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Charra [1.4K]
3 years ago
7

HELP, thanks, If a triangle has a side length of 8 inches, and another side length of 5 inches, which of the

Mathematics
1 answer:
Anna71 [15]3 years ago
4 0

Answer:

a. 7

Step-by-step explanation:

So a triangle a two sides that will always sum up to be greater than the third side.

Meaning:

a + b > c

a + c > b

b + c > a

So here, we're given a and b:

8 + 5 > c

8 + c > 5

5 + c > 8

Plug in the values:

13 > 7

15 > 5

12 > 8

These are all true, so it's probably 7.

13 > 2

10 > 5

7 > 8 This one is not true, so not 2.

13 > 13 This is not true, so not 13.

If not 13, we can't go any higher and so it can't be 15.

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ok so he uses 3/2 for 3/5 so first lets see how much it will take for 1/5 of the  room so

3/2 dived by 3=1/2

then lets multiply this by 5 to make 1

1/2*5=2.5

so it will take 2 1/2 so d

Hope This Helps!!!

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3 years ago
the rental rate for a hedge trimmer has a one time fee of $20 plus $30 for each day of the rental. the graph shows points repres
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so its 20 fixed or constant. so that +20

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do f(x)= 30x + 20

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4 years ago
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Step-by-step explanation:

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3 years ago
Emily wants to put trim around a square window and measures 3 ft on each side. How many feet of t r i m does she need?
Alexeev081 [22]

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2 years ago
Suppose a simple random sample of size n is drawn from a large population with mean mu and standard deviation sigma. The samplin
Sidana [21]

Answer:

X \sim N(\mu,\sigma)  

Where \mu the mean and \sigma  the deviation

Since the distribution for X is normal then we can conclude that the distribution for the sample mean \bar X is given by:

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

\mu_{\bar X} = \mu

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

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 variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(\mu,\sigma)  

Where \mu the mean and \sigma  the deviation

Since the distribution for X is normal then we can conclude that the distribution for the sample mean \bar X is given by:

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

\mu_{\bar X} = \mu

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

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