If scores on an exam follow an approximately normal distribution with a mean of 76.4 and a standard deviation of 6.1 points, then the minimum score you would need to be in the top 2% is equal to 88.929.
A problem of this type in mathematics can be characterized as a normal distribution problem. We can use the z-score to solve it by using the formula;
Z = x - μ / σ
In this formula the standard score is represented by Z, the observed value is represented by x, the mean is represented by μ, and the standard deviation is represented by σ.
The p-value can be used to determine the z-score with the help of a standard table.
As we have to find the minimum score to be in the top 2%, p-value = 0.02
The z-score that is found to correspond with this p-value of 0.02 in the standard table is 2.054
Therefore,
2.054 = x - 76.4 ÷ 6.1
2.054 × 6.1 = x - 76.4
12.529 = x - 76.4
12.529 + 76.4 = x
x = 88.929
Hence 88.929 is calculated to be the lowest score required to be in the top 2%.
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The answer is B.
Hope it helps!
Answer:
5/4
Step-by-step explanation:
(y₂ - y₁) / (x₂ - x₁)
(4,9) (-8, -6)
plug in
(-6 - 9) / (-8 - 4)
solve within parentheses
-15/-12
simplify
5/4
The scale factor that Thea uses to go from Rectangle Q to Rectangle R is equal to 6.
<h3>What is the scale factor from rectangle Q to rectangle R?</h3>
In geometry, the scale factor is a ratio of the resulting length to the initial length. Since the area of the square is equal to the square of its side length, then the scale factor is equal to:
k² = A' / A
k = √(A' / A)
Where:
- k - Scale factor
- A' - Area of the rectangle R.
- A - Area of the rectangle Q.
If we know that A = 2 and A' = 72, then the scale factor is:
k = √(72 / 2)
k = √36
k = 6
Then, the scale factor that Thea uses to go from Rectangle Q to Rectangle R is equal to 6.
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Answer:
8
Step-by-step explanation:
15 x (1/3+1/5) make common denominators and add
15 x (5/15+3/15) add fractions
15/1 x (8/15) 15 and 15 cancel out and become 1
1/1 x 8/1
1 x 8
8