(a) the interquartile range
Would be 72 - 88
the med would be 84
(b) I may be wrong but my guess would be
88% above all the seasons because the scale that was shown went to 100 and that would cause each win to be 1%. the bottom would have to equal out to 12% because that is what we would need to make it equal out to 100.
Hope this helps
To solve for proportion we make use of the z statistic.
The procedure is to solve for the value of the z score and then locate for the
proportion using the standard distribution tables. The formula for z score is:
z = (X – μ) / σ
where X is the sample value, μ is the mean value and σ is
the standard deviation
when X = 70
z1 = (70 – 100) / 15 = -2
Using the standard distribution tables, proportion is P1
= 0.0228
when X = 130
z2 = (130 – 100) /15 = 2
Using the standard distribution tables, proportion is P2
= 0.9772
Therefore the proportion between X of 70 and 130 is:
P (70<X<130) = P2 – P1
P (70<X<130) = 0.9772 - 0.0228
P (70<X<130) = 0.9544
Therefore 0.9544 or 95.44% of the test takers scored
between 70 and 130.
Answer:
The answer is 4.
Step-by-step explanation:
Vertical Angles are congruent.
You plug in 4 for x and you multiply the two to get you 112.
112-2=110, making the two angles the same angle measure.
Hope this helps :)
