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olga2289 [7]
3 years ago
6

What is subtracting fractions from mixed numbers for grade 4

Mathematics
1 answer:
Svetach [21]3 years ago
6 0

Answer:

see below

Step-by-step explanation:

So when you have mixed numbers, you have to turn them into regular fractions. How you change them into regular fractions is multiplying the whole number by the denominator (the bottom number in the fraction) and then adding that number to the numerator (the top number in the fraction). So, for example you have

3\frac{2}{3}

3 x 3 = 9

9 + 2 = 11

so, 3\frac{2}{3} = \frac{11}{3}

After you change the mixed fractions that you have into normal fractions, then you can normally subtract the two fractions. If they have the same denominator, then you just subtract the numerators like normal numbers and leave the denominators alone.

\frac{11}{3} - \frac{2}{3} = \frac{9}{3}

If there are different denominators, then you have to multiply one or more fractions by any number so that both fractions have the same denominator.

For example,

\frac{11}{3} - \frac{7}{6}

you have to multiply \frac{11}{3} by two, so that the denominator will equal six. This is the only way that you will be able to add or subtract the two fractions.

2(\frac{11}{3}) = \frac{22}{6}

\frac{22}{6} - \frac{7}{6} = \frac{22-7}{6} = \frac{15}{6}

Then, if you are asked to, this fraction can be simplified if you divide both the denominator and the numerator by \frac{3}{3}, so

\frac{15}{6} divided by \frac{3}{3} is equal to \frac{5}{2}.

This works because \frac{3}{3} = 1.

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Answer:

8.6% probability that none of the households are tuned to Lindsay and Tobias.

There is a 91.4% probability that at least one household is tuned to Lindsay and Tobias.

There is a 32.2% probability that at most one household is tuned to Lindsay and Tobias.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they are watching the TV show, or they are not. This means that we use the binomial probability distribution to solve this problem.

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 combinatios 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.

In this problem we have that:

n = 11, p = 0.20

Find the probability that none of the households are tuned to Lindsay and Tobias.

This is P(X = 0).

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

P(X = 0) = C_{11,0}.(0.20)^{0}.(0.80)^{11} = 0.086

8.6% probability that none of the households are tuned to Lindsay and Tobias.

Find the probability that at least one household is tuned to Lindsay and Tobias.

Either no household is tuned, or at least one is tuned. The sum of the probabilities of these events is decimal 1. So

P(X = 0) + P(X \geq 1) = 1

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.086 = 0.914

There is a 91.4% probability that at least one household is tuned to Lindsay and Tobias.

Find the probability that at most one household is tuned to Lindsay and Tobias.

This is

P(X \leq 1) = P(X = 0) + P(X = 1)

P(X = 1) = C_{11,1}.(0.20)^{1}.(0.80)^{10} = 0.236

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.086 + 0.236 = 0.322

There is a 32.2% probability that at most one household is tuned to Lindsay and Tobias.

8 0
3 years ago
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