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wolverine [178]
3 years ago
8

Add a meeting six people each one another by shaking hands. If each person must shake hands once with every other person in the

room, how many handshakes occur in the meeting
A) 10
B) 12
C) 15
D) 16
Mathematics
1 answer:
Phantasy [73]3 years ago
8 0

The number of handshakes is the number of ways 6 objects can be selected 2 at a time.

.. C(6,2) = 6·5/(2·1) = 15


The appropriate choice is

... C) 15


_____

Consider that each person can shake hands with 5 others. That's a total of 6*5 = 30 handshakes, but it counts both A shaking with B and B shaking with A. If each pair of persons only shakes hands once, the number of handshakes is 30/2 = 15.

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What is the total cost of three shirts 22.50 and two ties 14.90 with a 5% sales tax
olasank [31]

Answer:

$39.37 or $102.165 sorry i didn't understand the question.

Step-by-step explanation:

This is the total if 3 shirts= 22.5 and 2 ties=14.9 + 1.87= 39.37

but if 3 x 22.5= 67.5 and 2 x 14.9= 29.8 + 4.865= 102.165

5 0
3 years ago
Read 2 more answers
Plz use the graph help asap show work
Diano4ka-milaya [45]
Question 9.)
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7 0
3 years ago
Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n 2 if heads comes up
Artyom0805 [142]

Answer:

In the long run, ou expect to  lose $4 per game

Step-by-step explanation:

Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n^2 if heads comes up first on the nth toss.

Assuming X be the toss on which the first head appears.

then the geometric distribution of X is:

X \sim geom(p = 1/2)

the probability function P can be computed as:

P (X = n) = p(1-p)^{n-1}

where

n = 1,2,3 ...

If I agree to pay you $n^2 if heads comes up first on the nth toss.

this implies that , you need to be paid \sum \limits ^{n}_{i=1} n^2 P(X=n)

\sum \limits ^{n}_{i=1} n^2 P(X=n) = E(X^2)

\sum \limits ^{n}_{i=1} n^2 P(X=n) =Var (X) + [E(X)]^2

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+(\dfrac{1}{p})^2        ∵  X \sim geom(p = 1/2)

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+\dfrac{1}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p+1}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-p}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-\dfrac{1}{2}}{(\dfrac{1}{2})^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{4-1}{2} }{{\dfrac{1}{4}}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{3}{2} }{{\dfrac{1}{4}}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ 1.5}{{0.25}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =6

Given that during the game play, You pay me $10 , the calculated expected loss = $10 - $6

= $4

∴

In the long run, you expect to  lose $4 per game

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