Using the normal distribution, it is found that there are 68 students with scores between 72 and 82.
<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.
In this problem, the mean and the standard deviation are given, respectively, by:
The proportion of students with scores between 72 and 82 is the <u>p-value of Z when X = 82 subtracted by the p-value of Z when X = 72</u>.
X = 82:
Z = 1
Z = 1 has a p-value of 0.84.
X = 72:
Z = 0
Z = 0 has a p-value of 0.5.
0.84 - 0.5 = 0.34.
Out of 200 students, the number is given by:
0.34 x 200 = 68 students with scores between 72 and 82.
More can be learned about the normal distribution at brainly.com/question/24663213
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The lcm is 72 hence the fractions are 15/72 & 28/72
Answer:
Looks complicated!
Step-by-step explanation:
Answer:
-16
Step-by-step explanation:
The absolute value of 2 is 2 so -8*2=-16
Answer:
- monthly payment: $304.15
- total interest: $3249.00
Step-by-step explanation:
Using the time-value-of-money functions on your calculator, you can find the monthly payment to be $304.15.
The calculator function requires several inputs. There are 12 payments per year for 5 years, for a total of 60 monthly payments. As the problem statement tells you, the present value is 15,000, the interest rate is 8%. You want the remaining amount due after 5 years (the future value) to be zero. There are 12 payments per year, and interest is compounded 12 times per year.
After finding the monthly payment, you can find the total amount repaid by multiplying it by 60 (the number of months). Then the interest paid is the difference between that ($18249.00) and the initial loan amount ($15,000.00).
... $18,249 -15,000 = $3,249.00
The attachment shows what these calculations look like on a TI-84 calculator.