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lina2011 [118]
3 years ago
5

9(3+c+4) what is the answer

Mathematics
1 answer:
zzz [600]3 years ago
4 0

Answer:

the answer is 9c+63

Step-by-step explanation:

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Eric runs 6 miles in 44 min at the same rate how many minutes would he take to run 9 miles?
faltersainse [42]
66 minutes or 1 hour and 6 minutes

First divide 6 into 44 then multiply that number by 9 (:
5 0
4 years ago
Multiply & simplify: (2a - 3)(2a + 1)
RoseWind [281]
I hope this helps you



2a.2a+1.2a-3.2a-3.1


4a^2+2a-6a-3


4a^2-4a-3
3 0
3 years ago
Factor completely
bixtya [17]
The answer is (x+8)(x-4)
5 0
3 years ago
Need help ASAP.... what is number 8<br><br> Also #9<br><br> And 10-14
Setler79 [48]

The answer to #9 is 2.16%

7 0
3 years ago
The number of cars running a red light in a day, at a given intersection, possesses a distribution with a mean of 2.4 cars and a
liberstina [14]

Answer:

The sampling distribution of the sample mean is:

\bar X\sim N(\mu_{\bar x}=2.4,\ \sigma_{\bar x}=0.40)

Step-by-step explanation:

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

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

\mu_{\bar x}=\mu

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

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

Let <em>X</em> = number of cars running a red light in a day, at a given intersection.

The information provided is:

E(X)=\mu=2.4\\SD(X)=\sigma=4\\n=100

The sample selected is quite large, i.e. <em>n</em> = 100 > 30.

The Central limit theorem can be used to approximate the sampling distribution of the sample mean number of cars running a red light in a day, by the Normal distribution.

The mean of the sampling distribution of the sample mean is:

\mu_{\bar x}=\mu=2.4

The standard deviation of the sampling distribution of the sample mean is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{4}{\sqrt{100}}=0.40

The sampling distribution of the sample mean is:

\bar X\sim N(\mu_{\bar x}=2.4,\ \sigma_{\bar x}=0.40)

8 0
3 years ago
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