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r-ruslan [8.4K]
3 years ago
12

490 as a product of its prime factors

Mathematics
1 answer:
Katena32 [7]3 years ago
8 0

Answer:

2×5×7²

Step-by-step explanation:

You will start by using the factor tree.

490

^

2 245

^

5 49

^

7 7

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What does it mean to have a skewed distribution? What causes
IrinaK [193]

Answer:

Explained below.

Step-by-step explanation:

A) A skewed distribution in a dataset is when the median is not equal to the mean in such a manner that the bell curve is tilted to the left or right.

B) If in a data set, there are outliers which are extremely large or extremely small in comparison to other values in that same dataset, then we can say that such a curve will be pulled towards the outlier and thus the distribution is skewed.

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4 0
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
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