Please help me with this math problem!! NO LINKS PLEASE!! Will give brainliest if correct!! :)

How are percent diagrams used to solve real-world problems

fgiga [73]

One fourth of the students in Mrs. Singh’s music class chose a guitar as their favorite musical instrument. There are 24 students in Mrs. Singh’s music class. How many students chose a guitar as their favorite musical instrument? There are students in Mrs. Singh’s music class. Divide the number of students equally into 4 sections. Fill in the number in each section students % ---- %--- So, students chose guitar as their favorite musical instrument.

3 0

3 years ago

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Find out the number of combinations and the number of permutations for 8 objects taken 6 at a time. Express your answer in exact

umka2103 [35]

Solution:

The permutation formula is expressed as

$\begin{gathered} P^n_r=\frac{n!}{(n-r)!} \\ \end{gathered}$

The combination formula is expressed as

$\begin{gathered} C^n_r=\frac{n!}{(n-r)!r!} \\ \\ \end{gathered}$

where

$\begin{gathered} n\Rightarrow total\text{ number of objects} \\ r\Rightarrow number\text{ of object selected} \end{gathered}$

Given that 6 objects are taken at a time from 8, this implies that

$\begin{gathered} n=8 \\ r=6 \end{gathered}$

Thus,

Number of permuations:

$\begin{gathered} P^8_6=\frac{8!}{(8-6)!} \\ =\frac{8!}{2!}=\frac{8\times7\times6\times5\times4\times3\times2!}{2!} \\ 2!\text{ cancel out, thus we have} \\ \begin{equation*} 8\times7\times6\times5\times4\times3 \end{equation*} \\ \Rightarrow P_6^8=20160 \end{gathered}$

Number of combinations:

$\begin{gathered} C^8_6=\frac{8!}{(8-6)!6!} \\ =\frac{8!}{2!\times6!}=\frac{8\times7\times6!}{6!\times2\times1} \\ 6!\text{ cancel out, thus we have} \\ \frac{8\times7}{2} \\ \Rightarrow C_6^8=28 \end{gathered}$

Hence, there are 28 combinations and 20160 permutations.

7 0

1 year ago

Dont get it wrong, if its wrong i fail, explain how you got your answer

damaskus [11]

Answer: 50.24

Area of a circle is pi times r^2

The r is 1/2 the d

So 8/2 = 4

Pi times 4^2= 50.24

7 0

3 years ago

PLSS HELP ME WITH THIS! 20 POINTS

Svetradugi [14.3K]

Answer:

C)

Step-by-step explanation:

Every $x$ is 6 times bigger than $y$

3 0

2 years ago

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A school has 1800 students and 1800 light bulbs, each with a pull cord and all in a row. All the lights start out off. The first

Gnom [1K]

Solution-

A school has 1800 students and 1800 light bulbs, each with a pull cord and all in a row.

As all the lights start out off, in the first pass all bulbs will be turned on.

In the second pass all the multiples of 2 will be off and rest will be turned on.

In the third pass all the multiples of 3 will be off, but the common multiple of 2 and 3 will be on along with the rest. i.e all the multiples of 6 will be turned on along with the rest.

In the fourth pass 4th light bulb will be turned on and so does all the multiples of 4.

But, in the sixth pass the 6th light bulb will be turned off as it was on after the third pass.

This pattern can observed that when a number has odd number of factors then only it can stay on till the last pass.

1 = 1

2 = 1, 2

3 = 1, 3

4 = 1, 2, 4

5 = 1, 5

6 = 1, 2, 3, 6

7 = 1, 7

8 = 1, 2, 4, 8

9 = 1, 3, 9

10 = 1, 2, 5, 10

11 = 1, 11

12 = 1, 2, 3, 4, 6, 12

13 = 1, 13

14 = 1, 2, 7, 14

15 = 1, 3, 5, 15

16 = 1, 2, 4, 8, 16

so on.....

The numbers who have odd number of factors are the perfect squares.

So calculating the number of perfect squares upto 1800 will give the number of light bulbs that will stay on.

As, $\sqrt{1800} =42.42$ , so 42 perfect squared numbers are there which are less than 1800.

∴ 42 light bulbs will end up in the on position. And there position is given in the attached table.

7 0

3 years ago