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Rzqust [24]
2 years ago
11

How to simply this ratio

Mathematics
2 answers:
arlik [135]2 years ago
6 0

Answer:

75 : 3 : 6250

Step-by-step explanation:

We use 3 to simplify the ratio to 75 : 3 : 6,250.

elena-s [515]2 years ago
6 0

Answer: 75:3:6250

Step-by-step explanation:

All have prime factors of 3

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NikAS [45]

Answer:

the answer is 32 cups of water  and 12 cups drink mix.

Step-by-step explanation:

   

8 0
3 years ago
A game is played with the following rules. Two ordinary number cubes, each with six sides labeled 1 to 6, are rolled. After each
Marysya12 [62]
Answer: - 0.28


Explanation:

1) Expected value: is the weighted average of the values, being the probabilities the weight.

That is: ∑ of prbability of event i × value of event i.

In this case: (probability of getting 2 or 12) × (+6) + (probability of gettin 3 or 11) × (+2) + (probability of any other sum) × (-1).


2) Sample space:

Sum              Points awarded
    
1+ 1 = 2              +6

1 + 2 = 3             +2

1 + 3 = 4             -1

1 + 4 = 5             -1

1 + 5 = 6             -1

1 + 6 = 7             -1

2 + 1 = 3             +2

2 + 2 = 4             -1

2 + 3 = 5             -1

2 + 4 = 6             -1

2 + 5 = 7             -1

2 + 6 = 8             -1

3 + 1 = 4             -1

3 + 2 = 5             -1

3 + 3 = 6             -1

3 + 4 = 7             -1

3 + 5 = 8             -1

3 + 6 = 9             -1

4 + 1 = 5             -1

4 + 2 = 6             -1

4 + 3 = 7             -1

4 + 4 = 8            -1

4 + 5 = 9            -1

4 + 6 = 10          -1

5 + 1 = 6            -1

5 + 2 = 7            -1

5 + 3 = 8            -1

5 + 4 = 9            -1

5 + 5 = 10          -1

5 + 6 = 11          +2

6 + 1 = 7            -1

6 + 2 = 8            -1

6 + 3 = 9            -1

6 + 4 = 10          -1

6 + 5 = 11          +2

6 + 6 = 12          +6


2) Probabilities

From that, there is:
- 2/36 probabilities to earn + 6 points.
- 4/36 probabilites to earn + 2 points
- the rest, 30/36 probabilities to earn - 1 points


3) Expected value = (2/36)(+6) + (4/36) (+2) + (30/36) (-1) = - 0.28
4 0
3 years ago
Read 2 more answers
Lucy made $323 for 17 hours of work.
saveliy_v [14]
19 minutes I think just divide 323 and 17
3 0
3 years ago
Help please?????????
kvv77 [185]
6x+4 (H)
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4 0
3 years ago
Is anybody else here to help me ??​
Akimi4 [234]

Answer:

\cot(x)+\cot(\frac{\pi}{2}-x)

\cot(x)+\tan(x)

\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}

\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}]

\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}]

\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}]

\csc(x)[\frac{1}{\cos(x)}]

\csc(x)[\sec(x)]

\csc(x)[\csc(\frac{\pi}{2}-x)]

\csc(x)\csc(\frac{\pi}{2}-x)

Step-by-step explanation:

I'm going to use x instead of \theta because it is less characters for me to type.

I'm going to start with the left hand side and see if I can turn it into the right hand side.

\cot(x)+\cot(\frac{\pi}{2}-x)

I'm going to use a cofunction identity for the 2nd term.

This is the identity: \tan(x)=\cot(\frac{\pi}{2}-x) I'm going to use there.

\cot(x)+\tan(x)

I'm going to rewrite this in terms of \sin(x) and \cos(x) because I prefer to work in those terms. My objective here is to some how write this sum as a product.

I'm going to first use these quotient identities: \frac{\cos(x)}{\sin(x)}=\cot(x) and \frac{\sin(x)}{\cos(x)}=\tan(x)

So we have:

\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}

I'm going to factor out \frac{1}{\sin(x)} because if I do that I will have the \csc(x) factor I see on the right by the reciprocal identity:

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

\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

Now I need to somehow show right right factor of this is equal to the right factor of the right hand side.

That is, I need to show \cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)} is equal to \csc(\frac{\pi}{2}-x).

So since I want one term I'm going to write as a single fraction first:

\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)}

Find a common denominator which is \cos(x):

\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}

\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}

\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}

By  the Pythagorean Identity \cos^2(x)+\sin^2(x)=1 I can rewrite the top as 1:

\frac{1}{\cos(x)}

By the quotient identity \sec(x)=\frac{1}{\cos(x)}, I can rewrite this as:

\sec(x)

By the cofunction identity \sec(x)=\csc(x)=(\frac{\pi}{2}-x), we have the second factor of the right hand side:

\csc(\frac{\pi}{2}-x)

Let's just do it all together without all the words now:

\cot(x)+\cot(\frac{\pi}{2}-x)

\cot(x)+\tan(x)

\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}

\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}]

\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}]

\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}]

\csc(x)[\frac{1}{\cos(x)}]

\csc(x)[\sec(x)]

\csc(x)[\csc(\frac{\pi}{2}-x)]

\csc(x)\csc(\frac{\pi}{2}-x)

7 0
3 years ago
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