Answer:
<RQP = 70°
Step-by-step explanation:
<RQP = 1/2 (140)
<RQP = 70°
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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 score that separates the lower 5% of the class from the rest of the class is 55.6.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question:

Find the score that separates the lower 5% of the class from the rest of the class.
This score is the 5th percentile, which is X when Z has a pvalue of 0.05. So it is X when Z = -1.645.


The score that separates the lower 5% of the class from the rest of the class is 55.6.
Answer:
The LCM of 10, 12, and 18 is 180.
HOPE THIS HELPED!!!!!!!!!!!!XDDDDDDDD