Rectangle ABCD is translated (x + 2 y - 3) and then rotated 180° about the origin. Complete the table to show the locations of A
'. B', C', and D' after both transformations.
B
0
3
2
D
A-51) A' ?
B (-5,3) B 2
C (-1.3) C" ?
D--11)
D' ?
A(-2,-3), B' (0-3), C (0.1), D'(-2.1)
B
0
3
2
D
A-51) A' ?
B (-5,3) B 2
C (-1.3) C" ?
D--11)
D' ?
A(-2,-3), B' (0-3), C (0.1), D'(-2.1)
1 answer:
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Answer:
mean = 78.4
median = 77.5
mode = 75
This is Right - skewed (positive skewness) distribution
Step-by-step explanation:
<u>Mean:-</u>
The mean (average) is found by adding all of the numbers together and dividing by the number of items and it is denoted by x⁻
mean =
mean (x⁻ ) = 78.4
The mean of the given data = 78.4
<u>Median:</u>
The median is found by ordering the set from lowest to highest and finding the exact middle.
64 ,75, 80, 98
The middle term of the given data set =
<u>Mode :</u>
The mode is the most common repeated number in a data set.
64 ,75, 75, 80, 98
in data the most common number = 75
<u>Conclusion</u>:-
mean = 78.4
median = 77.5
mode = 75
This is Right - skewed (positive skewness) distribution
Answer: 0.813
Step-by-step explanation:
Let A be the event describes the number of residents believed that the amount of violent television programming had increased over the past 10 years.
& B be the event describes the number of residents believed that the amount of violent television programming had decreased over the past 10 years.
Given : n(A)=721 ; n(B)=454 ; n(A∩B)=362
We know that ,
i.e.
Also, the total number of U.S. residents surveyed n(S)= 1,000
Then, the proportion of the 1,000 U.S. residents believed that either the amount of violent programming had increased or the overall quality of programming had decreased over the past 10 years will be :-
Hence, the required answer = 0.813
Answer:
Step-by-step explanation:
Hello!
You need to construct a 95% CI for the population mean of the length of engineering conferences.
The variable has a normal distribution.
The information given is:
n= 84
x[bar]= 3.94
δ= 1.28
The formula for the Confidence interval is:
x[bar]±*(δ/n)
Lower bound(Lb): 3.698
Upper bound(Ub): 4.182
Error bound: (Ub - Lb)/2 = (4.182-3.698)/2 = 0.242
I hope it helps!