Answer:
x = 22/25
Step-by-step explanation:
Given
2 1/5 x
---------- = -----------
10 4
Using cross products
2 1/5 * 4 = x*10
Change 2 1/5 to an improper fraction
2 1/5 =(5*2+1)/5 = 11/5
11/5 *4 = 10x
44/5 =10x
Multiply each side by 1/10 to isolate x
44/5 *1/10 = 10x * 1/10
44/50 = x
Divide top and bottom by 2
22/25 = x
He will receive $13.70.
You would have to multiply 2.25 by 2.80 to get 6.30, which is the total cost due. Subtract 6.30 from 20 to get your answer, 13.70, or the change Mr. Stein will receive.
Answer:
see below
Step-by-step explanation:
The first step is to combine like terms
4x -12 + 2x
6x -12
Then determine the number of terms
2 terms so it is a binomial
Answer:
The explicit function is: 
And the number of lights after 33 weeks will be 252.
Step-by-step explanation:
Given that:
Total street lights = 153
Let x be the number of weeks
Then the number lights after x weeks will be 3x
This is a linear function where the y-intercept is 153 and slope is 3.
It can be written as: y = mx+b
The function is:
Putting the values for x will give us the number of total lights after that number of weeks.
To find, how many street lights were there at the end of 33rd week,
Putting x = 33
Hence.
The explicit function is:
And the number of lights after 33 weeks will be 252.
That outlier will appear as a single blip of data, but it will be far
away from the central part of the histogram. It will not be that visible
if there is a lot of data, since 1 outlying data point will just show a
very low height compared to the central data near the mean/median/mode,
where there will be higher frequencies and higher bars of data.
If these are the choices:
<span>A.The outlier will appear as a tall bar near one side of the distribution.
B.The outlier will appear as a bar far from all of the other bars with a height that corresponds to a frequency of 1.
C.Since a histogram shows frequencies, not individual data values, the
outlier will not appear. Instead, the outlier increases the frequency
for its class by 1.
D.The outlier will appear as the tallest bar near the center of the distribution.
</span>
Then the most accurate answer will be B, since it's just a single value with a bar corresponding to a frequency of 1.
