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NemiM [27]
3 years ago
12

A company with n employees publishes an internal calendar, where each day lists the employees having a birthday that day. What i

s the probability there is at least one day in a year when nobody has a birthday?
Mathematics
1 answer:
lutik1710 [3]3 years ago
5 0

Answer: 1-(\frac{364}{365})^n

Step-by-step explanation:

Binomial probability formula :-

P(x)=^nC_xp^x(1-p)^x, where P(x) is the probability of getting success in x trials, n is the total number of trials and p is the probability of getting success in each trial.

We assume that the total number of days in a particular year are 365.

Then , the probability for each employee to have birthday on a certain day :

p=\dfrac{1}{365}

Given : The number of employee in the company = n

Then, the probability there is at least one day in a year when nobody has a birthday is given by :-

P(x\geq1)=1-P(x

Hence, the probability there is at least one day in a year when nobody has a birthday =1-(\frac{364}{365})^n

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-114=-5x-2(4x+18)<br> ????????
SVETLANKA909090 [29]

Primeiro multiplica o -2 com tudo que tem dentro do parenteses (chuveirinho)  -114 = -5x -2(4x+18)

-2 . 4x = -8x        +     -2 . 18 = -36        =   -8x-36


Agora soma os números com incógnitas (no caso x)

-114 = - 5x - 8x -36    


Joga a incógnita do outro, tornando positiva

-114 = -13x - 36            

13x -114 = -36


e o 114 do outro lado, ficando positiva também  

13x -114 = -36

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x = 78 /13

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8 0
3 years ago
This make no sense to me can someone help and explain it please
slava [35]

Answer:

x + 1

Step-by-step explanation:

They want you to write a generic expression to represent the total (adding) between a number (which they want you to write as 'x') and 1.

4 0
2 years ago
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Solve the system using substitution.
Iteru [2.4K]
Q=2p+1
subsitute 2p+1 for q in other eqaution


4p+2q=8
4p+2(2p+1)=8
4p+4p+2=8
8p+2=8
minus 2 both sides
8p=6
divide both sides by 8
p=6/8
p=3/4
sub back
q=2p+1
q=2(3/4)+1
q=6/4+1
q=3/2+1
q=3/2+2/2
q=5/2


p=3/4
q=5/2
8 0
3 years ago
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