A partial proof was constructed given that MNOP is a parallelogram. Parallelogram M N O P is shown. By the definition of a paral
lelogram, MN ∥ PO and MP ∥ NO. Using MP as a transversal, ∠M and ∠P are same-side interior angles, so they are supplementary. Using NO as a transversal, ∠N and ∠O are same-side interior angles, so they are supplementary. Using OP as a transversal, ∠O and ∠P are same-side interior angles, so they are supplementary. Therefore, __________________ because they are supplements of the same angle. Which statement should fill in the blank in the last line of the proof? ∠M is supplementary to ∠O ∠N is supplementary to ∠P ∠M ≅ ∠P ∠N ≅ ∠P
2 answers:
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Answer:
a)
b)
c)
Step-by-step explanation:
Data given
Assuming the following data:
Method A: 18.0, 18.0, 18.0, 20.0, 22.0, 22.0, 22.5, 23.0, 24.0,24.0,25.0,25.0, 25.0,25.0,26.0,26.4
Method B: 18.8, 18.9, 18.9, 19.6, 20.1, 20.4, 20.4, 20.4, 20.4,20.5, 21.2. 22.0,22.0, 22.0,22.0,23.6
Method C: 20.2, 20.5, 20.5, 20.7, 20.8, 20.9, 21.0, 21.0,21.0,21.0, 21.0, 21.5,21.5,21.5,21.5,21.6
Method D: 20.0, 20.0, 20.0,-20.0, 20.2, 20.5, 20.5, 20.7, 20.7,20.7, 21.0, 21.1, 21.5, 21.6, 22.1,22.3.
Part a
The sample mean is defined as:
If we apply this formula for the four methods we got:
Part b
Since the total number of points for each method is 16 and this is an even number we need to calculate the median as the average between the 8th and the 9th position of the data ordered from the smallest to the largest. If we do this we have that:
Part c
The trimmed mean by 20% means that we need to calculate removing the 20% from each of the tails, the 20% of 16 is 3.2, so we need to remove 3 observations from both ends of the data like this:
Method A: 20.0, 22.0, 22.0, 22.5, 23.0, 24.0,24.0,25.0,25.0, 25.0
Method B: 19.6, 20.1, 20.4, 20.4, 20.4, 20.4,20.5, 21.2. 22.0,22.0
Method C: 20.7, 20.8, 20.9, 21.0, 21.0,21.0,21.0, 21.0, 21.5,21.5
Method D: 20.0 20.2, 20.5, 20.5, 20.7, 20.7,20.7, 21.0, 21.1, 21.5.
If we calculate the mean for each method we got:
1/10 is the same as saying 10 times less
Refer to the diagram below
The digit as the first decimal place is worth 1/10
The digit as the second decimal place is worth 1/100 which is 1/10 worth the digit as the first decimal place
Example of two decimal numbers:
2.56 and 2.68
The digit 6 in 2.56 is 6/100
The digit 6 in 2.68 is 6/10
The digit 6 in 2.56 is 1/10 as much as the digit 6 in 2.68
Refer to the diagram below
The digit as the first decimal place is worth 1/10
The digit as the second decimal place is worth 1/100 which is 1/10 worth the digit as the first decimal place
Example of two decimal numbers:
2.56 and 2.68
The digit 6 in 2.56 is 6/100
The digit 6 in 2.68 is 6/10
The digit 6 in 2.56 is 1/10 as much as the digit 6 in 2.68
Answer:
We conclude that a web-based company are not exceeding their goal of 90%.
Step-by-step explanation:
We are given that a web-based company has a goal of processing 90 percent of its orders on the same day they are received.
434 out of the next 471 orders are processed on the same day.
Let p = <u><em>proportion of orders processing on the same day they are received.</em></u>
SO, Null Hypothesis, : p
0.90 {means that they are not exceeding their goal of 90%}
Alternate Hypothesis, : p > 0.90 {means that they are exceeding their goal of 90%}
The test statistics that would be used here <u>One-sample z test for</u> <u>proportions</u>;
T.S. = ~ N(0,1)
where, = sample proportion of orders that are processed on the same day =
= 0.92
n = sample of orders = 471
So, <u><em>the test statistics</em></u> =
= 1.599
The value of z test statistics is 1.599.
<u>Now, at 0.025 significance level the z table gives critical value of 1.96 for right-tailed test.</u>
Since our test statistic is less than the critical value of z as 1.599 < 1.96, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.
Therefore, we conclude that a web-based company are not exceeding their goal of 90%.