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3 years ago

How many kilobytes are in in 1 terabyte if a terabyte is 2^40 and kilobyte is 2^10

Mathematics

1 answer:

Tresset [83]3 years ago

8 0

Answer:

There are $2^{30}$ kilobytes in a Terabyte

Step-by-step explanation:

If one Terabyte is $2^{40}$ bytes and one kilobyte is $2^{10}$ bytes, then, by simple quotient we can find how many kilobytes are in one Terabyte:

$\frac{2^{40}}{2^{10}} = 2^{30}$

Notice that in the numerator we have 40 factors of "2", while in the denominator we have 10 factors of "2". So we can cancel just 10 common in denominator and numerator and end up with 30 factors of "2" left in the numerator.

On can also use the "properties of exponents" to simplify exponents with the same base.

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The coordinates of the endpoints of AB and CD are A(2,

Ratling [72]

Answer:

Option 1: CD is a perpendicular bisector of AB

Step-by-step explanation:

Let us find out the slopes of various line segments and the Distances and then we will draw the conclusions accordingly.

Formula to find slope

$m= \frac{y_2-y_1}{x_2-x_1}$

Formula to Find Distance between two points

$D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

mAB ( represents , Slope of AB )

1. $mAC= \frac{3-2}{2-5}=\frac{1}{-3}=-\frac{1}{3}$

2. $mBC=\frac{2-1}{5-8}=\frac{1}{-3}=-\frac{1}{3}$

3. $mCD=\frac{5-2}{6-5}=\frac{3}{1}=3$

4. $AC=\sqrt{(3-2)^2+(2-5)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

5. $BC=\sqrt{(2-1)^2+(5-8)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

mAC = mBC , and C is common point , hence these three are collinear points making a straight line whole slope is $-\frac{1}{3}$

$mAB=-\frac{1}{3}$

$mCD=3$

$mAB \times mCD = -\frac{1}{3} \times 3 = -1$

Hence CD ⊥ AB

Also

From Point 4 and point 5 above , we see that

AC = CB

Hence CD bisect AB at C, also CD ⊥ AB

There fore

CD is a perpendicular bisector of AB

Therefor option 1 is true

4 0

3 years ago

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc

Bingel [31]

Answer:

We need a sample size of at least 719

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, 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.

How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when $M = 0.3, \sigma = 4.103$ . So

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

$0.3 = 1.96*\frac{4.103}{\sqrt{n}}$

$0.3\sqrt{n} = 1.96*4.103$

$\sqrt{n} = \frac{1.96*4.103}{0.3}$

$(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}$

$n = 718.57$

Rouding up

We need a sample size of at least 719

6 0

3 years ago

Appreciate answers and any help!

Savatey [412]

To find the expected value of the distribution, we multiply each outcome by it's probability. Doing this, we get that the expected value of defects on a skateboard is of $\frac{4}{25}$ .

Outcomes and probabilities:

0 defects, 9/10 probability

1 defect, 1/20 probability

2 defects, 1/25 probability

3 defects, 1/100 probability.

Expected value:

$E(X) = 0\frac{9}{10} + \frac{1}{20} + 2\frac{1}{25} + 3\frac{1}{100} = \frac{1}{20} + \frac{2}{25} + \frac{3}{100} = \frac{5 + 8 + 3}{100} = \frac{16}{100}$