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gogolik [260]
3 years ago
7

Is it true that the 6 in 26.495 is 10 times the value of the 6 in 17.64

Mathematics
2 answers:
Andreyy893 years ago
8 0
Yes because if you multiply 0.6 by 10, it would equal 6. So yes, the 6 in 26.495 IS 10 times the value of the 6 in 17.64.
andrezito [222]3 years ago
7 0
Yup!  The 6 in 26.495 is in the ones place, whereas the 6 in 17.64 is in the tenths place.  One is equal to ten tenths, so the ones place has ten times the value.  Thus, the statement is true.
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Let A={1'3'5'6}<br> B={0'7}<br> check whether A and B are disjoint sets
Setler79 [48]

Given:

The sets are A=\{1,3,5,6\} and B=\{0,7\}.

To find:

Whether A and B are disjoint sets or not.

Solution:

Two sets are called disjoint sets if there are no common elements between them.

We have,

A=\{1,3,5,6\}

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3 years ago
The Blair family was one of the first to come to the original 13 colonies (now part of the USA). They had 4 children. Assuming t
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Answer:

31.25% probability that the Blair family had at least 3 girls

Step-by-step explanation:

For each children, there are only two possible outcomes. Either it was a girl, or it was not. The probability of a child being a girl is independent of any other children. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

They had 4 children.

This means that n = 4

The probability of a child being a girl is 0.5

This means that p = 0.5

Probability of at least 3 children:

P(X \geq 3) = P(X = 3) + P(X = 4)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{4,3}.(0.5)^{3}.(0.5)^{1} = 0.25

P(X = 4) = C_{4,4}.(0.5)^{4}.(0.5)^{0} = 0.0625

P(X \geq 3) = P(X = 3) + P(X = 4) = 0.25 + 0.0625 = 0.3125

31.25% probability that the Blair family had at least 3 girls

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