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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Answer:
x=−2 and y=−1
Step-by-step explanation:
<u>Problem:</u>
Solve y=x2;y=−x−3
<u>Steps:</u>
I will solve your system by substitution.
y=1/2x;y=−x−3
Step: Solve y= 1/2x for y:
Step: Substitute 1/2 x for y in y=−x−3:
y=−x−3
1/2x= =−x−3
1/2x+x=−x−3+x(Add x to both sides)
3/2x = -3
3/2x/3/2 = -3/3/2 (Divide both sides by 3/2)
x=−2
Step: Substitute −2 for x in y=1/2x:
y=1/2x
y=1/2(-2)
y=−1(Simplify both sides of the equation)
<u>Answer:</u>
x=−2 and y=−1
Answer: 200
Step-by-step explanation:
First I would do it like this.
12.5
X. 16
—————-
6 times 5 is 0, carry the 3, 6 times 2 is 12 and add 3 from the other which is 15, put the 1 from the 15 to the one next to the 12. 6 times 1 is 6 plus 1 is 7 which now you have 750. 5 times 1 is 5, 2 times 1 is 2, 1 times 1 is 1. It should look like this now.
12.5
X 16
—————
750
+ 1250
——————-
Which is 8750 but wait. Where is the decimal? Well, there is only one number behind the decimal so it is 875.0 or 875.
Answer:
rational
Step-by-step explanation:
