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Radda [10]
3 years ago
9

What is the length of the diagonal of a square with a side that measures 11 centimeters

Mathematics
1 answer:
timofeeve [1]3 years ago
4 0
A square has 4 equal sides
use pythagroeas theorem since a square has 4 right angles

a^2+b^2=c^2
a=b because the sides are equal (11)

11^2+11^2=c^2
242=c^2
square root both sides
11√2=c

the diagonal legnth is 11√2 centimiters
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Explain each step in the process of finding the area of a circle given the circumference.
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Step-by-step explanation

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A system of equations is given below. y=1/2x-3 and -1/2x-3. Which of the following statements best describes the two lines?
pantera1 [17]

"They have different slopes but the same y-intercept, so they have one solution" is the statement which best describes the two lines.

Answer: Option D

<u>Step-by-step explanation:</u>

Given equations:

           y=\left(\frac{1}{2} \times x\right)-3

           y=\left(-\frac{1}{2} \times x\right)-3

As we know that the slope intercept form of a line is  

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So, from equation 1 and equation 2 we can see that

              m_{1}=\frac{1}{2} \quad \text { and } c_{1}=-3

              m_{2}=-\frac{1}{2} \text { and } c_{2}=-3

So, from the above expressions, we can say that both lines have different slopes but have same y – intercept with one common solution when x = 0.

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Please answer this correctly without making mistakes I want ace expert and genius people to answer this correctly without making
sergejj [24]

Answer: 38

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Plugging in -1 for y and -19 for z gets you this:

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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
2 years ago
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