All categories

1 year ago

Please fill out the graphs in the images

Mathematics

1 answer:

Igoryamba1 year ago

4 0

The relationship between angles depends on the positioning of the angles.

<h3>What are angles?</h3>

In mathematics, the place where we have two lines joining and touching each other is what we call an angle. There are several ways in which angles could be related to each other as a function of the way that are positioned.

The relationships between the angles in the figure as follows;

<1 and <2 are linear pair - True
<3 and <5 re vertical angles - True
<3and <4 are complimentary angles - False
<4 and <5 are supplementary angles - True

Learn more about angles: brainly.com/question/13954458

#SPJ1

You might be interested in

PLS HELP WILL MARK BRAINLIEST IM SO CLOSE TO FINISHING

Tom [10]

The answer is the 3 one

3 0

3 years ago

Read 2 more answers

Find m <2 and M<4 kekekskskkskskskskskksksksksks

adoni [48]

Answer:

ok this is a good question

Step-by-step explanation:

i don't even know it

5 0

3 years ago

A square garden has area 1764meter find it's length and perimeter<br><br>

Aneli [31]

Answer:

$area \: of \: a \: squre = {a}^{2} \\ {a}^{2} = 176 \\ a = 13.26m \\ perimeter = 4a = 4 \times 13.26 = 53.04m^2 \\ thank \: you$

5 0

3 years ago

Read 2 more answers

How many different right triangles are there with a hypotenuse of lenght 5 cm

Sveta_85 [38]

Answer:

Draw a horizontal line across a page of paper, somewhere in the middle. Mark a point A on the line, toward the left margin.

Spread open a compass to make good sized circle, but so that if the point of the compass is on the point you drew, the pencil fits on the page, both below the top edge and to the left of the right edge of the page. Draw the arc, from roughly the 12 O’Clock position over and down to the intersection of the line segment, at the 3 O’Clock position.

Call the opening of your compass, the radius of the arc you just drew, “5 units”.

Pick any point on the arc between the 12 and 3 positions, B. Drop a line down from that point B, perpendicular to the original horizontal line. Label the point that it intersects the horizontal line, C.

ABC is a right triangle with hypotenuse 5. Do this again with a point a little closer to the 3 O’Clock position. It’s another right triangle with hypotenuse 5.

Indeed, as you get closer to the right, along the arc, the height of the triangle declines, but the width of the triangle increases. The hypotenuse remains 5 in all cases. We picked points on the circle. We could have picked points on the horizontal line first, and raised perpendicular lines until they intersected the circle. Each point forms a distinct right triangle.

There are as many possible right triangles as there are possible points in a line segment.

Step-by-step explanation:

6 0

3 years ago

Which of the following is not a property of a chi-square distribution?

laiz [17]

Answer:

c) Is not a property (hence (d) is not either)

Step-by-step explanation:

Remember that the chi square distribution with k degrees of freedom has this formula

$\chi_k^2 = \matchal{N}_1^2 + \matchal{N}_2^2 + ... + \, \matchal{N}_{k-1}^2 + \matchal{N}_k^2$

Where N₁ , N₂m .... $N_k$ are independent random variables with standard normal distribution. Since it is a sum of squares, then the chi square distribution cant take negative values, thus (c) is not true as property. Therefore, (d) cant be true either.

Since the chi square is a sum of squares of a symmetrical random variable, it is skewed to the right (values with big absolute value, either positive or negative, will represent a big weight for the graph that is not compensated with values near 0). This shows that (a) is true

The more degrees of freedom the chi square has, the less skewed to the right it is, up to the point of being almost symmetrical for high values of k. In fact, the Central Limit Theorem states that a chi sqare with n degrees of freedom, with n big, will have a distribution approximate to a Normal distribution, therefore, it is not very skewed for high values of n. As a conclusion, the shape of the distribution changes when the degrees of freedom increase, because the distribution is more symmetrical the higher the degrees of freedom are. Thus, (b) is true.

6 0

3 years ago