Which statement is true?
- The probability that a randomly selected silver medal was awarded to Great Britain is StartFraction 17 Over 99 EndFraction.
- The probability that a randomly selected medal won by Russia was a bronze medal is StartFraction 32 Over 103 EndFraction.
- The probability that a randomly selected gold medal was awarded to China is StartFraction 88 Over 137 EndFraction.
- The probability that a randomly selected medal won by the United States was a silver medal is StartFraction 104 Over 339 EndFraction.
Answer:
(A)The probability that a randomly selected silver medal was awarded to Great Britain is 17/99.
Step-by-step explanation:
The table is given below:

We calculate the probabilities given in the statements.
(A) The probability that a randomly selected silver medal was awarded to Great Britain
= 17/99
(B)The probability that a randomly selected medal won by Russia was a bronze medal
=32/82
(C)The probability that a randomly selected gold medal was awarded to China
=38/137
(D)The probability that a randomly selected medal won by the United States was a silver medal
=29/104
We can see that only the first statement is true.
Answer:
Variant c
Step-by-step explanation:
c²=4*9
c=6
You can apply phyphagor
b²=6²+9²=36+81=117

Do you want a general solution or from 0 <= x < 2pi?
tan^2(2x) - 1 = 0
tan^2(2x) = 1
Take the square root of both sides,
tan(2x) = +/- 1
Two equations:
tan(2x) = 1
tan(2x) = -1
Solve each equation.
tan(2x) = 1, 2x = {pi/4, 5pi/4, 9pi/4, 13pi/4},
x = { pi/8, 5pi/8, 9pi/8, 13pi/8 }
tan(2x) = -1, 2x = { 3pi/4, 7pi/4, 11pi/4, 15pi/4 },
x = { 3pi/8, 7pi/8, 11pi/8, 15pi/8}
So for solutions within [0, 2pi),
x = {pi/8, 3pi/8, 5pi/8, 7pi/8, 9pi/8, 11pi/8, 13pi/8, 15pi/8 }
If we multiply the bottom equation by 2 and move x to the right, it becomes:
4y = 2x-38
Now we can substitute it for the 4y in the top equation:
3x + (2x-38) = -23 => 5x = -23+38 => 5x = 15 => x=3
Then 4y = 2*3-38 => y = -8
So the solution is (3,-8)