Answer:
I wish I had the answer for you since I feel the same way when I have a teast n iono the answer its soo stressing hope you have u get a good grade by finding the answer some how.
SRRY
Answer: The correct answer is option C: Both events are equally likely to occur
Step-by-step explanation: For the first experiment, Corrine has a six-sided die, which means there is a total of six possible outcomes altogether. In her experiment, Corrine rolls a number greater than three. The number of events that satisfies this condition in her experiment are the numbers four, five and six (that is, 3 events). Hence the probability can be calculated as follows;
P(>3) = Number of required outcomes/Number of possible outcomes
P(>3) = 3/6
P(>3) = 1/2 or 0.5
Therefore the probability of rolling a number greater than three is 0.5 or 50%.
For the second experiment, Pablo notes heads on the first flip of a coin and then tails on the second flip. for a coin there are two outcomes in total, so the probability of the coin landing on a head is equal to the probability of the coin landing on a tail. Hence the probability can be calculated as follows;
P(Head) = Number of required outcomes/Number of all possible outcomes
P(Head) = 1/2
P(Head) = 0.5
Therefore the probability of landing on a head is 0.5 or 50%. (Note that the probability of landing on a tail is equally 0.5 or 50%)
From these results we can conclude that in both experiments , both events are equally likely to occur.
To solve you first need to pick out the most important information,
~Spent 20 minutes
~Each prob takes 2 minutes
Lets have y be the total time spent, and x the number of probs
(y/2) = x
Hope this helped!! ^-^
Answer:
7.30167%
Step-by-step explanation:
Usando la fórmula de puntuación z
z = (x-μ) / σ, donde x es la puntuación bruta, μ es la media de la población y σ es la desviación estándar de la población
Para x <0.20 pulgadas
z = 0.20 - 0.25 / 0.02
z = -2.5
Valor de probabilidad de Z-Table:
P (x <0.20) = 0.0062097
Para x> 0.28 pulgadas
z = 0.28 - 0.20 / 0.02
z = 1.5
Valor de probabilidad de Z-Table:
P (x <0.28) = 0.93319
P (x> 0.28) = 1 - P (x <0.28) = 0.066807
La probabilidad de que se produzcan tornillos defectuosos cuando el tornillo se considera defectuoso si su diámetro es inferior a 0.20 pulgadas o superior a 0.28 pulgadas es
P (x <0.20) + P (x> 0.28)
= 0.0062097 + 0.066807
= 0.0730167
Conversión a porcentaje
= 0.0730167 × 100
= 7.30167%
El porcentaje de tornillos defectuosos producidos es
7.30167%
