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e-lub [12.9K]
3 years ago
9

What is the common endpoint called?​

Mathematics
2 answers:
vaieri [72.5K]3 years ago
7 0
I think its called vertex (of the angle)
erastova [34]3 years ago
5 0
The answer is vertex
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The question about triangles is below.
rusak2 [61]

Answer:

A.

Step-by-step explanation:

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3 years ago
Write and simplify an expression that represents the perimeter of the parallelogram. ​
Ira Lisetskai [31]
The perimeter of the parallelogram is 14x + 8. You add 3x+9+4x-5+3x+9+4x-5 to get the answer.
8 0
3 years ago
How tall is a tree which is 15 feet shorter than a pole that is 3 times as tall as the tree?
Citrus2011 [14]
<h2>Height of tree is 7.5 feet</h2>

Step-by-step explanation:

Let t be the height of pole.

Given that the tree is 5 feet shorter than a pole.

Height of pole = t + 15

Also given that the pole is 3 times as tall as the tree.

Height of pole = 3t

So we have

              t + 15 = 3t

                 2t = 15

                   t = 7.5 feet

Height of tree = 7.5 feet

7 0
3 years ago
Drag and drop the responses in to the boxes to correctly complete the statement
Roman55 [17]
Sum,shorter,longest.
5 0
3 years ago
The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78
____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

m-t_{a/2,n-1}\frac{s}{\sqrt{n} } ≤ μ ≤ m+t_{a/2,n-1}\frac{s}{\sqrt{n} }

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and t_{a/2,n-1} is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and t_{0.05,9} by 1.8331, we get that the 90% confidence interval for the mean speed is:

74.2-(1.8331)\frac{5.3083}{\sqrt{10} } ≤ μ ≤ 74.2+(1.8331)\frac{5.3083}{\sqrt{10} }

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0
3 years ago
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