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3 years ago

A group of school children consist of 25 boys and 18 girls. how many ways are there:

Mathematics

1 answer:

Serhud [2]3 years ago

8 0

Answer:

Step-by-step explanation:

1. Number of boys in the group = 25

Number of girls in the group = 18

Total children = 25 + 18 = 43

Number of ways to arrange the children in a way = 43!

2. If we consider all the boys as an individual then number of ways children can be arranged = 19!

Number of ways boys can sit next to each other = 25!

So the number of ways can be arranged = 19!×25!

3. Number of ways boys can sit next to each other = 25!

Number of ways girls can sit next to each other = 19!

Then number of ways to arrange the children in a row with all boys next to each other and all the girls next to each other will be = 2 × 18! × 25!

4. 1. To choose a chess team if anyone can be chosen

= $^{43}C_{6}$

= 6096454

4. 2. Exactly 2 girls must be chosen then number of ways

= $^{18}C_{2}\times ^{25}C_{4}=1935450$

4. 3. At least two boys must be chosen

= $^{25}C_{2}\times ^{18}C_{4}+^{25}C_{3}\times ^{18}C_{3}+^{25}C_{4}\times ^{18}C_{2}+^{25}C_{5}\times ^{18}C_{1}+^{25}C_{6}$

= 5863690

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Please solve the problem with steps

Debora [2.8K]

Answer:

Infinite series equals 4/5

Step-by-step explanation:

Notice that the series can be written as a combination of two geometric series, that can be found independently:

$\frac{3^{n-1}-1}{6^{n-1}} =\frac{3^{n-1}}{6^{n-1}} -\frac{1}{6^{n-1}} =(\frac{1}{2})^{n-1} -\frac{1}{6^{n-1}}$

The first one: $(\frac{1}{2})^{n-1}$ is a geometric sequence of first term ( $a_1$ ) "1" and common ratio (r) " $\frac{1}{2}$ ", so since the common ratio is smaller than one, we can find an answer for the infinite addition of its terms, given by: $Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{2} } =\frac{1}{\frac{1}{2} } =2$

The second one: $\frac{1}{6^{n-1}}$ is a geometric sequence of first term "1", and common ratio (r) " $\frac{1}{6}$ ". Again, since the common ratio is smaller than one, we can find its infinite sum:

$Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{6} } =\frac{1}{\frac{5}{6} } =\frac{6}{5}$

now we simply combine the results making sure we do the indicated difference: Infinite total sum= $2-\frac{6}{5} =\frac{10-6}{5} =\frac{4}{5}$

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3 years ago

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Rewrite standard form equation in graphing form and then sketch the graph

Aleonysh [2.5K]

Türkiye cumhuriyeti ccc

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3 years ago

A cylindrical can without a top is made to contain 25 3 cm of liquid. What are the dimensions of the can that will minimize the

Basile [38]

Answer:

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

Step-by-step explanation:

Given that, the volume of cylindrical can with out top is 25 cm³.

Consider the height of the can be h and radius be r.

The volume of the can is V= $\pi r^2h$

According to the problem,

$\pi r^2 h=25$

$\Rightarrow h=\frac{25}{\pi r^2}$

The surface area of the base of the can is = $\pi r^2$

The metal for the bottom will cost $2.00 per cm²

The metal cost for the base is =$(2.00× $\pi r^2$ )

The lateral surface area of the can is = $2\pi rh$

The metal for the side will cost $1.25 per cm²

The metal cost for the base is =$(1.25× $2\pi rh$ )

$=\$2.5 \pi r h$

Total cost of metal is C= 2.00 $\pi r^2$ + $2.5 \pi r h$

Putting $h=\frac{25}{\pi r^2}$

$\therefore C=2\pi r^2+2.5 \pi r \times \frac{25}{\pi r^2}$

$\Rightarrow C=2\pi r^2+ \frac{62.5}{ r}$

Differentiating with respect to r

$C'=4\pi r- \frac{62.5}{ r^2}$

Again differentiating with respect to r

$C''=4\pi + \frac{125}{ r^3}$

To find the minimize cost, we set C'=0

$4\pi r- \frac{62.5}{ r^2}=0$

$\Rightarrow 4\pi r=\frac{62.5}{ r^2}$

$\Rightarrow r^3=\frac{62.5}{ 4\pi}$

⇒r=1.71

Now,

$\left C''\right|_{x=1.71}=4\pi +\frac{125}{1.71^3}>0$

When r=1.71 cm, the metal cost will be minimum.

Therefore,

$h=\frac{25}{\pi\times 1.71^2}$

⇒h=2.72 cm

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

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a spinner is divided into nine equal sections numbered 1 through 9. predict how many times out of 270 spins the spinner is most

evablogger [386]

I'd say around 150 times

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There were 65 people waiting in two lines to buy a snowball ice drink they were joined in the line by 11 more people a total of

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Answer:

67 people were still left waiting in line.

Step-by-step explanation:

If there are 65 plus 11 more, there'd be 76 people in line.

I guess the fact that it says "waiting in two lines" can be completely discarded because bottom line is 9 people got their drinks- which means that "waiting in two lines" means nothing.

76 - 9 = 67 people still left waiting in line.

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