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Paha777 [63]
4 years ago
15

33/55 in simplest form

Mathematics
1 answer:
Korolek [52]4 years ago
5 0
3/5 or 0.6 should be your answer

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How many numbers of at mose three different digits can be formed from the integer 1,2,3,4,5,6.​
Olin [163]

Answer:

The answer is 1920, right? Here is how-

The numbers must contain at least three digit and can go above that which means the answer has 3 4 5 and 6 digits all together.

For three digits , there will be 6*5*4 = 120

For 4 Digits, There will be 6*5*4*3=360 possible numbers

For 5 digits , there will be 6*5*4*3*2 = 6! = 720

For 6 digits, there will be 6*5*4*3*2*1 = 6! = 720

<Note- the digits are non repeating as the question said it to be “different digits” >

Now add them up = 120+360+2*720 = 1920

Step-by-step explanation:

4 0
3 years ago
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What is the domain and range of f(x)= |x-6|
marissa [1.9K]
Domain: possible x values
Range: possible y values

Solution: domain: all real numbers
Range: y greater than 0
7 0
3 years ago
Can u help me x/4=-2
GenaCL600 [577]
-8/4=-2

x= -8

hope this helps!
3 0
3 years ago
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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
Find the following measure for this figure.
bezimeni [28]
We are asked to find the area of the base or the side with the dimensions 4 and 2. The length would be the longest side and the width is the shorter side. The area is calculated as the product of length and width.

Area of base = 4 x 2 = 8 square units. 

Therefore, the correct answer is B.
3 0
3 years ago
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