Answer:
<h2>use the required formulas please !</h2>
The answer to your question is 13<2/3
In a throw of 2 fair dice, there are 6*6=36 equiprobability outcomes.
To get a sum of 5, there are 4 ways, (1,4),(2,3),(3,2),(4,1) with probability of 4/36=1/9
To get at least one 5, there are 6+6-1=11 outcomes (note (5,5) has been counted in both, so subtracted from sum). The probability is 11/36
Since the two events are mutually exclusive (once we have a five, the sum can no longer be 5), we can add the probabilities to get the probability of one event or the other.
P(sum of 5 OR at least one 5)=1/9+11/36=4/36+11/36=15/36=5/9
(n+11)(n-4) is your answer
Responder:
42 bombones
Explicación paso a paso:
Dado que :
Número de bombones = 60
Cantidad consumida por Eba = 1/5 * 60 = 12
Cantidad comida por Ana = 1/2 * 60 = 30
Cantidad de chocolate que comieron Eba y Ana:
(12 + 30) = 42
De ahí que Ana y Eba se comieran 42 bombones
