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hram777 [196]
3 years ago
13

Express 0.484848..... in p/q form​

Mathematics
1 answer:
cupoosta [38]3 years ago
7 0
<h3>Answer:  16/33</h3>

It's in p/q form where p = 16 and q = 33.

=======================================================

Work Shown:

x = 0.484848.....

100x = 48.4848.....

I multiplied both sides by 100 to move the decimal over 2 spots. Both decimal values for x and 100x have an infinite string of "48"s repeated after the decimal point. When we subtract, those infinite strings will cancel out

100x - x = 99x

48.4848..... - 0.484848..... = 48

So after subtracting straight down, we have the new equation 99x = 48 which solves to x = 48/99

Divide both parts by the GCF 3 to fully reduce

48/3 = 16

99/3 = 33

Therefore, x = 48/99 = 16/33 = 0.484848...

I recommend using a calculator to confirm that 16/33 = 0.484848...

Side note: your calculator may round the last digit, but this is of course rounding error

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0.5757… as a fraction
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A football team has a probability of .75 of winning when playing any of the other four teams in its conference. If the games are
Alexeev081 [22]

Answer:

0.3164 = 31.64% probability the team wins all its conference games

Step-by-step explanation:

For each conference game, there are only two possible outcomes. Either the team wins it, or they lose. The probability of winning a game is independent of any other game. This means that the binomial probability distribution is used 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.

A football team has a probability of .75 of winning when playing any of the other four teams in its conference.

The probability means that p = 0.75, and four games means that n = 4

If the games are independent, what is the probability the team wins all its conference games?

This is P(X = 4). So

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

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3 years ago
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