The required equation is y = -9
Step-by-step explanation:
Step 1 :
Given the line l is perpendicular to the y axis
The equation of the y axis is x = 0
So any line perpendicular to the y axis will have equation as y = k , where k is a constant value for any value of x
Step 2 :
Its given that the line is passing through the point (0.-9). Here the y co ordinate is -9. Hence the perpendicular line has a constant y co ordinate of y = -9 for any value of x
So the required equation is y = -9
Step 3 :
Answer :
The required equation is y = -9
First we will compute the h+k and then multiply the result by 2.
To add polynomials, we add terms whose variables are alike, for example:
we add the coefficients of x^2 together, the coefficients of x together and so on.
Therefore:
h + k = x^2 + 1 + x - 2 = x^2+x-1
Now, we will multiply this answer by 2 to get the final answer:
2(h+k) = 2(x^2+x-1) = 2x^2 + 2x -2
Answer:B
Step-by-step explanation:
He charged the man $25.75 for each hour and he worked for 3 hours so multiply $25.75 x 3 = $77.25
x=6
(6x+4) = (8x-8)
∠3 and ∠6 have to be equal since they are opposite interior angles (these angles are always equal to each other). Opposite interior angles prove that two lines are parallel if they are equal, so you just set them equal to each other and solve for x.
(6x+4) = (8x-8)
-6x -6x
4 = (2x-8)
+8 +8
12 = 2x
/2 /2
6 = x
CHECK
(6x+4) = (8x-8)
(6[6]+4) = (8[6]-8)
(36+4) = (48-8)
40 = 40
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.
