Help so please answer fast............

What is 2e 8f in simplest form

Furkat [3]

It is all the way simplfyed all the way, I hope that that answers your question☻

6 0

2 years ago

Help! I’ll mark you brainly and give extra points!!

irina1246 [14]

Answer:

Slope: 2

y-intercept: (0, -4)

x y

0 -4

2 0

Hope this helps! Can I have Brainliest please?

See attachment below:

8 0

2 years ago

Read 2 more answers

How many students tired out for the volleyball team is this a statistical or no

vodomira [7]

That question is not a statistical question since there will be only one answer. If it were to be statistical it would have multiple answers an example of that type of question would be, Why did you decide to try out for the volleyball team? This is statistical because some could say they joined for fun, credits, to do something active, etc.. there would be multiple answers. Your question, How many students tried out for the volleyball team isn't statistical since it will have one answer such as, 26 students, 15 student, 2 students, etc...

Hope this helped!

4 0

3 years ago

Use the distributive property to remove the parentheses.<br> (v-2)9

Sunny_sXe [5.5K]

Answer:

9v-18

Step-by-step explanation:

.........................................

5 0

2 years ago

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

Oksana_A [137]

Answer:

$n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}$

Step-by-step explanation:

Lets divide it in cases, then sum everything

Case (1): All 5 numbers are different

In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.

The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.

${n \choose 5 } = \frac{n!}{5!(n-5)!}$

Case (2): 4 numbers are different

We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4 ${n \choose 4} .$

We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.

The total cardinality of this case is $4 * {n \choose 4} .$

Case (3): 3 numbers are different

As we did before, we pick 3 elements from a set of n. The amount of possibilities is ${n \choose 3} .$

Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have ${3 \choose 2 } = 3$ possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of $6 * {n \choose 3}$

Case (4): 2 numbers are different

We pick 2 numbers from a set of n, with a total of ${n \choose 2}$ possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.

The total amount of possibilities for this case is

$4 * {n \choose 2}$

Case (5): All numbers are the same

This is easy, he have as many possibilities as numbers the set has. In other words, n

Conclussion

By summing over all 5 cases, the total amount of possibilities to form 5-tuples of integers from 1 through n is

$n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}$

I hope that works for you!

4 0

3 years ago