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brilliants [131]
3 years ago
8

The number of souvenir coeffe mugs (in hundreds) that Ace Novelty can produce monthly is given by the production function:

Mathematics
1 answer:
koban [17]3 years ago
3 0

Answer:

step 1:

go online

Step-by-step explanation:

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PLEASE HELP THIS IS DUE BY TOMORROW
Bezzdna [24]

Answer:b

Step-by-step explanation:

4 0
3 years ago
Please help me please !
Morgarella [4.7K]

Hi there!  

»»————- ★ ————-««

I believe your answer is:  

Option C

»»————- ★ ————-««  

Here’s why:

⸻⸻⸻⸻  

\text{\underline{The Slope Formula Is:}}\\\\m=\frac{y_2-y_1}{x_2-x_1}\\\\(x_1,y_1)\text{ and } (x_2,y_2)\text{ are two points given.}\\\\\text{We are given the points: } (3,5) \text{ and } (9,2).\\\\\text{\underline{The formula for the points should be:}}\\\\m=\frac{5-2}{3-9}, \text{where } (9,2) \text{ is }(x_1,y_1)\text{ and } (3,5) \text{ is } (x_2,y_2).

⸻⸻⸻⸻

»»————- ★ ————-««  

Hope this helps you. I apologize if it’s incorrect.  

7 0
3 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
If the length of the cord is l, determine the coordinates
LekaFEV [45]
Yes it determine the length
3 0
3 years ago
Qual o valor de 2 3 -1 3 ?
ra1l [238]
El valor de 23-13 es 10
7 0
3 years ago
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