tino4ka555 [31]

Answer:

well all of these look like a way so we have to use elimination method

A : random number tables : well it has random numbers so X out

B: PHONE NUMBERS: well phone numbers are random so X out

C: USing the internet : totally X out

D: books of random numbers: X out

so none of the above i guess

7 0

3 years ago

When the price of a dozen eggs is raised by$2.4, 15 fewer eggs can be bought with $60

boyakko [2]

Where is the question? Is it a true or false?

6 0

3 years ago

Read 2 more answers

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 distribution of sample means. 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 normally 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 normal standard distribution, 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 distribution of sample means. 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 standardized value and it tells us that the sample with mean 89.87 is 1.56 standard deviations above the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any cumulative standard normal table 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 complement probability 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

Suppose you have a 6-face unfair dice with numbers 1,2,3,4,5,6 on each of its faces. If the probability distribution of throwing

kotegsom [21]

Answer:

D is correct

Step-by-step explanation:

Here, we want to select which of the options is correct.

The correct option is the option D

Since the die is unfair, we expect that the probability of each of the numbers turning up

will not be equal.

However, we should also expect that if we add the chances of all the numbers occurring together, then the total probability should be equal to 1. But this does not work in this case;

In this case, adding all the probabilities together, we have;

1/12 + 1/12 + 1/12 + 1/12 + 1/12 + 1/2

= 5(1/12) + 1/2 = 5/12 + 1/2 = 11/12

11/12 is not equal to 1 and thus the probability distribution cannot be correct

4 0

3 years ago

Select the equivalent expression. 2^-4 = ?

mrs_skeptik [129]

Answer:

1/16

Step-by-step explanation:

This means 1/2x2x2x2 = 1/16

7 0

2 years ago