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lyudmila [28]
2 years ago
9

What is the value of x? A. 6 B. 6 square root of 2 C. 12 D. 12 square root of 2

Mathematics
2 answers:
____ [38]2 years ago
8 0
You would use the sine function
sin(x)=opposite/hypotenuse
sin(45)=6√2/x (cross multiply)
x*sin(45)=6√2
x*√2/2=6√2 (multiply both sides by 2)
x*√2=12√2 (divide both sides by √2)
x=12 so C

Hope this helps
Aleksandr [31]2 years ago
6 0
C because 6 times 2 equals 12
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Answer:

a) 39 b) 17

Step-by-step explanation:

12x - 9

12(4) - 9

48 - 9

39

d/15 + 13

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17

3 0
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What is the solution for −40 = 10c? <br> A. c = 1 <br> B. C = -1 <br> C. c = -4 <br> D. c = 4
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Answer:

C. c = -4

Step-by-step explanation:

you must get c by itself. you have to divide both sides by 10. this gets rid of the 10c and keeps both sides equal

-40/10 = c

c = -4

5 0
3 years ago
What is the slope of (-3,5) and (2,0)
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is there anymore to the question?

Step-by-step explanation:

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If one pound of water has a mass of 453.6 grams, what would be the mass of 2.2 pounds of water?
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8 0
3 years ago
student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
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