As a wedding gift, Jonathan and Claire received $10,000 cash from Claires's
1 answer:
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Answer:
81.8%
Step-by-step explanation:
Mean =
Standard deviation =
Now we are supposed to find out what percent of the numbers fall between 35 and 50
Substitute the values
Now for P(35<x<50)
Substitute x = 35
Substitute x = 50
So, P(-1<z<2)
P(z<2)-P(z<-1)
=0.9772-0.1587
=0.8185
=
=81.8%
Hence 81.8% percent of the numbers fall between 35 and 50
Answer:
D
Step-by-step explanation:
Using the Cosine rule to find AC
AC² = BC² + AB² - (2 × BC × AB × cosB )
= 18² + 12² - ( 2 × 18 × 12 × cos75° )
= 324 + 144 - 432cos75°
= 468 - 111.8
= 356.2 ( take the square root of both sides )
AC = ≈ 18.9
-----------------------------------------
Using the Sine rule to find ∠ A
=
( cross- multiply )
18.9 sinA = 18 sin75° ( divide both sides by 18.9 )
sinA = , then
∠ A = (
) ≈ 66.9°
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.