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mash [69]
3 years ago
5

Raymond bought 3/4 of dozen rolls. How many rolls did he buy?

Mathematics
2 answers:
Juliette [100K]3 years ago
6 0
Hey there! A dozen rolls = 12 rolls, of means multiply, we are going to multipy 3/4 by 12, make 12 a fraction, 12=12/1. Now, we can multiply 3/4 by 12/1. 3/4×12/1=36/4. Simplify 36/4 to 9. So, Raymond bought 9 rolls.
Sindrei [870]3 years ago
5 0
There are 12 rolls in a dozen.
1/4 of a dozen is 3, 2/4 is 6, so essentially 3/4 would be 9
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Suppose there are 200 lockers and 200 students. Kayla reasons that since there are 10
Rus_ich [418]

Answer:

Kayla's reasoning is not correct.

Step-by-step explanation:

The locker problem is as follows:

Imagine you are at a school that has student lockers. There are 200 lockers, all shut and unlocked, and 200 students. Suppose the first student goes along the row and opens every locker. The second student then goes along and shuts every other locker beginning with number 2. The third student changes the state of every third locker beginning with number 3. (If the locker is open the student shuts it, and if the locker is closed the student opens it). The fourth student changes the state of every fourth locker beginning with number 4. Imagine that this continues until the 200 students have followed the pattern with the 200 lockers. At the end, which lockers will be open and which will be closed? Why?

Solution:

So from the information we know that the first student goes along the row and opens every locker.

Then the second student shuts every other locker, i.e. locker numbers 2, 4, 6, 8, 10, ..., 196, 198 and 200.

Then the third students changes the state of every third locker, i.e. he/she closes an open locker and opens a closed locker.

So the open lockers are: 1, 5, 6, 12,...

Then the fourth students changes the state of every fourth locker.

So the open lockers are: 1, 4, 5, 6, 8,....

So, on we will observe that the open lockers have a perfect square number such as, 1, 4, 9, 16,....

Consider that the pattern is as follows:

Student 1 opens the locker, Student 2 closes it, Student 3 opens it, person 4 Student and so on.

This is because the square numbers always have an odd number of factors, which leads them to be open at the end.

Take any locker number, 40, for example. Its state (open or closed) is changed for every student whose number in line is a factor of the locker number.

Student      Locker 40 status

     1                    Open

     2                   Close

     4                   Open

     5                   Close

     8                   Open

     10                  Close

    20                  Open

    40                  Close

Like all other lockers numbered with non-square numbers, it ends up closed after all the students have gone through the line because it has an even number of factors.

Consider the locker number 16:

Student      Locker 16 status

     1                    Open

     2                   Close

     4                   Open

     8                   Close

     16                  Open

Thus, we can conclude that all the doors with square numbers on them will remain open because all square numbers have an odd number of factors and the doors with non-square numbers on them will remain close because they have even number of factors.

There will be a total of 14 lockers open.

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169 and 196

So, if there are 10 lockers open in the first 100 lockers there must be only 4 other lockers opened in the next 100.

5 0
3 years ago
Plssss help me on this one. What’s is the interquartile range of the data?
Natasha2012 [34]

Answer:

The interquartile range is from 22-29

Step-by-step explanation:

3 0
3 years ago
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If you roll one die and flip one coin, what is the probability of rolling a 2 and flipping a head? Why?
Nezavi [6.7K]

Answer:

Probability of rolling a 2 and flipping a head will be \frac{1}{12}    

Step-by-step explanation:

If we roll one die then probability to get any one side is \frac{1}{6}

Therefore, probability to get 2 by rolling the die will be P(A) = \frac{1}{6}

Now we flip a coin then getting head or tale probability is \frac{1}{2}

Or probability to get head by flipping the coin P(B) = \frac{1}{2}

Probability of happening both the events (rolling a 2 and flipping a head) will be denoted by

P(A∩B) = P(A)×P(B)

           = \frac{1}{6}\times \frac{1}{2}

           = \frac{1}{12}

Therefore, probability of rolling a 2 and flipping a head will be \frac{1}{12}        

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3 years ago
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Archimedes is the father of mathematics.
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mrs_skeptik [129]
3 because 20/6=3.3333333333 so you would have 3 because you can split the pictures
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