5 Girls + 3 Boys =8 (sample size)
Probability of choosing 1 boy P(1 B) = 3/8
Probability of NOT CHOOSING ANY BOY = 1-3/8 =5/8
Now apply the binomial probability to choose 3 Boys:
8C3(3/8)³(5/8)⁵ ==>P(All boys) = 0.28
Using the normal distribution, the z-score that separates the bottom 70% from the top 30% is z = 0.525.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean and standard deviation is given by:
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The z-score that separates the bottom 70% from the top 30% is <u>z with a p-value of 0.7</u>, hence z = 0.525.
More can be learned about the normal distribution at brainly.com/question/24537145
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Answer:
Step-by-step explanation:
The formula for determining the area of a trapezoid is expressed as
Area = 1/2(a + b)h
Where
a and b are the length of The bases are the 2 sides of the trapezoid which are parallel with one another.
h represents the height of the trapezoid.
From the information given,
a = 9 units
b = 6 units
If It has an area of 120 square units, then
120 = 1/2(9 + 6)h
Cross multiplying by 2, it becomes
120 × 2 = (9 + 6)h
240 = 15h
h = 240/15 = 16 units
Answer:
4x⁻¹
Step-by-step explanation:
4x + 20 25 - x²
_______ ÷ _______ ↷
x² - 5x [x - 5]²
4[x + 5] [-5 + x][5 + x]
_______ ÷ _______
x[x - 5] [x - 5][x - 5]
So what we did here was factor out ALL the Greatest Common Factors [GCFs] in each denominator and numerator, and since we are dividing vertically AND horizontally, according to this, ALL [x - 5] and [x + 5] cancel each other out and all you are left with is this:
4x⁻¹
OR
This, according to the Negative Exponential Rule [Reverse], is how you rewrite it as a fraction. You bring the denominator to the numerator while ALTERING THE INTEGER SYMBOL FROM POSITIVE TO NEGATIVE:
b⁻ⁿ = 1\bⁿ
I hope this helps you out alot, and as always, I am joyous to assist anyone at any time.