A right triangle has a vertex at point M and a height of 5 units. The base of the triangle is on MN←→− and is 3 units long. One
of vertices of the triangle is at M.
Use the Polygon tool to draw the triangle.
Each segment on the grid represents 1 unit.
Use the Polygon tool to draw the triangle.
Each segment on the grid represents 1 unit.
2 answers:
Answer:
We are given a triangle having vertex at point M.
The base of the triangle is MN which is 3 units long.
Now, the height of the triangle is given as 5 units.
So, plotting a line starting at M, which is 5 units long towards the left or right will give us the height of the triangle.
Moreover, as the base is 3 units long.
We will have that the point N is 3 units down or up from the point M.
Using Pythagoras theorem, get that the length of the third side is 5.8 units.
After joining these points and constructing straight lines, we will get the following right angles triangle having base 3 units and height 5 units.
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Using the hypergeometric distribution, it is found that there is a 0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.
In this problem, the bulbs are chosen without replacement, hence the <em>hypergeometric distribution</em> is used to solve this question.
<h3>What is the hypergeometric distribution formula?</h3>The formula is:
The parameters are:
- x is the number of successes.
- N is the size of the population.
- n is the size of the sample.
- k is the total number of desired outcomes.
In this problem:
- There are 12 bulbs, hence N = 12.
- 3 are defective, hence k = 3.
The third defective bulb is the fifth bulb if:
- Two of the first 4 bulbs are defective, which is P(X = 2) when n = 4.
- The fifth is defective, with probability of 1/8, as of the eight remaining bulbs, one will be defective.
Hence:
0.2182 x 1/8 = 0.0273.
0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.
More can be learned about the hypergeometric distribution at brainly.com/question/24826394