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SVETLANKA909090 [29]
3 years ago
7

Which graph represents the solution set of the inequality Negative 9 greater-than x?

Mathematics
2 answers:
Kamila [148]3 years ago
7 0

Answer:

de answer is A

Step-by-step explanation:

hope i helped

ICE Princess25 [194]3 years ago
5 0

Answer:

a

Step-by-step explanation:

just took the unit test review and got all the questions right hope it helped

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Find the value of x in the following figure
tiny-mole [99]

Answer:

8

Step-by-step explanation:

3 0
3 years ago
How do I solve for x?
Marina CMI [18]
Multiply both sides by 2
7 0
2 years ago
Which of the following are important properties of the arithmetic mean? Check all that apply. Multiple select question. The mean
vovikov84 [41]

Answer:

All of the values in the data are used in calculating the mean.

The sum of the deviations is zero.

There is only one mean for a set of data.

Step-by-step explanation:

Required

True statement about arithmetic mean

(a) False

The mean can be equal to, greater than or less than the median

(b) True

The arithmetic mean is the summation of all data divided by the number of data; hence, all values are included.

(c) True

All mean literally represent the distance of each value from the average; so,  when each value used in calculating the mean is subtracted from the calculated mean, then the end result is 0. i.e.\sum(x - \bar x) = 0

(d) True

The mean value of a distribution is always 1 value. When more values are added to the existing values or some values are removed from the existing values, the mean value will change.

(e) False

Nominal data are not numerical or quantitative data; hence, the mean cannot be calculated.

3 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
you scored 72% on the Unit 1 exam and then 90% on the unit 2 exam. what was your percentage acorr increase
iVinArrow [24]
The difference between 90% and 72% is 18%
6 0
3 years ago
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