There are statistical analyses beyond simple descriptive measures, statistical inference, and differences tests including ________, which determine whether a stable relationship exists between two variables.

Since the obtained Chi square (23.45) is greater than the critical Chi square (9.488), we can conclude that there is a significant relationship between marital status and church attendance.

A Chi square test is used to determine whether there is a significant association between two categorical variables. In this case, the variables are marital status and church attendance.

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It is important to base design decisions on objective criteria such as user research, usability testing, and data analysis, rather than subjective biases or unfounded beliefs about numbers.

Not supported by any empirical evidence or logical reasoning. There is no inherent power or superiority associated with even numbers over odd numbers in the context of presenting or designing a product or message.

While some people may have personal preferences or cultural associations with even or odd numbers, the power or effectiveness of a product or message is determined by various factors such as its content, design, target audience, and context of use.

Therefore, it is important to base design decisions on objective criteria such as user research, usability testing, and data analysis, rather than subjective biases or unfounded beliefs about numbers.

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The fraction in lowest terms is 2/3. Option C

First, we need to know that a fraction is described as the part of a whole variable, a whole number or a whole element.

The different  fractions in mathematics are listed thus;

From the information given, we have that;

a+1/b

Such that a = 5 and b = 9

Now, substitute the values, we get;

5 + 1/9

Add the values of the numerator

6/9

Divide the values

2/3

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The mean estimation of cholesterol change  for the population of aquarobic program participants using 95% confidence falls between 8.17 and 11.83 mg/dL.

To estimate the mean cholesterol change using 95% confidence, we need to use a confidence interval. The formula for a confidence interval is:

Mean cholesterol change ± (t-value * standard error)

We can use a t-distribution with n-1 degrees of freedom, where n is the number of participants in the aquarobic program. We can assume that the sample is randomly selected and independent, and that the population of cholesterol changes follows a normal distribution.

To find the t-value, we need to use a t-table or calculator with the appropriate degrees of freedom and confidence level. For 95% confidence and n=sample size, the t-value is:

t-value = 2.306

To calculate the standard error, we can use the formula:

standard error = standard deviation / sqrt(n)

If the standard deviation is not given, we can use the sample standard deviation instead. We can assume that the sample standard deviation is a good estimate of the population standard deviation.

Once we have the standard error, we can substitute it into the confidence interval formula along with the t-value and the mean cholesterol change. This will give us the 95% confidence interval for the mean cholesterol change.

For example, if the mean cholesterol change is 10 mg/dL and the standard deviation is 3 mg/dL, and there were 20 participants in the aquarobic program, then the 95% confidence interval would be:

10 ± (2.306 * (3 / sqrt(20)))
10 ± 1.83

The confidence interval would be (8.17, 11.83). This means that we can be 95% confident that the true mean cholesterol change for the population of aquarobic program participants falls between 8.17 and 11.83 mg/dL.

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Set theory is a branch of mathematics that deals with the study of sets, which are collections of objects. A subset is a set that contains only elements that belong to a larger set, called the universal set. In this context, let S, T, X, and Y be subsets of some universal set.

Assuming that S [T X [Y, we can conclude that if a belongs to T, then a belongs to S or a belongs to X or a belongs to Y. This is because T is a subset of S [T X [Y, and any element in T must belong to at least one of these sets.

Assuming that S \T D, we can conclude that if a belongs to T, then a does not belong to S. This is because S \T is the set of elements that belong to S but not to T, and D is the empty set, meaning that there are no elements in the set. Therefore, if a belongs to T, it cannot belong to S \T, and so it must not belong to S.

Assuming that X S, we can use all three assumptions to prove that T Y. Suppose that a belongs to T. Then, using assumption (i), we know that a belongs to S or a belongs to X or a belongs to Y. But since a cannot belong to S (using assumption (ii)), we must have either a belongs to X or a belongs to Y. But since X is a subset of S (using assumption (iii)), we know that if a belongs to X, then a belongs to S. Therefore, we must have a belongs to Y. This holds for any element a in T, so we can conclude that T Y.

