Plz help me, plzzzzz (plz no trolling)

Mathematics

1 answer:

Phantasy [73]3 years ago

8 0

1=104 2=76 3=104 4=76 5=104 6=76 7=104 8=76

You might be interested in

PLZ HELP WILL MARK BRANLIEST (No dum answers or I will delete your question - seriously don't be annoying)

Vsevolod [243]

Answer:

A. tan(2π/3) = $-\sqrt{3}$

4 0

3 years ago

3. Find BC<br><br> 6. find m< A

sammy [17]

Answer:

Step-by-step explanation:

7 0

2 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:

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