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12345 [234]
2 years ago
9

A fair sortition trial is carried out, and one of the candidates is assigned the number 32,041. If each digit can be chosen from

0-4, and if each of the possible sequences is assigned to a candidate, how many candidates are there?
Mathematics
2 answers:
erik [133]2 years ago
6 0
The answer to this question is 3,125
Kitty [74]2 years ago
3 0

Answer: 3125


Step-by-step explanation:

Given: A fair sortition trial is carried out, and one of the candidates is assigned the number = 32,041

If each digit can be chosen from 0-4, and if each of the possible sequences is assigned to a candidate, then the number of candidates (repetition of digits allowed)=5\times5\times5\times5\times5

=3125

Hence, the number of candidates there are = 3125.




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Consider the following random sample from a normal population: 14, 10, 13, 16, 12, 18, 15, and 11. What is the 95% confidence in
seraphim [82]

Answer:

13.625-2.365\frac{2.669}{\sqrt{8}}=11.393  

13.625+2.365\frac{2.669}{\sqrt{8}}=15.857  

So on this case the 95% confidence interval would be given by (11.393;15.857)  

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Data: 14, 10, 13, 16, 12, 18, 15, 11

We can calculate the sample mean and deviation with the following formulas:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=13.625 represent the sample mean  

\mu population mean (variable of interest)  

s=2.669 represent the sample standard deviation  

n=8 represent the sample size  

Calculate the confidence interval

The confidence interval for the mean is given by the following formula:  

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:  

df=n-1=8-1=7  

Since the confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,7)".And we see that t_{\alpha/2}=2.365

Now we have everything in order to replace into formula (1):  

13.625-2.365\frac{2.669}{\sqrt{8}}=11.393  

13.625+2.365\frac{2.669}{\sqrt{8}}=15.857  

So on this case the 95% confidence interval would be given by (11.393;15.857)  

3 0
1 year ago
How do i evaluate 256 1/4 .? this is algebra 2.
Mariulka [41]
Look it up on photomath
6 0
2 years ago
2x+5y=-6 , wht is the x intercept, what is the y intercept
Arlecino [84]
These are the <span>xx</span> and <span>yy</span> intercepts of the equation <span><span>2x−5y=6</span><span>2x-5y=6</span></span>.x-intercept: <span><span>(3,0)</span><span>(3,0)</span></span>y-intercept: <span>(0,−<span>65</span><span>)

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7 0
2 years ago
In a large population, 61 % of the people have been vaccinated. if 4 people are randomly selected, what is the probability that
muminat
In a large population, 61% of the people are vaccinated, meaning there are 39% who are not. The problem asks for the probability that out of the 4 randomly selected people, at least one of them has been vaccinated. Therefore, we need to add all the possibilities that there could be one, two, three or four randomly selected persons who were vaccinated.

For only one person, we use P(1), same reasoning should hold for other subscripts.

P(1) = (61/100)(39/100)(39/100)(39/100) = 0.03618459
P(2) = (61/100)(61/100)(39/100)(39/100) = 0.05659641
P(3) = (61/100)(61/100)(61/100)(39/100) = 0.08852259
P(4) = (61/100)(61/100)(61/100)(61/100) = 0.13845841

Adding these probabilities, we have 0.319761. Therefore the probability of at least one person has been vaccinated out of 4 persons randomly selected is 0.32 or 32%, rounded off to the nearest hundredths.
8 0
1 year ago
Cameron wondered if the average score on a final exam was different between those who texted on a regular basis during the lectu
Margarita [4]

Answer:

c. A two-tailed test should be performed since the alternative hypothesis states that the parameter is not equal to the hypothesized value.

Step-by-step explanation:

Let p1 be the average score on a final exam who texted on a regular basis during the lectures for a particular class

And p2 be the average score on a final exam who did not texted at all during the lectures for a particular class

According to the Cameron's point of interest, null and alternative hypotheses are:

H_{0}: p1 = p2

H_{a}: p1 ≠ p2

Two tailed test should be performed since the alternative hypothesis states that the parameter is not equal to the hypothesized value.

3 0
2 years ago
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