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The total number of positive integers with exactly three proper divisors and each divisor less than 50 is [tex]7+18=\boxed{25}$.[/tex]

To have exactly three proper divisors, a number must be of the form [tex]p_1^2$ or $p_1p_2$[/tex], where [tex]$p_1$[/tex] and [tex]$p_2$[/tex] are prime numbers.

For the case of [tex]p_1^2$,[/tex] there are only 7 prime numbers less than 50. So, there are 7 possible numbers of the form [tex]p_1^2$.[/tex]

For the case of [tex]$p_1p_2$[/tex], we can choose 2 prime numbers from the 7 available prime numbers, which can be done in [tex]$\binom{7}{2} = 21$[/tex] ways. However, we must exclude the numbers that are squares of primes, which are already counted in the previous case. There are 3 such numbers: [tex]2^2$, $3^2$,[/tex]and $5^2$. So, there are [tex]$21-3=18$[/tex] possible numbers of the form [tex]p_1p_2$.[/tex]

Therefore, the total number of positive integers with exactly three proper divisors and each divisor less than 50 is [tex]7+18=\boxed{25}$.[/tex]

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The decision regarding the hypothesis that the training was effective in improving customer relationships would depend on the results of the hypothesis test and the associated p-value.

To determine the decision regarding the hypothesis that the training was effective in improving customer relationships, a hypothesis test would need to be conducted. The null hypothesis, denoted as H0, would be that the training had no effect on improving customer relationships. The alternative hypothesis, denoted as Ha, would be that the training was effective in improving customer relationships.

Assuming a 0.05 significance level, if the p-value associated with the hypothesis test is less than or equal to 0.05, then the null hypothesis can be rejected in favor of the alternative hypothesis. This would indicate that there is evidence to suggest that the training was effective in improving customer relationships.

On the other hand, if the p-value is greater than 0.05, then the null hypothesis cannot be rejected. This would indicate that there is insufficient evidence to suggest that the training was effective in improving customer relationships.

Therefore, the decision regarding the hypothesis that the training was effective in improving customer relationships would depend on the results of the hypothesis test and the associated p-value.

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The probability that John is late for work is 0.26, or 26%.

To calculate the probability that John is late for work, we need to use the law of total probability, which states that the probability of an event can be found by considering all possible ways that the event can occur.

Let B be the event that John takes the bus, and S be the event that he takes the subway. Let L be the event that he is late for work.

We know that P(B) = 0.3, P(S) = 0.7, P(L|B) = 0.4, and P(L|S) = 0.2. We want to find P(L), the probability that John is late for work.

Using the law of total probability, we have:

P(L) = P(L|B)P(B) + P(L|S)P(S)

= 0.4 x 0.3 + 0.2 x 0.7

= 0.12 + 0.14

= 0.26

Therefore, the probability that John is late for work is 0.26, or 26%.

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When a line segment is drawn from a point on a circle's circumference, through the center to another point on the circle's circumference, the distance of the line segment is equal to the diameter of the circle.

The diameter is defined as any straight line passing through the center of a circle, connecting two points on the circumference. It is the longest chord of the circle and is also the length of a circle's widest point.

The diameter is an important parameter of a circle, as it determines the circle's size and area. It is related to the circle's radius, which is half the length of the diameter. The diameter is also used in various formulas for calculating the circumference, area, and other properties of circles.

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The probability that her trip will take longer than 80 minutes  is 0.20.

The probability that the trip takes longer than 80 minutes is equal to the area of the uniform distribution to the right of 80. Since the distribution is uniform, the area to the right of 80 is equal to the proportion of the total area that is to the right of 80.

The total area under the uniform distribution is equal to the length of the interval, which is (90 - 40) = 50.

The area to the right of 80 is equal to (90 - 80) = 10.

So the probability that the trip takes longer than 80 minutes is:

10 / 50 = 0.2

Therefore, the answer is 0.20.

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We can add the two integrals to get the total distance traveled:

≈ 1,262.63 meters (rounded to two decimal.

