Answer:
D would be your answer. I used my algebra calculator and it shows you step by step how to do the problem
Answer:
∠A = ∠C = 60°
∠B = ∠D = 120°
Step-by-step explanation:
In quad. DPBQ, by angle sum property we have
∠PDQ + ∠DPB + ∠B + ∠BQD = 360°
60° + 90° + ∠B + 90° = 360°
∠B = 360° – 240°
Therefore, ∠B = 120°
But ∠B = ∠D = 120° opposite angles of parallelogram
As, AB || CD opposite sides of a parallelogram
∠B + ∠C = 180° sum of adjacent interior angles is 180°
120° + ∠C = 180°
∠C = 180° – 120° = 60°
Hence ∠A = ∠C = 60° Opposite angles of parallelogram are equal
Answer:
(
2
x
−
6
)
2
+
4
(
2
x
−
6
)
+
3
=
0
Simplify the left side.
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(
2
x
−
6
)
2
+
8
x
−
21
=
0
Use the quadratic formula to find the solutions.
−
b
±
√
b
2
−
4
(
a
c
)
2
a
Substitute the values
a
=
4
,
b
=
−
16
, and
c
=
15
into the quadratic formula and solve for
x
.
16
±
√
(
−
16
)
2
−
4
⋅
(
4
⋅
15
)
2
⋅
4
Simplify.
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x
=
4
±
1
2
The final answer is the combination of both solutions.
x
=
5
2
,
3
2
Step-by-step explanation:
For the first option, the range is a measure of variability which measures the spread of the data set from the least value to the greatest value, but it does not take into account the variability of the other data values of the data set. The range is easily affected by the presence of outliers (data points that are away from other data points). Thus the range is regarded as a weak measure of variability and is not used when other measures of variability are available. Thus, that the range of the two data sets are equal does not mean that the data sets have the same variability. Therefore, the first option is not the correct answer.
For the second option, the median is not a measure of variability. Thus, that a data set has a greater median than another data set does not mean that the data set would have a greater variability. Therefore, the second option is not the correct answer.
For the third option, the inter-quartile range (IQR) is a better measure of variability than the range because it takes into account more data points than the range. Now, because, the the IQR of Team 2 is less than the IQR of Team 1, this shows that Team 1 have greater variability than Team 2 and thus the conclusion of the coaches are inaccurate. Therefore, the third option is the correct answer.
For the fourth option, the mean absolute deviation, MAD, is a better measure of variability than the IQR because it takes into account all the points of the data set. While IQR measures variability with respect to the median, MAD measures variability with respect to the mean. Because we are told that the data sets are not symmetrical, the median will be a better measure of the center than the mean, thus the IQR will present a better measure of the variability of the data sets. Thus, though the MAD for Team 2 was calculated to be a larger number than the MAD for Team 1, the information can be misleading in arriving at a conclusion on which data set has more variability because the data sets are not symmetrical. Therefore, the fourth option is not the correct answer.
The order would be 1/5,5/8,2/3.
