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vfiekz [6]
3 years ago
6

1.

Mathematics
1 answer:
mr_godi [17]3 years ago
8 0
Plug the points into the equation.
Ex. y=3x-5
(X,y)
1. (5,10): 10=3(5)-5
15-5=10; point 1 is valid
-1=3(-2)-5
-1=-6-5
-1 does not equal -11
Point 2 is not valid
Same process for letter B
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If AB= 4 units and DC=4 units , what is the length of BC?<br>​
larisa [96]

answer 4

Step-by-step explanation:

if ab 4

dc 4

assuming a square bc 4

4 0
2 years ago
5.476 rounded to the hundredths place
Murrr4er [49]
5.476 rounded to the hundredths place = 5.48
5 0
3 years ago
Read 2 more answers
(3)/(2x+3)=(2)/(10x+4)
RoseWind [281]

Answer:

x = -3/13

Step-by-step explanation:

cross multiply

4x + 6 = 30x + 12

combine like terms

-6 = 26x

x = -6/26 = -3/13

7 0
3 years ago
A total of 12 players consisting 6 male and 6 female badminton players are attending a training camp
abruzzese [7]

Step-by-step explanation:

<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>

The order doesn't matter (AB is the same as BA), so use combinations.

For the first pair, there are ₁₂C₂ ways to choose 2 people from 12.

For the second pair, there are ₁₀C₂ ways to choose 2 people from 10.

So on and so forth.  The total number of combinations is:

₁₂C₂ × ₁₀C₂ × ₈C₂ × ₆C₂ × ₄C₂ × ₂C₂

= 66 × 45 × 28 × 15 × 6 × 1

= 7,484,400

<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!.

The total number of possible schedules is:

₁₂C₆ × 6!

= 924 × 720

= 665,280

<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>

If at most 4 girls are selected, that means there's either 0, 1, 2, 3, or 4 girls.

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

7 0
3 years ago
Find an acute angle whose sine is 0.52
Shalnov [3]
<span> Answer: 90° degrees </span>
5 0
3 years ago
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