A furniture manufacturer produces chairs and sofas. Each chair requires 10 yards of fabric, and each sofa requires 20 yards of f
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Answer:
Step-by-step explanation:
Hello!
X: number of absences per tutorial per student over the past 5 years(percentage)
X≈N(μ;σ²)
You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.
The formula for the CI is:
X[bar] ± *
⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.
Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4
X[bar]= 10.41
S= 3.71
[10.41±1.645*]
[9.26; 11.56]
Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.
I hope this helps!
Answer:
The answer in the procedure
Step-by-step explanation:
Let
A1 ------> the area of the first square painting
A2 ----> the area of the second square painting
D -----> the difference of the areas
we have
case 1) The area of the second square painting is greater than the area of the first square painting
The difference of the area of the paintings is equal to subtract the area of the first square painting from the area of the second square painting
D=A2-A1
case 2) The area of the first square painting is greater than the area of the second square painting
The difference of the area of the paintings is equal to subtract the area of the second square painting from the area of the first square painting
D=A1-A2