Sample space is the set of possible outcomes of an experiment. 1. <span>Tossing a coin three times There are possible outcomes. Tossing the coin for the first time it can be a haid (H) or a tail (T). So, we can have this outcomes: HHH HHT HTH THH TTH THT TTT total: 6 outcomes 2. </span><span>The order that the top 5 students will receive their diplomas. Here we need to find the number of permutations: 5!=5*4*3*2*1=120 3. </span><span>Tossing a pair of dice The possible outcomes are the following: (a,b), where a is the first dice and b is the second dice </span>(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6) , total 36 possible outcomes.
We can only have a right triangle if and only if a^2+b^2 = c^2 is a true equation. The 'c' is the longest side, aka hypotenuse. The legs 'a' and 'b' can be in any order you want.
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For choice A,
a = 2
b = 3
c = sqrt(10)
So,
a^2+b^2 = 2^2+3^2 = 4+9 = 13
but
c^2 = (sqrt(10))^2 = 10
which is not equal to 13 from above. Cross choice A off the list.
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Checking choice B
a = 3
b = 5
c = sqrt(34)
Square each equation
a^2 = 3^2 = 9
b^2 = 5^2 = 25
c^2 = (sqrt(34))^2 = 34
We can see that
a^2+b^2 = 9+25 = 34
which is exactly equal to c^2 above. This confirms the answer.
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Let's check choice C
a = 5, b = 8, c = 12
a^2 = 25, b^2 = 64, c^2 = 144
So,
a^2+b^2 = c^2
25+64 = 144
89 = 144
which is a false equation allowing us to cross choice C off the list.
