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Bingel [31]
4 years ago
5

One day 10% of the students enrolled in the school were absent. There were 1,620 students present on that day. How many students

were absent?
Mathematics
2 answers:
Sidana [21]4 years ago
7 0
10% = 0.1
So, 1,620 * 0.1 = 162
162 students were absent. 
Sounds like a terrible school. 
Lorico [155]4 years ago
4 0
You might want to add or subtract 1,620 and 10% to find the total then make sure to put the 10% into a 0.10
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Calculate the average rate of change for the given function, from x = −3 to x = 7.
emmainna [20.7K]

Answer:

The average rate of change for the given function, from x = −3 to x = 7 will be:

  • Average rate = -5

Step-by-step explanation:

Given the table

x                      f(x)

-3                     50

3                      10

7                       0

As the interval is from x = -3 to x = 7.

so

at x₁ = -3, f(x₁) = 50

at x₂ = 7,  f(x₂) = 0

Using the formula to determine the average rate of change at which the total cost increases will be:

Average rate = [f(x₂) - f(x₁)] / [ x₂ - x₁]

                       = [0 - 50] / [7-(-3)]

                      = [-50] / [7+3]

                      = -50 / 10

                      = -5

Therefore, the average rate of change for the given function, from x = −3 to x = 7 will be:

  • Average rate = -5
6 0
3 years ago
A random sample of 81 automobiles traveling on a section of an interstate showed an average speed of 60 mph. The distribution of
myrzilka [38]

Answer:

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(60,13.5)  

Where \mu=60 and \sigma=13.5

And for this case we select a sample size of n= 81. Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

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

And the standard error of the mean would be:

\sigma_{\bar X} =\frac{13.5}{\sqrt{81}}= 1.5

4.1.5

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(60,13.5)  

Where \mu=60 and \sigma=13.5

And for this case we select a sample size of n= 81. Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

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

And the standard error of the mean would be:

\sigma_{\bar X} =\frac{13.5}{\sqrt{81}}= 1.5

4.1.5

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