The probability of a spinner landing on yellow is 1/4 what does 3/4 represent? A. yellow B. blue C. yellow D. purple and blue
D. Purple and blue.
If the probability of landing on yellow is 1/4, then the probability of NOT landing on yellow (landing on any of the other colors) is 3/4. Therefore, 3/4 represents the combined probability of landing on purple and/or blue.
Marty can go to one movie on either Friday or saturday. He will be able to choose either a comedy or an action movie. Which list shows all the possible outcomes of one movie on one day.
A. Comedy, Friday
Comedy, Saturday
Action, Friday
Action, Saturday
B. Comedy, action
Action, comedy
Friday, Saturday
Saturday, Friday
C. Action, Friday
Action, Saturday
D. Comedy, Friday
Action, Friday
A. Comedy, Friday
Comedy, Saturday
Action, Friday
Action, Saturday
These are all the possible outcomes of Marty going to one movie on one day, where he can choose between a comedy or an action movie and can go either on Friday or Saturday.
a health study of $540 adults found that only 7 and 20 adults get enough sleep each night based on the results of the sample how many adults got enough sleep
A. 27 B. 35 c. 89 D.202
We can start by setting up a proportion where the number of adults who get enough sleep is compared to the total number of adults in the population:
x / 540 = 7/20
To solve for x, we can cross-multiply to get:
20x = 7 * 540
20x = 3780
x = 3780 / 20
x = 189
Therefore, based on the results of the sample, 189 adults got enough sleep. None of the answer choices given match this result, so it's possible that there was an error in the question or answer choices.
Two coins are flipped what is the probability that both coins land on heads type your answer is a fraction
The probability of getting heads on one coin flip is 1/2.
To find the probability of getting heads on both coin flips, we multiply the probability of each event. So:
P(heads on first coin AND heads on second coin) = P(heads on first coin) * P(heads on second coin)
P(heads on first coin AND heads on second coin) = (1/2) * (1/2)
P(heads on first coin AND heads on second coin) = 1/4
Therefore, the probability of both coins landing on heads is 1/4.
Daniel is selecting a sock from his drawer at random. If Daniel dose this 60 times, what is reasonable prediction for the number of times he will get a striped sock?
We can't provide an exact prediction of the number of times Daniel will get a striped sock, but we can provide a reasonable prediction based on probability.
Let's say that Daniel's drawer has 20 socks in it and 4 of those socks are striped. Then the probability of him selecting a striped sock at random is 4/20 or 1/5.
If he selects a sock 60 times, we can predict that he will get a striped sock about 1/5 of the time. To find out exactly how many times that would be, we multiply the probability of getting a striped sock by the number of times he selects a sock:
(1/5) * 60 = 12
Therefore, a reasonable prediction is that Daniel will get a striped sock about 12 times out of the 60 times he selects a sock. However, if the drawer has a different number of socks or a different number of striped socks, this prediction will change.
Daniel is selecting a sock from his drawer at random. If Daniel dose this 60 times, what is reasonable prediction for the number of times he will get a striped sock?
there are 6 white and 2 striped socks
In this case, the probability of selecting a striped sock at random is 2/8, which can be simplified to 1/4.
Using this probability, we can make a reasonable prediction for the number of times Daniel will get a striped sock by multiplying it by the number of times he selects a sock:
(1/4) * 60 = 15
Therefore, a reasonable prediction is that Daniel will get a striped sock about 15 times out of the 60 times he selects a sock, assuming the drawer contains 6 white socks and 2 striped socks.
Michael has a set of five cards shown below. Michael will randomly select a card, not replace it, and then select another card. What is the probability that he selects a prime number and then a 1?
1 2 3 4 5
There is only one card that has a value of 1 and it is not a prime number, so the probability of selecting a 1 first is 0.
There are three prime numbers in the set: 2, 3, and 5. If Michael selects one of these prime numbers first, he will be left with four cards in the set, one of which is a 1. So the probability of selecting a prime number and then a 1 is:
P(select prime first) * P(select 1 second | prime selected first)
The probability of selecting a prime number first is 3/5.
If Michael selects a prime number first, there are 4 cards left in the set, one of which is a 1. So the probability of selecting a 1 second given that a prime number was selected first is 1/4.
Putting it all together:
P(select prime first) * P(select 1 second | prime selected first) = (3/5) * (1/4) = 3/20
Therefore, the probability of Michael selecting a prime number first and then a 1 is 3/20.
Amos is going to randomly select a donut from a box containing the donuts shown in the table Two statements are made. Which statement(s) are true?
1. Amos is three times as likly yo select a filled donut than a glazed donut.
2. Amos is equally likely to select a filled or sprinkled donut as a glazed donut.
A. 1 only
B. 1 and 2
C. 2 only
D. Neither 1 or 2
type quantity
glazed 6
filled 2
sprinkled 4
To determine which statement(s) are true, we need to calculate the probability of selecting each type of donut.
The total number of donuts in the box is:
6 (glazed) + 2 (filled) + 4 (sprinkled) = 12
So, the probability of selecting a glazed donut is 6/12 or 1/2.
The probability of selecting a filled donut is 2/12 or 1/6.
The probability of selecting a sprinkled donut is 4/12 or 1/3.
1. Amos is three times as likely to select a filled donut than a glazed donut. This statement is false since the probability of selecting a filled donut is 1/6 and the probability of selecting a glazed donut is 1/2. 1/6 is not three times 1/2.
2. Amos is equally likely to select a filled or sprinkled donut as a glazed donut. This statement is also false since the probability of selecting a glazed donut is 1/2, but the probability of selecting either a filled or a sprinkled donut is 1/6 + 1/3 = 1/2.
Therefore, the answer is D. Neither 1 or 2.