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3 years ago

(4 tens 3 ones) x 10

1 answer:

6 0

To solve this problem, we must first convert all of the numbers represented by words into their numeral equivalents so we can simplify. To do this, we must remember that the ones place is the first to the left of the decimal point, and the tens place is one place farther to the left. If we have 4 tens and 3 ones, this is the same as 43 (we put the digit 4 in the tens place and the digit 3 in the ones place). If we replace our words with our new numerical value, we get:

43 * 10

To solve this, we just multiply these two numbers together, which gives us:

430

Therefore, your answer is 430.

Hope this helps!

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3 years ago

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3 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