Answer:
A score of 150.25 is necessary to reach the 75th percentile.
Step-by-step explanation:
Problems of normal distributions 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.
A set of test scores is normally distributed with a mean of 130 and a standard deviation of 30.
This means that 
What score is necessary to reach the 75th percentile?
This is X when Z has a pvalue of 0.75, so X when Z = 0.675.




A score of 150.25 is necessary to reach the 75th percentile.
Answer:
Quadratic binomial
Step-by-step explanation:
4x(x+1) - (3x-8)(x+4)
Open parenthesis
(4x^2+4x) - (3x^2+12x-8x-32)
(4x^2+4x - 3x^2-12x+8x+32)
Collect like terms
4x^2-3x^2+4x+8x-12x+32
=x^2+32
Quadratic binomial
Answer:
Use the slope formula (y2-y1)÷(x2-x1)
Step-by-step explanation:
(-5-(-3))÷2-3
= (-2.5,-1)
Answer:
P(25 < x < 37) = 0.77
Step-by-step explanation:
Given - If a Variable has a normal distribution with mean 30 and standard deviation 5
To find - find the probability that the variable will be between 25 and 37.
Proof -
Given that,
Mean, μ = 30
S.D, σ = 5
Now,
~ N(0,1)
Now,
P(25 < x < 37)
= 
= P(1 < z < 1.4)
= P(z < 1.4) - P(z < -1)
= 0.9192 - 0.1587
= 0.7605
≈ 0.77
∴ we get
P(25 < x < 37) = 0.77
Answer:
36 girls
Step-by-step explanation:
2:3
24 divided by 2 is 12
3 times 12 is 36