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vodomira [7]
2 years ago
6

Come up with two examples of ratio relationships that are interesting to you.

Mathematics
1 answer:
Likurg_2 [28]2 years ago
8 0

Answer:

1:3 is an example and for every two boy there are 5 girls is another example

Edited: I go out Five times as much as My sister does. My sister goes out only 3 times. Which is 15:1

I see 5 blue cars for every 6 red cars. Which is 5:6

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Between which two values would 75% of the data represented on a box plot lie? Check all that apply.
Inga [223]

Answer:

Hi, everybody, today I'm going to answer this question for you, I took the unit test review, I only got one wrong to be honest, I will tell you which one that was and it's answer, but for now let me answer this question for you. The guy below no offense but you were wrong, the real answer is

the lower quartile and the maximum

the minimum and the upper quartile

I was able to get these two right, and a bonus question with it's own answer.

Mrs. Barnes recorded the grades her students earned on the last chapter test in the frequency table below. The frequency table will be used to make a histogram.

Test Score

Number of Students

31 - 40

1

41 - 50

2

51 - 60

4

61 - 70

6

71 - 80

11

81 - 90

5

91 - 100

4

Which chart has the correct setup for Mrs. Barnes’ histogram?

A 2-column table with 4 rows titled Mrs. Barnes's Histogram. Column 1 has entries interval used, appropriate scale, vertical axis represents, horizontal axis represents. Column 2 has entries 10, 31 to 100, number of students, test scores.

A 2-column table with 4 rows titled Mrs. Barnes's Histogram. Column 1 has entries interval used, appropriate scale, vertical axis represents, horizontal axis represents. Column 2 has entries 10, 0 to 12, number of students, test scores.

A 2-column table with 4 rows titled Mrs. Barnes's Histogram. Column 1 has entries interval used, appropriate scale, vertical axis represents, horizontal axis represents. Column 2 has entries 7, 0 to 12, test scores, number of students.

A 2-column table with 4 rows titled Mrs. Barnes's Histogram. Column 1 has entries interval used, appropriate scale, vertical axis represents, horizontal axis represents. Column 2 has entries 7, 31 to 100, test scores, number of students.

The answer is b, I did a and got it wrong, and it told me the answer is b.

Step-by-step explanation:

8 0
3 years ago
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Given a function, describe how to find the average rate of change over a given interval.
icang [17]

Answer:

To find the average rate of change, we divide the change in the output value by the change in the input value. The Greek letter Δ (delta) signifies the change in a quantity; we read the ratio as “delta-y over delta-x” or “the change in y divided by the change in x”.

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Can someone help me on this question
dexar [7]
It’s an obtuse triangle, this is because none of the sides/angles are equal to one another. Hope this helped you
3 0
3 years ago
If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.
nydimaria [60]

Answer:

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}

Step-by-step explanation:

Given

\cos(\theta) = -\frac{2}{3}

\theta \to Quadrant III

Required

Determine \tan(\theta) \cdot \cot(\theta) + \csc(\theta)

We have:

\cos(\theta) = -\frac{2}{3}

We know that:

\sin^2(\theta) + \cos^2(\theta) = 1

This gives:

\sin^2(\theta) + (-\frac{2}{3})^2 = 1

\sin^2(\theta) + (\frac{4}{9}) = 1

Collect like terms

\sin^2(\theta)  = 1 - \frac{4}{9}

Take LCM and solve

\sin^2(\theta)  = \frac{9 -4}{9}

\sin^2(\theta)  = \frac{5}{9}

Take the square roots of both sides

\sin(\theta)  = \±\frac{\sqrt 5}{3}

Sin is negative in quadrant III. So:

\sin(\theta)  = -\frac{\sqrt 5}{3}

Calculate \csc(\theta)

\csc(\theta) = \frac{1}{\sin(\theta)}

We have: \sin(\theta)  = -\frac{\sqrt 5}{3}

So:

\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}

\csc(\theta) = \frac{-3}{\sqrt 5}

Rationalize

\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}

\csc(\theta) = \frac{-3\sqrt 5}{5}

So, we have:

\tan(\theta) \cdot \cot(\theta) + \csc(\theta)

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \tan(\theta) \cdot \frac{1}{\tan(\theta)} + \csc(\theta)

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)

Substitute: \csc(\theta) = \frac{-3\sqrt 5}{5}

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}

Take LCM

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}

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3 years ago
Can I get some help please
LenaWriter [7]
Dividing integers is the opposite of multiplying integers.
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