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Murrr4er [49]
3 years ago
5

I'm confused on how to simplify variable expressions...

Mathematics
2 answers:
Vitek1552 [10]3 years ago
4 0

Answer:

the answer is 3p

Step-by-step explanation:

gulaghasi [49]3 years ago
3 0
Basically, simplifying variables is the same as regular equations (like 5-3=2 and 5x-3x=2x)

So 7x-x is the same as 7x-1x which equals 6x

And -3p+6p is the same thing as -3+6 but with a p. With this in mind, the answer is 3p

Hope this helps
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The area of a circle is pi times the radius squared (A = π r²).

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Jaime bakes24 muffins and gives away 3. She also bakes 36 bagels and gives away 4. Writean expression using GROUPING SYMBOLS ILL
Sergeu [11.5K]

Answer:

24-3=21 , 36-4=32 ,  3+4=7 , 7 x 5 , which equals 35

Step-by-step explanation:

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3 years ago
A blueprint for a new house has a scale of 1:50. Here are the dimensions of two rooms. Calculate the dimensions of each room on
Lorico [155]

Answer:

a.) =30.72*50

b.)13.02*50

Step-by-step explanation:

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8 0
4 years ago
A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. ​(a) What must be true
valentina_108 [34]

Answer:

(a) The distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b) The value of P(\bar X is 0.7642.

(c) The value of P(\bar X\geq 69.1) is 0.3670.

Step-by-step explanation:

A random sample of size <em>n</em> = 10 is selected from a population.

Let the population be made up of the random variable <em>X</em>.

The mean and standard deviation of <em>X</em> are:

\mu=68\\\sigma=15

(a)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. <em>n</em> = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

\mu_{\bar x}=\mu=68

And the standard deviation of the distribution of the sample mean is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74

Thus, the distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b)

Compute the value of P(\bar X as follows:

P(\bar X

                    =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the value of P(\bar X is 0.7642.

(c)

Compute the value of P(\bar X\geq 69.1) as follows:

Apply continuity correction as follows:

P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)

                    =P(\bar X>69.6)

                    =P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})

                    =P(Z>0.34)\\=1-P(Z

Thus, the value of P(\bar X\geq 69.1) is 0.3670.

7 0
4 years ago
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