To find the total distance traveled by the referee, we need to integrate the absolute value of the velocity function over the given time interval.

distance = ∫[tex][2,6] |v(t)| dt[/tex]

The absolute value of the velocity function is given by:

[tex]|v(t)| = |4t^6 cos(2t+5)|[/tex]

So, we have:

[tex]distance = ∫[2,6] |4t^6 cos(2t+5)| dt[/tex]

We can simplify this integral by using the fact that the absolute value of a product is the product of the absolute values:

[tex]distance = ∫[2,6] 4t^6 |cos(2t+5)| dt[/tex]

Next, we split the integral into two parts, depending on whether the argument of the cosine function is positive or negative:

distance = [tex]∫[2,6] 4t^6 cos(2t+5) dt + ∫[2,6] 4t^6 cos(-2t-5) dt[/tex]

Simplifying the second integral using the identity cos(-x) = cos(x), we get:

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The probability of getting heads up is 5/6, or approximately 0.833.

There are three coins in the box, so the probability of picking any one of them at random is 1/3.

Let's consider the probability of getting heads up for each of the three coins:

For the two regular coins, the probability of getting heads up is 1/2, since they are fair coins.

For the fake two-headed coin, the probability of getting heads up is 1, since both sides are heads.

So, the overall probability of getting heads up depends on which coin is selected.

If you select one of the regular coins, the probability of getting heads up is 1/2.

If you select the fake two-headed coin, the probability of getting heads up is 1.

The probability of selecting one of the regular coins is 2/3, since there are two regular coins out of the total of three coins.

Therefore, the overall probability of getting heads up when you pick a coin at random and toss it is:

(2/3) x (1/2) + (1/3) x (1) = 2/6 + 3/6 = 5/6.  

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This  queue does generate the correct steady state distribution.

To determine the arrival rate of the queue, we need to use the balance equations, which state that the arrival rate λ must be equal to the departure rate μ.

In this case, the departure rate is given by the service rate s multiplied by the probability of there being at least one customer in the system, which is 1 - P(n=0):

μ = s(1 - P(n=0))

Using the binomial distribution, we can calculate the probability of there being no customers in the system:

P(n=0) = (1-p)^n = (1-p)^0 = 1

So the departure rate simplifies to:

μ = s(1 - P(n=0)) = s(1 - 1) = 0

This means that there are no customers leaving the system, and therefore no departure rate to balance against. To find the arrival rate, we need to consider the fact that the system is in a steady state, meaning that the number of arrivals must equal the number of departures.

Since there are no departures, the number of arrivals must also be zero. This implies that the arrival rate λ is also zero.

We can confirm that this queue generates the correct steady state distribution by verifying that the probability mass function (pmf) of the number of customers in the system matches the given (n,p) binomial distribution.

The probability of there being n customers in the system at steady state is given by:

P(n) = P(n arrivals before the first departure) x P(no arrivals during service) x P(n-1 arrivals before the second departure) x P(no arrivals during service) x ...

This simplifies to:

P(n) = λ/(s+λ) x (1-p)^n x λ/(s+λ) x (1-p)^(n-1) x λ/(s+λ) x (1-p)^(n-2) x ...

which can be written as:

P(n) = (λ/(s+λ))^n x (1-p)^(n(n-1)/2)

Comparing this with the (n,p) binomial distribution, we can see that the pmf matches if:

λ/(s+λ) = p

Taking the limit as n approaches infinity, we can see that the steady state distribution converges to the (n,p) binomial distribution. Therefore, this queue does generate the correct steady state distribution.

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For nonnormal populations, as the sample size (n) increases, the distribution of sample means approaches a normal distribution.

This is known as the central limit theorem. The central limit theorem states that as the sample size increases, the distribution of sample means becomes more and more normal, even if the original population is not normal. The central limit theorem is a fundamental concept in statistics, as it allows us to use the normal distribution as an approximation for many statistical analyses, regardless of the underlying distribution of the population.

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The hole size required for the meter to be bolted to the switchboard is 1.7565 inches in diameter, expressed in decimal form.

