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3 years ago

What is 3/4 x 2 2/9 in fraction form

Mathematics

2 answers:

Ivan3 years ago

8 0

Answer:

6/36 as simplified as 1/6 in fraction form

Tems11 [23]3 years ago

4 0

Answer:

$\frac{5}{3}$

Step-by-step explanation:

We have two fractions being multiplied.

$\frac{3}{4} * 2\frac{2}{9}$

Convert 2 2/9 into an improper fraction by multiplying the whole number by the denominator and add the numerator :

2 * 9

18 + 2

$\frac{3}{4} * \frac{20}{9}$

Remember that when multiplying fractions you have to multiply across :

$\frac{3*20}{4*9}$

$\frac{60}{36}$

Divide both sides by 6.

$\frac{5}{3}$

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Help plz show work I need to have all Aś this year Only do 18,20,24,22,28,26,30,32,

Veseljchak [2.6K]

Answer: see below

<u>Step-by-step explanation:</u>

The "point" on the inequality symbol points to the smaller number

< is the "less than" inequality symbol
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18) 3 = 3

20) $\sqrt6 < \sqrt{10}$

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26) 0.075 < 0.39

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6 0

3 years ago

A bag contains three red marbles, two green ones, one lavender one, four yellows, and five orange marbles. HINT [See Example 7.

kirill115 [55]

Answer:

There are 12 sets of four marbles include all the red ones.

Step-by-step explanation:

Given : A bag contains three red marbles, two green ones, one lavender one, four yellows, and five orange marbles.

To find : How many sets of four marbles include all the red ones?

Solution :

Number of red marbles = 3

Number of green marbles = 2

Number of lavender marbles = 1

Number of yellow marbles = 4

Number of orange marbles = 5

We have to form sets of four marbles include all the red ones,

For position of getting red ones we have three red marbles i.e. $^3C_3$

For the fourth one we have 12 choices i.e. $^{12}C_1$

Total sets of four marbles include all the red ones is

$=^3C_3\times ^{12}C_1$

$=1\times 12$

$=12$

Therefore, There are 12 sets of four marbles include all the red ones.

5 0

3 years ago

Suppose that 50% of all young adults prefer McDonald's to Burger King when asked to state a preference. A group of 12 young adul

ddd [48]

Answer:

a) 0.194 = 19.4% probability that more than 7 preferred McDonald's

b) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

c) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they prefer McDonalds, or they prefer burger king. The probability of an adult prefering McDonalds is independent from other adults. 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.

50% of all young adults prefer McDonald's to Burger King when asked to state a preference.

This means that $p = 0.5$

12 young adults were randomly selected

This means that $n = 12$

(a) What is the probability that more than 7 preferred McDonald's?

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

In which

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.121$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.054$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.016$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.003$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.000$

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.121 + 0.054 + 0.016 + 0.003 + 0.000 = 0.194$

0.194 = 19.4% probability that more than 7 preferred McDonald's

(b) What is the probability that between 3 and 7 (inclusive) preferred McDonald's?

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

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

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.054$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.121$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.193$

$P(X = 6) = C_{12,6}.(0.5)^{6}.(0.5)^{6} = 0.226$

$P(X = 7) = C_{12,7}.(0.5)^{7}.(0.5)^{5} = 0.193$

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.054 + 0.121 + 0.193 + 0.226 + 0.193 = 0.787$

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

(c) What is the probability that between 3 and 7 (inclusive) preferred Burger King?

Since $p = 1-p = 0.5$ , this is the same as b) above.

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

7 0

3 years ago

How do i solve this proportion? 3t/10 = 1/2

never [62]

Answer:

Multiply both sides of the equation

Step-by-step explanation:

5 0

3 years ago

he table shows the mean number of basketball goals made by four random samples of players from the school team during this year’

Strike441 [17]

The valid prediction for the mean of the population possible using these samples is C. Yes, the variation of the sample means is small

<h3>What is a sample mean?</h3>

It should be noted that a sample mean is an average of a set of data . Also, the sample mean can be used to calculate the standard deviation, central tendency, and the variance of a data set.

The sample mean can be applied to a variety of uses, including calculating population averages.

Here, the sample variance is a measure of the degree to which the numbers in a list are spread out. Here, the numbers given are close to each other.

If the numbers in a list are all close to the expected values, the variance will be small and if they are far away, the variance will be large.

Therefore, the correct option is C.

Learn more about mean on:

brainly.com/question/1136789

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3 0

2 years ago