Answer:
A) 34.13%
B) 15.87%
C) 95.44%
D) 97.72%
E) 49.87%
F) 0.13%
Step-by-step explanation:
To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:
Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:
So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:
P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)
= 0.5 - 0.1587 = 0.3413
It means that the PERCENT of scores that are between 90 and 100 is 34.13%
At the same way, we can calculated the percentages of B, C, D, E and F as:
B) Over 110
C) Between 80 and 120
D) less than 80
E) Between 70 and 100
F) More than 130
Answer:
The equation should be 2.5x+1.8=59.3
Step-by-step explanation:
First, subtract the 1.8 dollars from the 59.3. You will get 57.5. Divide it by 2.5 and you will get the number 21 as x.
The answer is (x,y) =(28/11, -21/11)
The value of n in given proportion is 16
<u><em>Solution:</em></u>
We have to find the value of "n" in the proportion
<em><u>Given proportion is:</u></em>
<em><u></u></em>
<em><u></u></em>
We can solve the above proportion by cross-multiplying
Multiply the numerator of the left-hand fraction by the denominator of the right-hand fraction
Multiply the numerator of the right-hand fraction by the denominator of the left-hand fraction
Set the two products equal to each other
Solve for the variable
Thus the value of n in given proportion is 16
2m+3=23
2m=20
m=10
let me know in the comments if you need further explanation :)
