The combined math and verbal scores for females taking the SAT-I test are normally distributed with a mean of 900 and a standard
1 answer:
Answer:
50% of females do not satisfy that requirement
Step-by-step explanation:
Problems of normally distributed samples are 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 problem, we have that:
If a college includes a minimum score of 900 among its requirements, what percentage of females do not satisfy that requirement
This is scores lower than 900, which is the pvalue of Z when X = 900.
So
has a pvalue of 0.5
50% of females do not satisfy that requirement
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Answer:
y = 52 - 8x
Step-by-step explanation:
If he painted for 3 hours at a rate of 8 square meters per hour, the area painted was:
If after 24 squared meters painted, there are still 28 left, the total area is:
The area still left to paint is initially 52 squared meters and decreases at a rate of 8 square meters per hour, the function that describes the relationship between the area left to paint (y) and number of hours (x) is:
y = 52 - 8x
Answer: |p-72% |≤ 4%
Step-by-step explanation:
Let p be the population proportion.
The absolute inequality about p using an absolute value inequality.:
, where E = margin of error,
= sample proportion
Given: A poll result of 72% with a margin of error of 4% indicates that p is most likely to be between 68% and 76% .
|p-72% |≤ 4%
⇒ 72% - 4% ≤ p ≤ 72% +4%
⇒ 68% ≤ p ≤ 76%.
i.e. p is most likely to be between 68% and 76% (.
Answer:
The total of ways the buildings can be painted are = 4 3
2
2
1 = 48 ways.
Step-by-step explanation:
i) color five buildings that are in a row.
ii) any two neighboring buildings must be different colors
iii) the first third and fifth buildings must be different colors.
iv) the number of ways to paint the fist building are 4.
v) the number of ways to paint the second building are 3.
vi) the number of ways to paint the third building are 2.
vii) the number of ways to paint the fourth building are 2.
viii) the number of ways to paint the last building are 1.
ix) therefore the total of ways the buildings can be painted are
= 4 3
2
2
1 = 48 ways.