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gtnhenbr [62]
3 years ago
13

If FG=10 and HJ = 10. Which additional facts guarantee that FGHJ is a parallelogram? check all that apply.

Mathematics
1 answer:
Kamila [148]3 years ago
5 0
Yes, if FJ || GH, and also if FG || HJ, since a parallelogram has two pairs of parallel sides.

now, we know FG = HJ = 10, but if it's a parallelogram, then the other two sides must also have the same length, so chances are FJ = GH = 15.
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<em>"A total of 12 players consisting 6 male and 6 female badminton players are attending a training camp."</em>

<em />

<em>"(a) During a morning activity of the camp, these 12 players have to randomly group into six pairs of two players each."</em>

<em>"(i) Find the total number of possible ways that these six pairs can be formed."</em>

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<em>"(ii) Find the probability that each pair contains players of the same gender only. Correct your final answer to 4 decimal places."</em>

We need to find the number of ways that 6 boys can be grouped into 3 pairs.  Using the same logic as before:

₆C₂ × ₄C₂ × ₂C₂

= 15 × 6 × 1

= 90

There are 90 ways that 6 boys can be grouped into 3 pairs, which means there's also 90 ways that 6 girls can be grouped into 3 pairs.  So the probability is:

90 × 90 / 7,484,400

= 1 / 924

≈ 0.0011

<em>"(b) During an afternoon activity of the camp, 6 players are randomly selected and 6 one-on-one matches with the coach are to be scheduled.</em>

<em>(i) How many different schedules are possible?"</em>

There are ₁₂C₆ ways that 6 players can be selected from 12.  From there, each possible schedule has a different order of players, so we need to use permutations.

There are 6 options for the first match.  After that, there are 5 options for the second match.  Then 4 options for the third match.  So on and so forth.  So the number of permutations is 6!.

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<em>"(ii) Find the probability that the number of selected male players is higher than that of female players given that at most 4 females were selected. Correct your final answer to 4 decimal places."</em>

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If 0 girls are selected, the number of combinations is:

₆C₆ × ₆C₀ = 1 × 1 = 1

If 1 girl is selected, the number of combinations is:

₆C₅ × ₆C₁ = 6 × 6 = 36

If 2 girls are selected, the number of combinations is:

₆C₄ × ₆C₂ = 15 × 15 = 225

If 3 girls are selected, the number of combinations is:

₆C₃ × ₆C₃ = 20 × 20 = 400

If 4 girls are selected, the number of combinations is:

₆C₂ × ₆C₄ = 15 × 15 = 225

The probability that there are more boys than girls is:

(1 + 36 + 225) / (1 + 36 + 225 + 400 + 225)

= 262 / 887

≈ 0.2954

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