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ss7ja [257]
3 years ago
6

Of the students attending a school party, 60% of the students are girls, and 40% of the students like to dance. After these stud

ents are joined by 20 more boy students, all of whom like to dance, the party is now 58% girls. How many students now at the party like to dance?
Mathematics
1 answer:
elena-s [515]3 years ago
7 0

Answer:

252 students at the party like to dance

Step-by-step explanation:

Given:

60% of the students are girls

40% of the students like to dance

20 more boy students

Percentage of the girls in the party = 58%

To Find:

How many students now at the party like to dance=?

Solution:

Let the number of girls be g.

Let the number of total people originally be t.

We know that

\frac{g}{t}=\frac{60}{100}

\frac{g}{t}=\frac{3}{5}----------------------------------(1)

We also know that

\frac{g}{t+20}=\frac{58}{100}

\frac{g}{t+20}=\frac{29}{50}-----------------------------(2)

We now have a system of equation from (1) and (2)

3t=5g

g = \frac{3t}{5} ------------------------(3)

50g=29t+580------------------------(4)

Substituting 3 in 4, we get

50(\frac{3t}{5})=29t+580

(\frac{150t}{5})=29t+580

30t =29t+580

30t -29t = 580

t =589

So originally there was 580 students

In this 580 students 40% liked to dance

Then the number of students liked to dance =

=> 40\% \text{of} 580

=>\frac{40}{100} \times 580

=>0.4 \times 580

=> 232

We also know that with these people, 20 boys joined, all of whom like to dance.

Now the number of students like to dance  = 232+20 = 252 students

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Domain: {-4, -2, 0, 2, 4}

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Step-by-step explanation:

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1

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Put Y=x^2+10x+18 in vertex form
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Rewrite in Vertex form and use this form to find the vertex (h,k)

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3 years ago
Read 2 more answers
15 POINTSSS ANSWER ASAP PLEASE!!!!! What is the equation, in point-slope form, of the line that is
Gnesinka [82]

Answer:

The Answer is : D

Step-by-step explanation:

First find the slope of the line that the equation should be parallel to. In this case it is 2/1 which simplified is 2.

Next insert the X (4) of the point (4,1) and solve to see if you get the Y(1).

y-1 = 2 (4-4)

y-1= 2 (0)

y-1= 0

y= 1

In this case D is correct.

TIP*

If you see the question ask you about a parallel formula, look at the slopes of them to see if they match up. Parallel formulas have the same slope, just a tip because you can see in the answer choices none of the equations have the same slope as the line on the graph except for D.

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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!

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