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3 years ago

The yearbook club had a meeting. The meeting had 12 people, which is two-thirds of the club. How many people are in the club?

Mathematics

1 answer:

finlep [7]3 years ago

4 0

Answer:

Step-by-step explanation:

2/3 of the club is 12 people

let c represent the club

2/3c = 12

c = 12 * 3/2

c = 36/2

c = 18 <=== the club has 18 people

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A population of endangered birds decreases by 3% each year.

denis-greek [22]

10,000 of the endangered birds would be left

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3 years ago

A company is choosing between two tanks for mixing chemicals. Both tanks have the same price so the company plans to choose the

Nataliya [291]

Find volume
volumecone=(1/3)(h)(pi)(r^2)

hemisphere is just a half sphere aka, 1/2 times volume of sphere so
hemisphere=(1/2)(4/3)(pi)(r^3)

fnd their volumes don't evaluate the true value till end, leav pi as pi
(or use pi key for graphing or scientific calculators)

cone
d/2=r
10/2=r=5
h=15
V=(1/3)(15)(pi)(5^2)
V=(5)(pi)(25)
V=125pi

hemisphere
V=(1/2)(4/3)(pi)(5.5^3)
V=(4/6)(pi)(166.375)
V=110.91666666666666666pi

which is bigger
125pi or 110.916666666666666666pi
125 pi
by how much?
125pi-110.91666666666666pi=14.08333333333pi
aprox pi=3.14
14.08333333*3.14=44.2217
rond
44

answer is CONICAL will hold 44 cubic feet more

7 0

3 years ago

13. How many distinct 4-letter groupings can be made with the letters from the word champion if letters may not be repeated?

ddd [48]

Answer:

N = 1680

Therefore, 1680 '4 letter groupings' can be formed from the word 'champion'

Step-by-step explanation:

Given the word 'champion' which is an 8 distinct Letter word. The number of 4 letter groupings that could be formed from it can be given by the permutation since in this case order of letters are important.

The number of distinct 4 letter groupings can be given as;

N = nPr = 8!/(8-4)!, where n= 8 and r = 4

N = 8P4 = 8!/(8-4)!

N= 8!/4!

N = 1680

Therefore, 1680 '4 letter groupings' can be formed from the word 'champion'

7 0

4 years ago

Consider all 5 letter "words" made from the full English alphabet. (a) How many of these words are there total? (b) How many of

VARVARA [1.3K]

Answer:

a) There are 11,881,336 of these words in total.

b) There are 7,893,600 of these words with no repeated letters.

c) 896,376 of these words start with an a or end with a z or both

Step-by-step explanation:

Our words have the following format:

L1 - L2 - L3 - L4 - L5

In which L1 is the first letter, L2 the second letter, etc...

There are 26 letters in the English alphabet.

(a) How many of these words are there total?

Each of L1, L2, L3, L4 and L5 have 26 possible options.

So there are $26^{5} = 11,881,336$ of these words total

(b) How many of these words contain no repeated letters?

The first letter can be any of them, so L1 = 26.

At the second letter, the first one cannot be repeated, so L2 = L1 - 1 = 25.

At the third letter, nor the first nor the second one can be repeated, so L3 = L1 - 2 = 24

This logic applies until L5

So we have

26-25-24-23-22

In total there are

$26*25*24*23*22 = 7,893,600$

of these words with no repeated letters.

(c) How many of these words start with an a or end with a z or both (repeated letters are allowed)?

$T = T_{1} + T_{2} + T_{3}$

$T_{1}$ is the number of words that start with an a and do not end with z. So we have

1 - 26 - 26 - 26 - 25

The first letter can only be a, and the last one cannot be z. So:

$T_{1} = 26^{3}*25 = 439,400$

$T_{2}$ is the number of words that start with any letter other than a and end with z. So we have

25 - 26 - 26 - 25 - 1

The first letter can be any of them, other than a, and the last can only be z. So:

$T_{2} = 26^{3}*25 = 439,400$

$T_{3}$ is the number of words that both start with a and end with z. So:

1 - 26 - 26 - 26 - 1

The first letter can only be a, and the last can only be z. The other three letters could be anything. So:

$T_{3} = 26^{3} = 17,576$

$T = T_{1} + T_{2} + T_{3} = 2*439,400 + 17,576 = 896,376$

896,376 of these words start with an a or end with a z or both

4 0

3 years ago

Solve the following equation; domain 0< x < 360; 3 cot x + sqrt 3 = 0

kondaur [170]

ANSWER

$x =120 \degree \: or \: 300 \degree$

EXPLANATION

We want to solve the trigonometric equation

$3 \cot(x) + \sqrt{3} = 0$

We group like terms to obtain,

$3\cot(x) = - \sqrt{3}$

Divide both sides of the equation by 3 to get,

$\cot(x) = - \frac{ \sqrt{3} }{3}$

We can reciprocate both sides to get,

$\tan(x) = - \frac{3}{ \sqrt{3} }$

We simplify the right hand side to get,

$\tan(x) = - \sqrt{3}$

Since the tangent ratio is negative, it implies that it is either in the second quadrant or fourth quadrant.

In the second quadrant,

$x = 180 \degree - arctan( \sqrt{3})$

$x = 180 \degree - 60 \degree = 120 \degree$

In the fourth quadrant,

$x = 360 \degree - arctan( \sqrt{3})$

$x = 360 \degree - 60 \degree$

$x = 300 \degree$

6 0

4 years ago