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Annette [7]
3 years ago
15

(5+10p) +13 a.28p b.15p+13 c.23p+5 d.10p+18

Mathematics
2 answers:
Ivan3 years ago
6 0
Now, here, we need to combine like terms.

5 and 13 are like terms.

5+13 = 18

Final answer:  10p + 18
Sliva [168]3 years ago
3 0
<span><span>5+10p</span>+13 <--remove the parentheses  
</span><span>And now just add like terms
</span><span>(10p)</span>+<span>(5+13)
</span><span>10p</span>+<span>18 <---answer</span>
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Algebra 1 Grade 9
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Since the price increases by $60 every hour, the constant rate would be $60. The total cost of the bill would be 50+60*6
( $410 )
3 0
4 years ago
Mr. Burns built a wooden platform for the Drama Club at the school where he teaches. The shape below represents the dimensions o
Tresset [83]

Answer:

108 square feet

Step-by-step explanation:

First you want to break up the "L" into two parts. These parts can be seens as a and b. You need to find out the length on the left hand side, and half of it is 6ft as seen, so if half of it is 6ft, then you need to find the other half, which is obviously 6ft. 6+6=12, and to find the area of the thinner rectangle, multiply 2*24=48. That would be the area of section a. Section b is the bottom section and what you want to do is multiply 10*6 because this is how you would find the area of section b, which would be 60. 60+48=108. I know I was not very descriptive but I still hope this somewhat helps. I also might be wrong because of the fact that it says surface area despite the fact that this shape is 2D.

3 0
3 years ago
It is known that IQ scores form a normal distribution with a mean of 100 and a standard deviation of 15. If a researcher obtains
sdas [7]

Answer:

X \sim N(100,15)  

Where \mu=100 and \sigma=15

We select a sample of n=16 and we are interested on the distribution of \bar X, since the distribution for X is normal then we can conclude that the distribution for \bar X is also normal and given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

Because by definition:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

E(\bar X) = \mu

Var(\bar X) = \frac{\sigma^2}{n}

And for this case we have this:

\mu_{\bar X}= \mu = 100

\sigma_{\bar X} = \frac{15}{\sqrt{16}}= 3.75

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the IQ scores of a population, and for this case we know the distribution for X is given by:

X \sim N(100,15)  

Where \mu=100 and \sigma=15

We select a sample of n=16 and we are interested on the distribution of \bar X, since the distribution for X is normal then we can conclude that the distribution for \bar X is also normal and given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

Because by definition:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

E(\bar X) = \mu

Var(\bar X) = \frac{\sigma^2}{n}

And for this case we have this:

\mu_{\bar X}= \mu = 100

\sigma_{\bar X} = \frac{15}{\sqrt{16}}= 3.75

6 0
3 years ago
It’s only 10 points but it’s all i got (:
weqwewe [10]

Answer:

Thx a lot

Step-by-step explanation:

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3 years ago
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nevsk [136]
You have to post a pic I don’t understand
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3 years ago
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