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scZoUnD [109]
3 years ago
12

You have a friendship club. there are 12 members in your club. The 12th day of each month, you have a friendship call. Each memb

er of the club talks to every other member that day by phone. this way everybody expresses his or her friendship with each other. How many conversations are made on that day? Explain
How many conversations occur each year? Show how you figured it out.

If the club adds four new friends, how many conversations will occur each month? Explain your solution.

Suppose another friendship club likes your friendship conversation idea, but wants to know how they can figure out how many conversations will occur given any number of members. Explain how they can figure out the total number of conversations for each month
Mathematics
1 answer:
aleksklad [387]3 years ago
5 0
That day they will have 132 conversations. There are 12 members and each one will be calling the remaining 11 members of the club which means each one will make 11 calls and that will be done by 12 members so 11x12= 132 conversations in total

12 months= 1 year
Every month they make 132 calls so 132x12 will be 1,584 calls per year.

If the club decides to add four new members to the club the conversations each month will rise up to 240.

If the other club likes the idea they can use the following equation
(x-1)(x)
X represents the number of members of the group the subtraction of 1 is when a member makes a call not including them selves so there are 12 members but the one making the call isn't included so there are 11 calls being made then being multiplied by the total number of members.

(x-1)(x)
(12-1)(12)
11(12)
132
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Find the 62nd term of the following sequence:<br> (-21, – 27, - 33, - 39,...)
iris [78.8K]

Answer:

-387

Step-by-step explanation:

First term (T1) = -21 = a

second term (t2) = - 27

common difference (d) = t2 - t1 = - 27 + 21 = - 6

Now

Tn = a + ( n - 1) d

T62 = - 21 + ( 62 - 1 ) - 6

= - 21 -366

= -387

8 0
3 years ago
I need help asap, please and thank you :))​
Troyanec [42]

Answer:

What is the value of x?

<em>x = 40 degrees.</em>

What is the measure of angle AFE?

<em>Angle AFE = 140 degrees</em>

What is the measure of angle BFD?

<em>Angle BFD = 140 degrees.</em>

Step-by-step explanation:

Angle AFC is 90 degrees.

To find angle x, we have to subtract angle BFC.

90 - 50 = 40 degrees.

Line BFE is 180 degrees.

To find angle AFE, we have to subtract angle x from 180 degrees.

We have solved x already.

180 - 40 = 140 degrees.

Angle BFD is just 90 degrees plus 50 degrees.

90 + 50 = 140 degrees.

6 0
3 years ago
Read 2 more answers
1948 Olympics
mash [69]

Answer:

For 1948 Men's :Interquartile range is 1.5, Median is 58.3

For 2012 Men's: Interquartile range is 0.315, Median is 47.86

You can infer that in 2012 the swimmers were better and it was more competitive as the interquartile range was lower and the median was also lower as well

Step-by-step explanation:

1948 Men's 100m

57.3, 57.8, 58.1, 58.3, 58.3, 59.3, 59.6,1:00.5

               ⬆                ⬆                 ⬆

         Low IQR      Median       High IQR

Low IQR is average of 57.8 and 58.1 = 57.95

High IQR  is average of 59.3 and 59.6 =  59.45

Median is average of 58.3 and 58.3 = 58.3

Interquartile range is High IQR- Low IQR

59.45 - 57.95=1.5

2012 Men's 100m

47.52, 47.53, 47.8, 47.84, 47.88, 47.92, 48.04, 48.44

                    ⬆                  ⬆                      ⬆

             Low IQR         Median          High IQR

Low IQR is average of 47.53 and 47.8 = 47.665

High IQR is average of 48.04 and 47.92 = 47.98

Median is average of 47.88 and 47.84 = 47.86

Interquartile range is High IQR-Low IQR

47.98-47.665=0.315

Read more on Brainly.com - brainly.com/question/14915771#readmore

8 0
3 years ago
Please help and quickly please and 20 points with brainelest <br><br><br><br><br> Thank you
uysha [10]

Answer: The answer is B. 34

Step-by-step explanation:

3(12−9)+5^2

=(3)(3)+5^2

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=9+25

=34

5 0
3 years ago
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Solve the trigonometric equation for all values 0 ≤ x &lt; 2π.
Vanyuwa [196]
Not really sure but hope this helps

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3 years ago
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