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marissa [1.9K]
3 years ago
14

Que por ciento de 20 es 6?

Mathematics
1 answer:
Mariulka [41]3 years ago
8 0

Answer:

30%

Step-by-step explanation:

x% de 20 es 6.

x% * 20 = 6

0.01x * 20 = 6

0.2x = 6

x = 6/0.2

x = 30

Respuesta: 30%

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You flip a coin 10 times and find the experimental probability of flipping tails to be 0.7. Does this seem reasonable? Explain.
Vedmedyk [2.9K]

No. I would say 50/50 chance. This answer probably isnt helpful :/

8 0
3 years ago
I Prove that Cos(90-θ)/1+sin(90-θ) +1+sin(90-θ) /cos(90-θ) =2
USPshnik [31]
Okay,

so first you gotta know that cos(90-x) is equal to sinx and sin(90-x) is equal to cosx!!
Now all you gotta do is replace the cos(90-x) to sinx in the numerator and sin(90-x) to cosx in the denomenator inorder to make the numerator all into sin and denomenator all into cos.
After that, open up the brackets and solve...
At the end you'll hopefully get something like this :( 1+sin90 ÷ 1+cos90 )
And since sin90 is 1 (put it in the calculator!) and cos90 is 0, you'll get 2÷1 which is equals to 2!!

Hope this helped!  :)
6 0
3 years ago
Read 2 more answers
Show me how to solve 6j-3 for j=4
aleksklad [387]

Answer:

21

Step-by-step explanation:

plug in 4 where you have the j

6(4)-3=

24-3=

21

6 0
3 years ago
Lcm of 8 and 15 Please!
FromTheMoon [43]

Since, 8 and 5 both are prime numbers u can simply multiply 8 and 15 to get your answer..

the ans is 120..

8 0
3 years ago
Which coefficient matrix represents a system of linear equations that has a unique solution ?
finlep [7]

Answer:

Option C

Step-by-step explanation:

We are given a coefficient matrix along and not the solution matrix

Since solution matrix is not given we cannot check for infinity solutions.

But we can check whether coefficient matrix is 0 or not

If coefficient matrix is zero, the system is inconsistent and hence no solution.

Option A)

|A|=\left[\begin{array}{ccc}4&2&6\\2&1&3\\-2&3&-4\end{array}\right] =0

since II row is a multiple of I row

Hence no solution or infinite

OPtion B

|B|=\left[\begin{array}{ccc}2&0&-2\\-7&1&5\\4&-2&0\end{array}\right] \\=2(10)-2(10)=0

Hence no solution or infinite

Option C

\left[\begin{array}{ccc}6&0&-2\\-2&0&6\\1&-2&0\end{array}\right] \\=2(36-2)=68

Hence there will be a unique solution

Option D

\left[\begin{array}{ccc}5&10&5\\4&1&4\\-1&-2&-1\end{array}\right] \\=2(10)-2(10)=0=0

(since I row is -5 times III row)

Hence there will be no or infinite solution

Option C is the correct answer



4 0
3 years ago
Read 2 more answers
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