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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A is NOT a function because for some values of x you get more than one y value. For example if x=2 you get 2 values for y.
b is a function because for each x you get ONLY one y
12.5 is bigger.
Because if we convert the fraction of 3/20 to normal numbers is equal to 0.15. <span>and</span> 12.5 is way bigger than 0.15 :3
Hope it helped.
Answer:
the answer would be no solution
Step-by-step explanation:
because, Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation
3/4y-(5/x)=0
and from there you may not do anything else.
Answer:
8 / sqrt(3)
Step-by-step explanation:
Since this is a right triangle, we can use trig functions
tan theta = opp / adj
tan A = BC / AC
tan 30 = BC / 8
8 tan 30 = BC
8 * sqrt(3)/3 = BC
8/3 sqrt(3) = BC
