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pashok25 [27]
3 years ago
7

Luis rolled a number cube 60 times. He rolled the number 6 four times. Which is most likely the cause of the discrepancy between

Luis’s experimental outcome and the predicted outcome?
A.He did not perform enough trials.
B.He performed too many trials.
C.He needs to add the number 6 to the number of times it occurred.
D.He needs to subtract the experimental outcome from the predicted outcome.
Mathematics
1 answer:
Ket [755]3 years ago
8 0
I think your answer is D sorry if wrong :( but i tried :) -hope it was right :) 
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Does a polygon usually have more sides or <br> more angles
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I've seen this question on Brainly before, and I always shake my head.

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5 0
3 years ago
Read 2 more answers
(5 points)
Schach [20]

Answer:

The original Slope is y=-1/3x-2

The perpendicular slope would be y=3x+b

Step-by-step explanation:

Original Slope:

You already have the y-interecpet, (0,-2), and you know two points (0,-2) and (-6,0), so you can find the slope with (y1-y2)/(x1-x2) so in this case, it would be (0-(-2))/(-6-0)= -1/3 and you can find y=1/3x-2.

Perpendicular Slope:

The slope of any perpendicular line would be the reciprocal of the original. So the reciprocal of -1/3 is 3. The b can be any real number because no matter the y-intercept the slope would still be perpendicular

3 0
3 years ago
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7 0
3 years ago
Identify the center and the radius of a circle that has a diameter with endpoints at (−5, 9) and (3, 5)
marusya05 [52]

Check the picture below, so the circle looks more or less like that one.

well, the center of it is simply the Midpoint of those two points, and its radius is simply half-the-distance between them.

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 3 -5}{2}~~~ ,~~~ \cfrac{ 5 + 9}{2} \right)\implies \left( \cfrac{-2}{2}~~,~~\cfrac{14}{2} \right)\implies \stackrel{center}{(-1~~,~~7)} \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[3 - (-5)]^2 + [5 - 9]^2}\implies d=\sqrt{(3+5)^2+(-4)^2} \\\\\\ d=\sqrt{8^2+16}\implies d=\sqrt{80}\implies d=4\sqrt{5}~\hfill \stackrel{\textit{half the diameter}}{\cfrac{4\sqrt{5}}{2}\implies \underset{radius}{2\sqrt{5}}}

8 0
2 years ago
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