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bekas [8.4K]
2 years ago
13

How would you solve 3/4 over 1/2 as a ratio

Mathematics
1 answer:
miskamm [114]2 years ago
6 0

Answer:

  3/2

Step-by-step explanation:

A couple of different methods are used for dividing fractions.

1. "Invert and multiply". Dividing by a number is the same as multiplying by its reciprocal:

  (3/4) / (1/2) = (3/4) × (2/1)

  = (3·2)/(4·1) = 6/4 = 3/2

__

2. Make the denominators the same and use the ratio of numerators.

  (3/4) / (1/2) = (3/4) / (2/4) = 3/2

_____

For these fractions, you can recognize that 3/4 is 1/4 more than 1/2, and that 1/4 is half of 1/2. That means 3/4 is half-again as much as 1/2, so is 1 1/2 = 3/2 times 1/2. This tells you the ratio (3/4) : (1/2) = 3/2.

__

Along the lines of the above, you can write the ratio as ...

  (3/4) : (1/2)

and multiply by 4 to get

  = (3/4)×4 : (1/2)×4 = 3 : 2

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Help Please!!!!!!!!!!!<br><br> Questions 9 and 10
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Answer:

Step-by-step explanation:

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1. ∠ACB ≅∠ECD ; vertical angles are congruent (A)

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2 years ago
245. 202, 254, 245, 276<br>Write the values from least to<br>greatest<br>Mean<br>Median:<br>Mode:​
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Answer:

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The median would be 245

The mode would be 245

Step-by-step explanation:

To find the mean, you add all the numbers together and divide that sum by the number of numbers. In this case, you add all the numbers together and you get 1222. Next, you divide that by 5 because there are 5 numbers. To find the median, you order the numbers from least to greatest and find the middle number. For this question, the answer would be 245. Lastly, to find the mode you look for the number that appears the most. In this case it's 245.

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3 years ago
Sample spaces For each of the following, list the sample space and tell whether you think the events are equally likely:
Schach [20]

Answer and explanation:

To find : List the sample space and tell whether you think the events are equally likely ?

Solution :

a) Toss 2 coins; record the order of heads and tails.

Let H is getting head and t is getting tail.

When two coins are tossed the sample space is {HH,HT,TH,TT}.

Total number of outcome = 4

As the outcome HT is different from TH. Each outcome is unique.

Events are equally likely since their probabilities \frac{1}{4} are same.

b) A family has 3 children; record the number of boys.

Let B denote boy and G denote girl.

If there are 3 children then the sample space is

{GGG,GGB,GBG,BGG,BBG,GBB,BGB,BBB}

The possible number of boys are 0,1,2 and 3.

Number of boys      Favorable outcome    Probability

           0                      GGG                        \frac{1}{8}

           1                    GGB,GBG,BGG          \frac{3}{8}

           2                   GBB,BGB,BBG           \frac{3}{8}

           3                       BBB                         \frac{1}{8}

Since the probabilities are not equal the events are not equally likely.

c)  Flip a coin until you get a head or 3 consecutive tails; record each flip.

Getting a head in a trial is dependent on the previous toss.

Similarly getting 3 consecutive tails also dependent on previous toss.

Hence, the probabilities cannot be equal and events cannot be equally likely.

d) Roll two dice; record the larger number

The sample space of rolling two dice is

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Now we form a table that the number of time each number occurs as maximum number then we find probability,

Highest number        Number of times         Probability

           1                                   1                     \frac{1}{36}

           2                                  3                    \frac{3}{36}

           3                                  5                    \frac{5}{36}

           4                                  7                    \frac{7}{36}

           5                                  9                    \frac{9}{36}

           6                                  11                    \frac{11}{36}

Since the probabilities are not the same the events are not equally likely.

4 0
3 years ago
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