Oscar played games vs number of points he scored is, C) positive, linear association.
Step-by-step explanation:
- no association is when points Oscar graph will remain between 8to10.
- number of games he scored his points remain the same which is mean.
- non linear is only when there is no straight line passing.
- Linear is either exponential or polynomial.
- Positive as the game increase he scoring abilities increases.
- Negative as the game increases his scoring decreases.
- Negative x axis will have more number of points.
- Negative y axis will high to low of the graph.
- Linear lines are best way to predict a data doesn't work will all data.
Answers:
33. Angle R is 68 degrees
35. The fraction 21/2 or the decimal 10.5
36. Triangle ACG
37. Segment AB
38. The values are x = 6; y = 2
40. The value of x is x = 29
41. C) 108 degrees
42. The value of x is x = 70
43. The segment WY is 24 units long
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Work Shown:
Problem 33)
RS = ST, means that the vertex angle is at angle S
Angle S = 44
Angle R = x, angle T = x are the base angles
R+S+T = 180
x+44+x = 180
2x+44 = 180
2x+44-44 = 180-44
2x = 136
2x/2 = 136/2
x = 68
So angle R is 68 degrees
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Problem 35)
Angle A = angle H
Angle B = angle I
Angle C = angle J
A = 97
B = 4x+4
C = J = 37
A+B+C = 180
97+4x+4+37 = 180
4x+138 = 180
4x+138-138 = 180-138
4x = 42
4x/4 = 42/4
x = 21/2
x = 10.5
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Problem 36)
GD is the median of triangle ACG. It stretches from the vertex G to point D. Point D is the midpoint of segment AC
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Problem 37)
Segment AB is an altitude of triangle ACG. It is perpendicular to line CG (extend out segment CG) and it goes through vertex A.
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Problem 38)
triangle LMN = triangle PQR
LM = PQ
MN = QR
LN = PR
Since LM = PQ, we can say 2x+3 = 5x-15. Let's solve for x
2x+3 = 5x-15
2x-5x = -15-3
-3x = -18
x = -18/(-3)
x = 6
Similarly, MN = QR, so 9 = 3y+3
Solve for y
9 = 3y+3
3y+3 = 9
3y+3-3 = 9-3
3y = 6
3y/3 = 6/3
y = 2
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Problem 40)
The remote interior angles (2x and 21) add up to the exterior angle (3x-8)
2x+21 = 3x-8
2x-3x = -8-21
-x = -29
x = 29
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Problem 41)
For any quadrilateral, the four angles always add to 360 degrees
J+K+L+M = 360
3x+45+2x+45 = 360
5x+90 = 360
5x+90-90 = 360-90
5x = 270
5x/5 = 270/5
x = 54
Use this to find L
L = 2x
L = 2*54
L = 108
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Problem 42)
The adjacent or consecutive angles are supplementary. They add to 180 degrees
K+N = 180
2x+40 = 180
2x+40-40 = 180-40
2x = 140
2x/2 = 140/2
x = 70
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Problem 43)
All sides of the rhombus are congruent, so WX = WZ.
Triangle WPZ is a right triangle (right angle at point P).
Use the pythagorean theorem to find PW
a^2+b^2 = c^2
(PW)^2+(PZ)^2 = (WZ)^2
(PW)^2+256 = 400
(PW)^2+256-256 = 400-256
(PW)^2 = 144
PW = sqrt(144)
PW = 12
WY = 2*PW
WY = 2*12
WY = 24
Roughly 1.7 percent of the bands are shorter than 3cm. We calculate the z score of the data point in standard distribution. By definition of z score, we use score minus mean divided by standard deviation. z=(3-6)/1.5=-2. A z score of -2 corresponds to approximately 1.7%, in other words, roughly 1.7 percent of data is less than 3cm.
Answer:
a. connect the point (0 , 3) with A
b. connect the origin (0 , 0) with B
c. For A: y = 0.5x +3
For B: y = 0.5x
Step-by-step explanation:
y = ax + b is the general rule for any straight line
a being the slope and b being the y intercept, a is given to be 0.5
y = 0.5x + b, substitute the coordinates of point A
4 = 0.5 *2 +b hence b = 4 - 0.5 *2 = 4 - 1 = 3
so y = 0.5 x + 3 is the equation of the line passing through A
since the second line that passes through B is parallel to the first, hence it has the same slope of 0.5
same procedure, substitute coordinates of B
2 = 0.5 * 4 + b hence b = 2 - 0.5 *4 = 2 - 2= 0
so y = 0.5 x is the equation of the line passing through B
Hi There!
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Full Question:
The box plots show the high temperatures in June and August for Denver in degrees Fahrenheit.
Which can you tell about the mean temperatures for these two months?
There is not enough information to determine the mean temperatures.
The mean temperature for August is higher than June's mean temperature.
The mean temperature for June is equal to the mean temperature for August.
The high interquartile range for August pulls the mean temperature above June's mean temperature.
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Interquartile Range Formula: Q3 - Q1
Interquartile Range for August: 10
Interquartile Range for June: 8
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Median = Mean
June: 82
August: 82
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Answer: The mean temperature for June is equal to the mean temperature for August.
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Hope This Helps :)
