Esto significa que debemos tener 58 páginas de plástico para contener las 517 tarjetas.´
<h3>¿Cuántas páginas se necesitarán para almacenar 517 tarjetas? </h3>
Sabemos que cada página puede almacenar hasta 9 cartas.
Entonces queremos ver cuantos grupos de 9 cartas hay en el conjunto de 517, para ver esto tomamos el cociente entre 517 y 9.
N = 517/9 = 57.44
Y no podemos tener un numero racional, así que debemos redondear al proximo número entero, que es 58.
Esto significa que debemos tener 58 páginas de plástico para contener las 517 tarjetas.
Sí quieres aprender más sobre cocientes:
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Answer:
3:1:2
Step-by-step explanation:
First convert g to kg Multiply each by 2 to get whole number
Answer: 6 hours
Explanation: You divide 330 miles by 55 mph.
Answer:
A darts player practices throwing a dart at the bull’s eye on a dart board. Her probability of hitting the bull’s eye for each throw is 0.2.
(a) Find the probability that she is successful for the first time on the third throw:
The number F of unsuccessful throws till the first bull’s eye follows a geometric
distribution with probability of success q = 0.2 and probability of failure p = 0.8.
If the first bull’s eye is on the third throw, there must be two failures:
P(F = 2) = p
2
q = (0.8)2
(0.2) = 0.128.
(b) Find the probability that she will have at least three failures before her first
success.
We want the probability of F ≥ 3. This can be found in two ways:
P(F ≥ 3) = P(F = 3) + P(F = 4) + P(F = 5) + P(F = 6) + . . .
= p
3
q + p
4
q + p
5
q + p
6
q + . . . (geometric series with ratio p)
=
p
3
q
1 − p
=
(0.8)3
(0.2)
1 − 0.8
= (0.8)3 = 0.512.
Alternatively,
P(F ≥ 3) = 1 − (P(F = 0) + P(F = 1) + P(F = 2))
= 1 − (q + pq + p
2
q)
= 1 − (0.2)(1 + 0.8 + (0.8)2
)
= 1 − 0.488 = 0.512.
(c) How many throws on average will fail before she hits bull’s eye?
Since p = 0.8 and q = 0.2, the expected number of failures before the first success
is
E[F] = p
q
=
0.8
0.2
= 4.
