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Law Incorporation [45]
3 years ago
9

When you are doing math homework do u need to show your work for every problem?

Mathematics
2 answers:
Alex17521 [72]3 years ago
6 0

Yes, it is very important otherwise some teachers may count the problem wrong.


Y_Kistochka [10]3 years ago
3 0
If the question is only fill in the blank kind, then, no.
But for the word problems and calculation questions, you usually need to show your work. hope that helps :)
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Plsss Help me plsssssssssssss
Mazyrski [523]

Answer:

If, a = 2 and b = 3

<h3>Jada</h3><h3>10a + 1b = 7a + 5b  - 3a + 4b \\ 10(2) + 1(3) = 7(2) + 5(3) - 3(2) + 4(3) \\ 20 + 3 = 14 + 15 - 6 + 12 \\ 23 = 35 \\</h3>

It's not equivalent.

<h3>Diego</h3><h3>4a + 9b = 7a + 5b - 3a + 4b \\ 4(2) + 9(3) = 7(2) + 5(3) - 3(2) + 4(3) \\ 8 + 27 = 14 + 15 - 6 + 12 \\ 35 = 35 \\</h3>

It's equivalent.

The correct equation is 4a + 9b.

5 0
3 years ago
Explain what is the better buy. 1 litre of milk for 75p or 2 pints of milk for 75p. make sure to explain why.
Soloha48 [4]

Answer:

1 litre of milk for 75p

Step-by-step explanation:

1 pint ≈ 473 ml

2 pints = 2×473ml =  946 ml

1 l = 1000 ml

7 0
3 years ago
Read 2 more answers
Which of the following equations describes the graph? Urgent lol
o-na [289]

Answer : I think 3rd is the answer

8 0
2 years ago
student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
Which two ratios represent qauntities that are proportional
Leokris [45]
Are there answer choices ?
6 0
3 years ago
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