The frequency of the second class is 6.
Since the class width is 6 and the lower limit of the first class is 50, this means the first class goes from 50-55. This would put the second class at 56-61. There are 6 data points in this set that would go in the second class.
Answer:
So i just used the calculator to help me find the two co exisiting lines and then worked my way up to find the nearest decimal to fit the equation for the missing parablar length which was parallel to find the cointerior angle which lead me to get a decimal number involving an ordinary form number.
Answer:
95% confidence interval for the mean number of months is between a lower limit of 6.67 months and an upper limit of 25.73 months.
Step-by-step explanation:
Confidence interval is given as mean +/- margin of error (E)
Data: 5, 15, 12, 22, 27
mean = (5+15+12+22+27)/5 = 81/5 = 16.2 months
sd = sqrt[((5-16.2)^2 + (15-16.2)^2 + (12-16.2)^2 + (22-16.2)^2 + (27-16.2)^2) ÷ 5] = sqrt(58.96) = 7.68 months
n = 5
degree of freedom = n-1 = 5-1 = 4
confidence level (C) = 95% = 0.95
significance level = 1 - C = 1 - 0.95 = 0.05 = 5%
critical value (t) corresponding to 4 degrees of freedom and 5% significance level is 2.776
E = t×sd/√n = 2.776×7.68/√5 = 9.53 months
Lower limit of mean = mean - E = 16.2 - 9.53 = 6.67 months
Upper limit of mean = mean + E = 16.2 + 9.53 = 25.73 months
95% confidence interval is (6.67, 25.73)
Answer:
<h3>324 ft²</h3>
Step-by-step explanation:
We have the triangle and the rectangle.
The formula of an area of a triangle:
b - base
h - height
The formula of an area of a rectangle:
l - length
w - width
We have b = 24 ft, h = 7 ft, l = 24 ft and w = 10 ft.
Substitute:
The area of the figure:
Substitute:
