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attashe74 [19]
3 years ago
6

Peter attempted to use the divide-center method to find the line of best fit on a scatterplot.

Mathematics
1 answer:
SpyIntel [72]3 years ago
8 0

Answer:

He had a different number of points above the line of best fit than below the line of best fit.

Step-by-step explanation:

This is the best likely answer to the question about Peter's attempt to use divide-center method to find the line of best fit on a scatterplot.

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Jo's collection contains US, Indian and British stamps. If the ratio of US to Indian stamps is 5 to 2 and the ratio of Indian to
Arisa [49]
Ratio of US to Indian stamps is 5:2
U/I = 5:2
= 25:10

Ratio of Indian to British stamps is 5:1
I/B = 5:1
= 10:2

Now , Ratio of US to British stamps 
U/B = 25:2

The answer is E.
4 0
2 years ago
Anton will be constructing a segment bisector with a compass and straightedge, while Maxim will be constructing an angle bisecto
Genrish500 [490]

The similarities are;

  • Compass and a straight edge required for both construction
  • Both construction includes a line drawn from the intersection of arcs to bisect a segment or an angle
  • The bases for the construction of both bisector are the ends of segment and the angle to be bisected
  • The width of the compass when drawing intersecting arcs, is more than half the width of the segment or angle being bisected

The differences are;

  • Two points of intersection of arcs are used in the segment bisector while only one is requited in an angle bisector
  • The bisecting line crosses the segment in a segment bisector, while it stops at the vertex of the angle being bisected in an angle bisector

The sources of the above equations are as follows;

The steps to construct a segment bisector are;

  • Place the needle of the compass at one of the ends of the line segment to be bisected
  • Widen the compass so as to extend more than half of the length of the segment to be bisected
  • Draw two arcs, one above, and the other below the line
  • Place the compass needle at the other end and with the same compass width draw arcs that intersects with the arcs drawn in the above step
  • Draw a line segment by placing the ruler on the points of intersection of the arcs above and below the line

The steps to construct an angle bisector are;

  • With the compass needle at the vertex, open the pencil end such that arcs can be drawn on the rays (lines) forming the angle
  • Draw an arc on both lines forming the angle
  • Place the compass needle at one of the intersection points and draw an arc in between the lines forming the angle
  • Repeat the above step with the same compass width from the other intersection point with the rays forming the angle
  • Join the point of intersection of the two arcs to the vertex of the angle to bisect the angle

Therefore, we have;

The similarities are;

  • A compass and a straight edge can be used for both construction
  • A straight line is drawn from the point of intersection of arcs to bisect the segment or the angle
  • The arcs are drawn from the ends of the segment or angle to be bisected
  • The width of the compass is more than half the width of the line or angle when drawing the arcs

The differences are;

  • In a segment bisector, the intersection point is above and below the line, while in an angle bisector only one pair of arcs are drawn to intersect above the line
  • The bisecting line passes through the segment being bisected, while the line stops at the vertex in an angle bisector

Learn more about the construction of segment and angle bisectors here;

brainly.com/question/17335869

brainly.com/question/12028523

7 0
2 years ago
The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min. If one such class is r
Lady bird [3.3K]

Answer: the probability that the class length is between 50.8 and 51 min is  0.1 ≈ 10%

Step-by-step explanation:

Given data;

lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min

hence, height = 1 / ( 52.0 - 50.0) = 1 / 2

now the probability that the class length is between 50.8 and 51 min = ?

P( 50.8 < X < 51 ) = base × height

= ( 51 - 50.8) × 1/2

= 0.2 × 0.5

= 0.1 ≈ 10%  

therefore the probability that the class length is between 50.8 and 51 min is  0.1 ≈ 10%  

5 0
2 years ago
point I is on line segment HJ. GIVEN IJ = 3× + 3, HI = 3× - 1, and HJ = 3× + 8, determine the numerical length of HJ.
Vedmedyk [2.9K]

Given, IJ = 3x + 3, HI = 3x - 1, and HJ = 3x + 8.

Since I is a point on line segment HJ, we can write

HJ=HI+IJ\begin{gathered} 3x+8=(3x-1)+(3x+3) \\ 3x+8=6x+2 \\ 8-2=6x-3x \\ 6=3x \\ 2=x \end{gathered}

Put x=2 in HJ=3x+8.

\begin{gathered} HJ=3\times2+8 \\ HJ=6+8 \\ =14 \end{gathered}

Therefore, the numerical length of HJ is 14.

4 0
1 year ago
Can someone please help me!!
Sauron [17]
I think that the answer could be D
8 0
3 years ago
Read 2 more answers
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