27 is what percent of 1875

Mathematics

2 answers:

Degger [83]3 years ago

8 0

Answer:

506.25

Step-by-step explanation:

d0: 27÷100×1875

[im s0rry if this is wr0ng i tried l0l]

vredina [299]3 years ago

4 0

Answer:

506.25

Step-by-step explanation:

27% is writen as 27/100

27/100 of 1875 = 27/100 x 1875

therefore, 506.25 is the answer.

If you are using a calculator, simply enter 27÷100×1875 which will give you 506.25 as the answer.

You might be interested in

Please factorise 16p²-9q² for me<br>

kirill115 [55]

Answer:

a ^ 2 - b ^ 2 = ( a + b ) ( a - b ) where a = 4p and b = 3q.

( 4p + 3q ) ( 4p - 3q )

7 0

3 years ago

) An instructor gives his class a set of 18 problems with the information that the next quiz will consist of a random selection

RSB [31]

Answer:

The probability the he or she will answer correctly is 1.5%

Step-by-step explanation:

In all, there are 18 problems. In this question, the order of which the problems are sorted for the quiz makes no difference. For example, if the question A of the quiz is P1 and question B P2, and question A P2 and question B P1, it is the same thing.

There are 18 problems and 9 are going to be selected. So, there is going to be a combination of 9 elements from a set of 18 elements.

A combination of n elements from a set of m objects has the following formula:

$C_{(m,n)} = \frac{m!}{n!(m-n)!}$

In this question, m = 18, n = 9. So the total number of possibilities is:

$T_{p} = C_{(18,9)} = \frac{18!}{9!(18-9)!} = 48620$

Now we have to calculate the number of desired outcomes. This number is a combination of 9 elements from a set of 13 elements(13 is the number of problems that the student has figured out how to do).

Now, m = 13, n = 9. The number of desired possibilities is:

$D_{p} = C_{(13,9)} = \frac{13!}{9!(13-9)!} = 715$