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2 years ago

Which ratio is equivalent to the ratio 28:48

Mathematics

2 answers:

marshall27 [118]2 years ago

6 0

Answer:

7:12

It is 7:12 because we divide by 4.

28 ÷ 4 = 7.

48 ÷ 4 = 12.

Now, you have a final answer 7:12.

Lorico [155]2 years ago

5 0

Answer:

7:12

Step-by-step explanation:

Divide each side by four.

Answer is 7:12

Hope this helps! :) Let me know if it does.

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We have to calculate the difference of the given polynomials, we follows as:

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After opening the brackets, the signs of all the terms changes as there is negative sign before the bracket.

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Combining all the like terms, we get as

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1) A family consisting of three persons—A, B, and C—goes to a medical clinic that always has a doctor at each of stations 1, 2,

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Answer:

Step-by-step explanation:

Let us record the station number 1, 2 or 3 for each family member A, B or C.

I am attaching a table containing total outcomes. Outcomes are presented along rows while the assigned station to each member is written along columns. For ease of understanding, 1 3 2 in the table should be interpreted as family member A being assigned to station 1, member B to station 3 and member C to station number 2, respectively.

From table it is clear that the total outcomes possible are 27.

We know that, probability can be defined as,

$PROBABITILY = \frac{NUMBER\;OF\;DESIRED\;OUTCOMES}{TOTAL\;NUMBER\;OF\;OUTCOMES}$

a) All Members Assigned to the Same Station.

Cases for all members being assigned to same station are as follows:

[1 1 1], [2 2 2], [3 3 3] (outcome number 1, 14 and 27 in the table).

Therefore,

$PROBABILITY\;(Case\;a) = \frac{3}{27}\\\\PROBABILITY\;(Case\;a) = 0.111$

b) At Most Two Members Assigned to the Same Station.

It means that maximum of 2 members can have the same station. Cases for this situation are as follows:

[1 1 2], [1 1 3], [1 2 1], [1 2 2], [1 3 1], [1 3 3], [2 1 1], [2 1 2], [2 2 1], [2 2 3], [2 3 2],

[2 3 3], [3 1 1], [3 1 3], [3 2 2], [3 2 3], [3 3 1], [3 3 2]

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$PROBABILITY\;(Case\;b) = \frac{18}{27}\\\\PROBABILITY\;(Case\;b) = 0.666$

c) All Members Assigned to a Different Station.

For this scenario, we have the following results:

[1 2 3], [1 3 2], [2 1 3], [2 3 1], [3 1 2], [3 2 1] (outcome number 6, 8, 12, 16, 20 and 22 in the table).

Therefore,

$PROBABILITY\;(Case\;c) = \frac{6}{27}\\\\PROBABILITY\;(Case\;c) = 0.222$

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