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Semmy [17]
2 years ago
9

How do we convert decimal numbers into binary numbers 2483?.

Mathematics
1 answer:
alisha [4.7K]2 years ago
8 0

Answer:

Places in binary coding

1 - 2 - 4 - 8 - 16 - 32 - 64 - 128 - 256 - 512 - 1024 - 2048

2483 - 2048 = 435      there is 1 in the 2048 position

435 - 256 = 179    there is 1 in the 256 position

179 - 128 = 51        there is 1 in the 128 position

51 - 32 = 19         there is 1 in the 32 position

19 - 16 = 3           there is 1 in the 16 position

3 - 2 = 1            there is 1 in the 2 position

1 - 1 = 0             there is 1 in the 1 position

10011011 0011     would be the binary result

Check:

2048 + 0 + 0 + 256 + 128 + 0 + 32 + 16 + 0 + 0 + 2 + 1 = 2483

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It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken if the desir
Ksju [112]

Answer:

We need a sample size of at least 75.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, we find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

The standard deviation is the square root of the variance. So:

\sigma = \sqrt{484} = 22

With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is

We need a sample size of at least n, in which n is found when M = 5. So

M = z*\frac{\sigma}{\sqrt{n}}

5 = 1.96*\frac{22}{\sqrt{n}}

5\sqrt{n} = 43.12

\sqrt{n} = \frac{43.12}{5}

\sqrt{n} = 8.624

(\sqrt{n})^{2} = (8.624)^{2}

n = 74.4

We need a sample size of at least 75.

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