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gregori [183]
3 years ago
13

Evaluate -8w + 3z when w = 2.2 and z = -9.1.

Mathematics
2 answers:
atroni [7]3 years ago
8 0
-8w+3z
w=2.2
z=-9.1

-8(2.2)+3(-9.1)
-17.6-27.3
-44.9

The answer is -44.9
kogti [31]3 years ago
7 0
-8w + 3z when w = 2.2, z = -9.1

Simply plug in everything.

Plug in 2.2 for w, and -9.1 for z :)

-8(2.2) + 3(-9.1)

Simplify.

-17.6 + (-27.3)

-17.6 - 27.3

Simplify.

-44.9

~Hope I helped!~
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Answer:

y=2x+3 is non-proportional. y=5x is proportional.

Step-by-step explanation:

y=2x+3 is not proportional because there is the +3. also, when it is graphed on a graph, the line does not pass through the origin/zero.

y=5x is proportional because it does not have any other values attached to the y value. when it is graphed, the lie goes through the origin/zero.

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3 years ago
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Phantasy [73]

The ratio of males to females is 1.4 : 1, so the fraction of the total that is made up of males is 1.4/(1.4 + 1) = 1.4/2.4 = 7/12.


There were (7/12)*475 ≈ 277 males at the party.

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7 0
3 years ago
You want to survey seventh-grade students about computer use. Which sample is more likely to be random? You ask seventh-graders
vova2212 [387]

Answer:

You ask seventh-graders leaving the cafeteria after lunch.

Step-by-step explanation:

Try and choose a sample with the student group that has nothing to do with what you're testing for. It will take a bit of "creative" thinking and guessing about the lives of students in each of these groups. We try to choose a good sample to get accurate or less-biased results.

<u>You ask seventh-graders entering a library on Friday night. </u>

Friday night, some students are quicker to leave school and start the weekend. The students who go to the library might be more studious and work can be done on the computer. Libraries also have computers available for people to use for gaming. <em>Your sample would have students who use the computer more.</em>

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<u>You ask seventh-graders leaving a school basketball game. </u>

Students who watch a basketball game usually do so by choice. We could assume that these students spend most of their free time playing sports, which are not done on the computer. <em>Your sample would contain students who use a computer less.</em>

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7 0
3 years ago
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4 0
3 years ago
Read 2 more answers
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
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