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The trigonometric function gives the ratio of different sides of a right-angle triangle. The given problems can be solved as given below.
<h3>What are Trigonometric functions?</h3>The trigonometric function gives the ratio of different sides of a right-angle triangle.
where perpendicular is the side of the triangle which is opposite to the angle, and the hypotenuse is the longest side of the triangle which is opposite to the 90° angle.
1st.) x = 5 /Sin(30°)
x = 10
!) sin(45°) = 4/x
x = 4/sin(45°)
x = 4√2
I) Cos(45°) = √3 / x
x = √3 / Cos(45°)
x = √6
E) Tan(60°) = 3√3 / x
x = 3√3 / 3
W) For isosceles right-triangle, the angle made by the legs and the hypotenuse is always 45°.
x = 45°
N) x² + x² = (7√2)²
x = 7
V) Tan(60°) = 7 / x
x = 7√3/3
K) x² + x² = (9)²
x = 9/√2
Y) Sin(60°) = 7√3/x
x = 14
M) Sin(30°) = x/11
x = 11/2
T) Sin(45°) = x/√10
x = √5
A) x + 2x + 90° = 180°
x = 30°
O) Sin(45°) = √2 / x
x = 2
R) Tan(30°) = x / 4
x = 4/√3 = 4√3 / 3
S) Sin(60°) = x / (10/3)
x = 5√3 / 3
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Answer:
The 1st statement i.e. "The probability that a randomly selected silver medal was awarded to Great Britain is StartFraction 17 Over 99 EndFraction" is true.
Step-by-step explanation:
<em>The</em><em> </em><em> table:</em>
Gold Silver Bronze Total
United States 46 29 29 104
China 38 27 23 88
Russia 24 26 32 82
Great Britain 29 17 19 65
Total 137 99 103 339
Let S be the sample space and E be the event and P(E) by the probability of a certain event.
1st Statement: The probability that a randomly selected silver medal was awarded to Great Britain is StartFraction 17 Over 99 EndFraction.
As the total number of silver medals are = n(S) = 99
The number of silver medals won by Great Britain = n(E) = 17
P(E) = n(E)/n(S) = 17/99
Hence, the probability that a randomly selected silver medal was awarded to Great Britain is 17/99.
So, the 1st statement i.e. "The probability that a randomly selected silver medal was awarded to Great Britain is StartFraction 17 Over 99 EndFraction" is true.
2nd Statement: The probability that a randomly selected medal won by Russia was a bronze medal is StartFraction 32 Over 103 EndFraction.
As the total number of medals won by Russian are = 82 = n(S) = 82
The number of bronze medals won by Russia are = n(E) = 32
P(E) = n(E)/n(S) = 32/82 = 16/41
Hence, the probability that a randomly selected medal won by Russia was a bronze medal is 16/41.
So, the 2nd statement i.e. "The probability that a randomly selected medal won by Russia was a bronze medal is StartFraction 32 Over 103 EndFraction" is NOT true.
3rd Statement: The probability that a randomly selected gold medal was awarded to China is StartFraction 88 Over 137 EndFraction.
As the total number of gold medals are = n(S) = 137
The number of gold medals won by China are = n(E) = 38
P(E) = n(E)/n(S) = 38/137
Hence, the probability that a randomly selected gold medal was awarded to China is 38/137.
So, the 3rd statement i.e. "The probability that a randomly selected gold medal was awarded to China is StartFraction 88 Over 137 EndFraction" is NOT true.
4th Statement: The probability that a randomly selected medal won by the United States was a silver medal is StartFraction 104 Over 339 EndFraction.
As the total number of medals won by United States are = n(S) = 104
The number of silver medals won by United States are = n(E) = 29
P(E) = n(E)/n(S) = 29/104 = 29/104
Hence, The probability that a randomly selected medal won by the United States was a silver medal is 29/104.
So, the 4th statement i.e. "The probability that a randomly selected medal won by the United States was a silver medal is StartFraction 104 Over 339 EndFraction." is NOT true.
Keywords: Probability, event
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recall that, log <--- with no apparent base, implies base10, so you can just plug that in your calculator
for the change of base rule, it doesn't really matter what base you use, so long is the same above and below, it just so happen, that we used base10 in this case, but could have been anything, same result.