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Svet_ta [14]
3 years ago
13

A number cube is rolled 25 times. A number less than four comes up 20 times. What is the experimental probability that a number

less than four is rolled?
Mathematics
1 answer:
sesenic [268]3 years ago
4 0

a number less than four came up 20 out of 25 rolls

 so the probability was 20/25 which reduces to 4/5

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The lowest temperature ever recorded is -89c. The highest temperature is 56.7c. What is the difference between the highest and t
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Answer:

The difference is 145.7

Step-by-step explanation:

89+56.7= 145.7

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Find the value of the variable using the given secant. Show your work.
storchak [24]

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x=3.5

Step-by-step explanation:

x×6=3×7

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What is the LCM of 2 numbers that have no common factors greater than 1
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7=7, 14, ,21, 28
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kolbaska11 [484]

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10017

Step-by-step explanation:

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In constructing a 95 percent confidence interval, if you increase n to 4n, the width of your confidence interval will (assuming
Damm [24]

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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.

7 0
4 years ago
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