Answer:
2/3, or 66.7%
Step-by-step explanation:
There are 2 options for the first digit (3 or 4).
After the first digit is selected, it cannot be repeated, so there are 2 options left for the second digit.
After the first two digits are selected, there is only 1 option for the third digit.
So there are 4 permutations greater than 300.
The total number of permutations is 3! = 6.
So the probability is 4/6 = 2/3.
Answer:
60 times will they ring together at the same second in one hour excluding the one at the end.
Step-by-step explanation:
Given : Five bells begin to ring together and they ring at intervals of 3, 6, 10, 12 and 15 seconds, respectively.
To find : How many times will they ring together at the same second in one hour excluding the one at the end?
Solution :
First we find the LCM of 3, 6, 10, 12 and 15.
2 | 3 6 10 12 15
2 | 3 3 5 6 15
3 | 3 3 5 3 15
5 | 1 1 5 1 5
| 1 1 1 1 1
So, the bells will ring together after every 60 seconds i.e. 1 minutes.
i.e. in 1 minute they rand together 1 time.
We know, 1 hour = 60 minutes
So, in 60 minute they rang together 60 times.
Therefore, 60 times will they ring together at the same second in one hour excluding the one at the end.
Answer:
2115 = 9:15 PM
0830 = 8:30 AM
Step-by-step explanation:
12 - hour conversion.
9514 1404 393
Answer:
(3, 1)
Step-by-step explanation:
We assume you want the solution to the system ...
The second equation gives a nice expression for x, so we can use that in the first equation.
2(y+2) -3y = 3 . . . . substitute for x in the first equation
2y +4 -3y = 3 . . . . . eliminate parentheses
-y = -1 . . . . . . . . . . . collect terms, subtract 4
y = 1 . . . . . . . . . . . . multiply by -1
x = 1 +2 = 3 . . . . . . substitute for y in the second equation
The solution is (x, y) = (3, 1).
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.
