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serg [7]
3 years ago
14

Factor as the product of two binomials. x^2 - 4 Thank you!!

Mathematics
2 answers:
Alika [10]3 years ago
8 0
(x-2)(x+2)
hope this helps!!
EastWind [94]3 years ago
3 0
I think the answer is (x-2 ) x ( x+2)
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the building manager put new door knobs on 4 doors in each apartment there are 3 apartments on each floor and 3 floors in the ap
barxatty [35]
(4) x (3) x (3)= (12) x (3)= 36
6 0
3 years ago
5) Mrs. Bell went bowling again. Her bowling scores were 100, 172, 85, and 92. She wants
Alex73 [517]

Answer: She needs to bowl 120 or more on her next game.

Step-by-step explanation:

8 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
On the diagram, two angles (&lt;1 and &lt;2) formed by the road, and the borders are congruent.
iren [92.7K]

Answer: they share a road lol

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Factor completely 56a2b3 − 35ab.
VMariaS [17]

I would start off by taking away 1a. That would make the problem be 56ab3-35b.I only took away 1 because each have at least 1a and is okay to do.


Next I would deal with the variable b. I would cross of 1 b. That's because both sides have at least 1b.  Now, it's shortened to be 56ab2-35.


Since you cannot take away anymore variables, you have to deal with 56 and 35. I start small with dividing each by 2. I am trying to see what the greatest number could be while making the numbers still be whole. That turns 56 into 28 when it's cut in half. The 35 now turns into 17.5.


I would assume your teacher would want the numbers to be whole. seeing as though when 35 is cut in half and makes a decimal number, I would leave them. What I mean by that is to leave the numbers as 56 and 35.


So, that means the answer is 56ab2-35.


I hope this helps!! (And makes sense)

5 0
3 years ago
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