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3 years ago

How many nanoseconds does it take for a computer to perform one calculation if it performs 6.7 × 10^{7} calculations per second?

Mathematics

1 answer:

crimeas [40]3 years ago

8 0

Answer:

the comuter does 1 calculation in 1.5* $10^{-17}$ nanosec

Step-by-step explanation:

First step: we need to write all terms in nanosec, as we know:

1[nanosec]= $10^{-9}$ [sec]

The computer does 6.7* $10^{7}$ [ $\frac{calculation}{sec}$ ]

if we multiplicate for $\frac{sec}{nanosec*10^{*9}}$ , from the units convertion.

6.7* $10^{7}$ [ $\frac{calculation}{sec}$ * $\frac{sec}{nanosec*10^{*9}}$

6.7* $10^{16}$ [ $\frac{calculation}{nanosec}$ ]

Now doing a simple three rule we have this:

6.7 * $10^{16}$ calculation in ⇒ 1 nanosec

1 calculation in ⇒ $\frac{1 nanosec* 1calculation}{6.7*10^{16} calculation }$

So, we found: the comuter does 1 calculation in 1.5* $10^{-17}$ nanosec

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Rosario and Enrqiue are in the same mathematics

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Answer:Enrqiue's average was higher than Rosario's average by 3

Step-by-step explanation:

Rosario's average= sum of 78, 77, 64, 86, and 70/ no of scores= 375/5= 75

Enrique's average=sum of 90, 61, 79, 73, and 87/ no of scores = 390/5=78

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4 years ago

Math-Limits and Derivative. Could anybody help me,please ?

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Step-by-step explanation:

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3 years ago

Katie starts a business selling lemonade.On monday she sells 4 cups.On tuesdays she sells 8 cups. On wensdays she sells 12 cups.

trapecia [35]

She will sell 18 cups on Friday

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4 years ago

Read 2 more answers

For this problem, for finding the slope, I need help with this.

Elena-2011 [213]

Answer: the slope is 1

Step-by-step explanation:

Slope = Rise over Run. Rise= y run= x

For this problem the rise is the difference between 3 and 1 and the run is the difference between 0 and -2. The difference between 3 and 1 is 2 and the difference between 0 and -2 is also too. Which makes it 2/2 which simplifies to 1.

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3 years ago

In constructing a 95 percent confidence interval, if you increase n to 4n, the width of your confidence interval will (assuming

Damm [24]

Answer:

about 50 percent of its former width.

Step-by-step explanation:

Let's assume that our parameter of interest is given by $\theta$ and in order to construct a confidence interval we can use the following formula:

$\hat \theta \pm ME(\hat \theta)$

Where $\hat \theta$ is an estimator for the parameter of interest and the margin of error is defined usually if the distribution for the parameter is normal as:

$ME = z_{\alpha} SE$

Where $z_{\alpha/2}$ is a quantile from the normal standard distribution that accumulates $\alpha/2$ of the area on each tail of the distribution. And $SE$ represent the standard error for the parameter.

If our parameter of interest is the population proportion the standard of error is given by:

$SE= \frac{\hat p (1-\hat p)}{n}$

And if our parameter of interest is the sample mean the standard error is given by:

$SE = \frac{s}{\sqrt{n}}$

As we can see the standard error for both cases assuming that the other things remain the same are function of n the sample size and we can write this as:

$SE = f(n)$

And since the margin of error is a multiple of the standard error we have that $ME = f(n)$

Now if we find the width for a confidence interval we got this:

$Width = \hat \theta + ME(\hat \theta) -[\hat \theta -ME(\hat \theta)]$

$Width = 2 ME (\hat \theta)$

And we can express this as:

$Width =2 f(n)$

And we can define the function $f(n) = \frac{1}{\sqrt{n}}$ since as we can see the margin of error and the standard error are function of the inverse square root of n. So then we have this:

$Width_i= 2 \frac{1}{\sqrt{n}}$

The subscript i is in order to say that is with the sample size n

If we increase the sample size from n to 4n now our width is:

$Width_f = 2 \frac{1}{\sqrt{4n}} =2 \frac{1}{\sqrt{4}\sqrt{n}} =\frac{2}{2} \frac{1}{\sqrt{n}} =\frac{1}{\sqrt{n}} =\frac{1}{2} Width_i$

The subscript f is in order to say that is the width for the sample size 4n.

So then as we can see the width for the sample size of 4n is the half of the wisth for the width obtained with the sample size of n. So then the best option for this case is:

about 50 percent of its former width.

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4 years ago