I need help with this pleas
1 answer:
You might be interested in
Answer:
12 ÷ 4
Step-by-step explanation:
Let's try to solve the given expression to see if it matches with any of the options guven..
(this is equivalent to 12 ÷ 4, which is given in the question)
Thus,
12 ÷ 4 equivalent to
<u>Solution-</u>
A school has 1800 students and 1800 light bulbs, each with a pull cord and all in a row.
As all the lights start out off, in the first pass all bulbs will be turned on.
In the second pass all the multiples of 2 will be off and rest will be turned on.
In the third pass all the multiples of 3 will be off, but the common multiple of 2 and 3 will be on along with the rest. i.e all the multiples of 6 will be turned on along with the rest.
In the fourth pass 4th light bulb will be turned on and so does all the multiples of 4.
But, in the sixth pass the 6th light bulb will be turned off as it was on after the third pass.
This pattern can observed that when a number has odd number of factors then only it can stay on till the last pass.
1 = 1
2 = 1, 2
3 = 1, 3
<u>4 = 1, 2, 4</u>
5 = 1, 5
6 = 1, 2, 3, 6
7 = 1, 7
8 = 1, 2, 4, 8
9 = 1, 3, 9
10 = 1, 2, 5, 10
11 = 1, 11
12 = 1, 2, 3, 4, 6, 12
13 = 1, 13
14 = 1, 2, 7, 14
15 = 1, 3, 5, 15
16 = 1, 2, 4, 8, 16
so on.....
The numbers who have odd number of factors are the perfect squares.
So calculating the number of perfect squares upto 1800 will give the number of light bulbs that will stay on.
As, , so 42 perfect squared numbers are there which are less than 1800.
∴ 42 light bulbs will end up in the on position. And there position is given in the attached table.
Using the Poisson distribution, we have that:
- There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.
- There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.
- There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
The parameters are:
- x is the number of successes
- e = 2.71828 is the Euler number
is the mean in the given interval.
For this problem, the mean is given as follows:
The probability of having exactly 10 days of precipitation in the month of April is P(X = 10), hence:
There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.
The probability of having less than three days of precipitation in the month of April is:
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
In which:
Then:
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0.00003 + 0.00019 = 0.00022
There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.
For more than 15 days, the probability is:
P(X > 15) = P(X = 16) + P(X = 17) + ... + P(X = 20)
Applying the formula for each of these values and adding them, we have that P(X > 15) = 0.2364, hence:
There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.
More can be learned about the Poisson distribution at brainly.com/question/13971530
#SPJ1