A gardener uses a coordinate grid to design a new garden. The gardener uses polygon WXYZ on the grid to represent the garden. Th
e vertices of this polygon are W(3,3), X(-3,3), Y(-3,-3), and Z(3,-3). Each grid unit represents one yard. What is the shape of the garden?
1 answer:
Answer:
A square
36 square yard.
Step-by-step explanation:
Given the following information:
W(3,3), X(-3,3), Y(-3,-3), and Z(3,-3)
So we need to graph those vertices and connect them in order.
(please have a look at the attached photo)
From the graph, it clear that the shape of the garden is a square.
Then, we need to calculate the length of the four sides
<u>Find the length of the side YZ. </u>
The length of side YZ = 3 + 3 = 6 yards because point Y and Z are on the same line y = -3 so we just need to find the distance between their x coordinate
<u>Find the length of the side WZ.</u>
The length of side WZ = 3 + 3 = 6 yards because point W and Z are on the same line x = 3 so we just need to find the distance between their y coordinate
=> the area of the square WXYZ:
= YZ*WZ
= 6*6
= 36 square yard.
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Answer:
1) Fail to reject the Null hypothesis
2) We do not have sufficient evidence to support the claim that the mean distance students traveled to school from their current residence was different for males and females.
Step-by-step explanation:
A university administrator wants to test if there is a difference between the distance men and women travel to class from their current residence. So, the hypothesis would be:
The results of his tests are:
t-value = -1.05
p-value = 0.305
Degrees of freedom = df = 21
Based on this data we need to draw a conclusion about test. The significance level is not given, but the normally used levels of significance are 0.001, 0.005, 0.01 and 0.05
The rule of the thumb is:
- If p-value is equal to or less than the significance level, then we reject the null hypothesis
- If p-value is greater than the significance level, we fail to reject the null hypothesis.
No matter which significance level is used from the above mentioned significance levels, p-value will always be larger than it. Therefore, we fail to reject the null hypothesis.
Conclusion:
We do not have sufficient evidence to support the claim that the mean distance students traveled to school from their current residence was different for males and females.
The two dot plots are missing, so i have attached it.
Answer:
The mean at the beginning of the school year was 9.5 miles and the mean at the end of the school year was 10.2 miles
Step-by-step explanation:
From the attached image, we are told to compare the means for each plot to the nearest tenth.
Mean = Σx/n
Now, from the image, total number of miles run by the 14 students at the beginning of the school year is;
(1 × 7) + (2 × 8) + (4 × 9) + (4 × 10) + (2 × 11) + (1 × 12) = 133
Mean of miles run at the beginning of the school year = 133/14 = 9.5 miles
Again, from the table, total miles run at the end of the school year = (2 × 8) + (2 × 9) + (4 × 10) + (3 × 11) + (3 × 12) = 143
Mean of miles run at the end of the school year = 143/14 = 10.2 miles
Thus;
The mean at the beginning of the school year was 9.5 miles and the mean at the end of the school year was 10.2 miles
