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Sholpan [36]
2 years ago
6

con 16 barras de mantequilla se prepararon 20 postres ?que fraccion del total de mantequilla corresponde a cada postre

Mathematics
1 answer:
MatroZZZ [7]2 years ago
5 0

De ahí que la fracción de la mantequilla total que corresponde a cada postre sea 1/4

Si con 16 barras de mantequilla se prepararon 20 postres, esto se puede expresar como:

16 pegatinas = 20 postres

Para obtener la fracción del total de mantequilla que corresponde a cada postre, podemos escribir;

x = 1 postre

Divide ambas expresiones

16 / x = 20/1

20 veces = 16 * 1

20x = 16

x = 16/20

x = 4/5

De ahí que la fracción de la mantequilla total que corresponde a cada postre sea 1/4

Obtenga más información aquí: brainly.com/question/1878884

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3 years ago
Suppose you have an experiment where you flip a coin three times. You then count the number of heads. a.)State the random variab
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Answer:

a) X=number of heads observed when flipped the coin 3 times

b) the probability distribution is

P(X=x) = 3/4 * (1/[(3-x)!*x!)])

or

P(X)=1/8 for x=0 and x=3 and P(X)=3/8 for x=1 and x=2

Step-by-step explanation:

the random variable will be X=number of heads observed when flipped the coin 3 times . Since the result from each flip is independent of the others , then X has a binomial probability distribution , such that

P(X=x)= n!/[(n-x)!*x!)*p^x * (1-p)^(n-x)

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n= number of times the coin is flipped = 3

p= probability of getting heads in a flip of the coin = 1/2 (assuming that the coin is fair)

therefore

P(X=x)= 3!/[(3-x)!*x!)*(1/2)^(3-x) * (1/2)^x = 3!/[(3-x)!*x!) * (1/2)³ = 3/4 * (1/[(3-x)!*x!)])

P(X=x)= 3/4 * (1/[(3-x)!*x!)])   , for x=[0,1,2,3]

for x=0 and x=3 → P(X)=3/4 * (1/[3!*0!)]) = 1/8

for x=1 and x=2 → P(X)=3/4 * (1/[2!*1!)]) = 3/8

we can verify that is correct since the sum of all the probabilities from x=0 to x=3 is  1/8 +  3/8+ 3/8+ 1/8 = 1

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