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ivanzaharov
13 days ago
10

The probability of it snowing in Eastern Canada tomorrow is 0.40.40, point, 4. The probability of it snowing in Western Canada t

omorrow is \dfrac25 5 2 ​ start fraction, 2, divided by, 5, end fraction. Which of these events is more likely?
Mathematics
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Determine if each scenario is either a permutation or combination. Do NOT solve these scenarios. a) An art gallery displays 125
PIT_PIT [12445]

Answer:

a) Permutaciones

b) combinaciones

Step-by-step explanation:

a)

Dado que el orden de las 15 obras de arte más importantes es relevante, comenzando por la más popular y luego las que ocupan los lugares 2, 3, 4, y así sucesivamente. Cuando nos ocupamos del "orden de colocación", el tema se refiere a permutaciones.

b)

De un total de 320 obras, se deben "seleccionar" 125 para ser exhibidas, el proceso de selección implica combinaciones en las que el orden de la selección no importa.

3 0
3 months ago
A credit card contains 16 digits between 0 and 9. However, only 100 million numbers are valid. If a number is entered randomly,
PIT_PIT [12445]

Answer:

P(valid) = 1 * 10^{-8}

Step-by-step explanation:

First, we need to calculate how many potential credit card numbers can be created using the digits 0 through 9.

We aim to create a number that consists of 16 digits from a total of 10 digits, therefore:

N = 10^{16}

The total count of possible numbers formed is 10^{16}. However, only 100 million of these are valid (100 * 10^6 = 10^8). Let n represent the valid number of combinations.

This leads us to the probability of selecting a valid number:

P(valid) = n / N

P(valid) = 10^{8} / 10^{16}\\\\P(valid) = 1 * 10^{-8}

8 0
4 months ago
John's average for making free throws in a basketball game is .80. In a one-and-one free throw situation (where he shoots a seco
babunello [11817]
The likelihood that John scores on the first attempt and misses the second is:
P(makes\ 1\ basket)=0.8\times(1-0.8)=0.16
4 0
3 months ago
Read 2 more answers
Zucchini weights are approximately normally distributed with mean 08 pound and standard deviation 0.25 pound. Which of the follo
Svet_ta [12734]
The correct answer is A. According to the question's details, we're provided with key statistics on zucchini weights, which suggest that the average turns out to be typically 0.8 pounds, while the standard deviation is noted as 0.25 pounds. The probability of a randomly selected zucchini weighing between 0.55 pounds and 1.3 pounds can be mathematically expressed. Observing the provided normal distribution options, option A aligns with our specified weight range.
8 0
2 months ago
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