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bulgar [2K]
3 years ago
7

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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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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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}

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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}

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

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

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

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