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quester [9]
3 years ago
14

1/3 ÷ 2 = ? in fractions

Mathematics
2 answers:
4vir4ik [10]3 years ago
7 0
2 is also the fraction 2/1, so the question is:
\frac{1}{3} / \frac{2}{1}
when dividing fractions, you take the second fraction flip the top number (numerator) with the lower number (denominator) and then change the division sign to multiplication:
\frac{1}{3} * \frac{1}{2}
= \frac{1}{6} or one sixth
Jlenok [28]3 years ago
6 0
1/3 ÷ 2 =

1/3 x 2/1. you do this by doing the reciprocal of 2, and then you do KCF ( keep, change, flip,)

keep 1/3 change ÷ to × and flip 2/1 to 1/2.

1/3 x 1/2 = 1/6.

hopefully this helps you
You might be interested in
Algebra identify the pattern. Then write the next three terms in each sequence. 63,58,53,48.....
vovikov84 [41]

Answer:43, 38, 33

Step-by-step explanation:

The pattern is subtracting 5

63 -5 = 58

58 - 5 = 53

53 - 5 = 48

48 - 5 = 43

43 - 5 = 38

38 - 5 = 33

8 0
3 years ago
⅘ of a packet of sweets are fizzy. If there are 40 sweets in the packet, how many are fizzy sweets?
Colt1911 [192]

Answer:

32 fizzies

Step-by-step explanation:

4/5 = .8

.8 x 40 = 32

- I found that 80% of 40 is 32 therefore 32 fizzies are in the packet of sweets.

- Hope this helps! If you need a further explanation please let me know.

4 0
3 years ago
At a competition with 6 runners, 2 medals are awarded for first and second
insens350 [35]

Answer:

30 ways

Step-by-step explanation:

The first medal can be awarded in 6 different ways since any of the runners could have it. Once the first medal is awarded, the second medal can only be given to any of the 5 runners available.

So there are 6x5=30 different ways to award the medals

3 0
3 years ago
At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population pro
Gnom [1K]

Answer:

A sample of 1068 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?

We need a sample of n.

n is found when M = 0.03.

We have no prior estimate of \pi, so we use the worst case scenario, which is \pi = 0.5

Then

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.03\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.03}

(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}

n = 1067.11

Rounding up

A sample of 1068 is needed.

8 0
2 years ago
What number comes next in this pattern 2,4,8,16,32
exis [7]
2+2 = 4
4+4 =8
8+8 = 16
16+16 = 32
32+32 = 64
and so on :)

5 0
3 years ago
Read 2 more answers
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