To find the hole size for the meter to be bolted to the switchboard, you need to add the given diameter difference of 1.32 inches to the diameter of the meter studs. The meter studs are 0.4365 inches in diameter. So, the calculation is as follows:

Hole size = Meter stud diameter + Diameter difference
Hole size = 0.4365 inches + 1.32 inches
Hole size = 1.7565 inches

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To ensure that only the original hole is patterned in the second direction and not the patterned holes, select the "Seed Only" or "Pattern Seed Only" check box when creating the pattern in your CAD software.

The check box selected to ensure that only the original hole is patterned in the second direction and not the patterned holes is typically called the "Seed Only" or "Pattern Seed Only" option.

Here is a step-by-step explanation:

1. In a CAD (Computer-Aided Design) software, you may have created an original hole and then used the "Pattern" or "Pattern Holes" feature to create a pattern of holes in one direction.

2. Now, you want to create a pattern of this original hole in the second direction, but you do not want to pattern the holes that were already created in the first direction.

3. In the pattern settings or options, you will find a check box called "Seed Only" or "Pattern Seed Only." This option is specifically designed for situations like yours, where you only want to pattern the original hole and not the other holes created by previous patterns.

4. Check the "Seed Only" or "Pattern Seed Only" option to make sure that the software knows to pattern only the original hole in the second direction.

5. Proceed with setting the pattern direction, distance, and the number of instances or copies of the hole you want in the second direction.

6. After confirming the settings, the software will create a pattern of the original hole in the second direction without patterning the previously created holes from the first direction.

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From the calculation, the population density of the place is 800 people/[tex]mile^2[/tex].

We know that we can be able to obtain the population density by the use of the formula;

Population density = Population of people/ Area of the place;

The area is obtained from;

Length = 30 miles

Width = 2/3 * 30 = 20 miles

Area of the city = 30 miles * 20 miles = 600 [tex]miles^2[/tex]

Final population after the evacuation = 555,000 - 75,000

= 480,000

Then the population density = 480,000/600 [tex]miles^2[/tex]

= 800 people/[tex]miles^2[/tex]

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95% confidence that the true mean of the violent crime rate in Seattle's census tracts lies between 20.43 and 33.69 per 1,000.

In the given study, Rountree and Warner (1999) observed a mean official violent crime rate of 27.06 per 1,000 in 100 census tracts over a two-year period in Seattle, WA. The standard deviation was 33.81. To construct a 95% confidence interval around this mean, we can use the formula:

[tex]CI = mean ± (Z-score * (standard deviation / √sample size))[/tex]

For a 95% confidence interval, the Z-score is 1.96. The sample size is 100 census tracts. So, we can calculate the interval as follows:

CI = 27.06 ± (1.96 * (33.81 / √100))

CI = 27.06 ± (1.96 * (33.81 / 10))

CI = 27.06 ± (1.96 * 3.381)

CI = 27.06 ± 6.63

The lower bound for this confidence interval is 27.06 - 6.63, which is approximately 20.43. Therefore, we can say with 95% confidence that the true mean of the violent crime rate in Seattle's census tracts lies between 20.43 and 33.69 per 1,000.

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The calculated dimensions of the paper are 6 inches by 10 inches

From the question, we have the following parameters that can be used in our computation:

Let the dimensions be

L = Length

W = Width

So, we have

L * W = 60

2 * (L + W) = 32

This gives

L * W = 60

(L + W) = 16

Let L = 10

So, we have

W = 6

Testing these values, we have

6 * 10 = 60 -- true

2 * (6 + 10) = 32 -- true

Hence, the dimensions of the paper are 6 inches by 10 inches

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It will take Jill and John [tex]$\frac{6}{5}$[/tex] hours or 1 hour and 12 minutes.

To solve the problem, we can use the formula:

[tex]$ \frac{1}{x} = \frac{1}{a} + \frac{1}{b}$[/tex]

where x is the time it takes for Jill and John to complete the job working together, and a and b are the times it takes for Jill and John to complete the job individually, respectively.

Substituting the given values, we get:

[tex]$ \frac{1}{x} = \frac{1}{2} + \frac{1}{3}$[/tex]

Simplifying this expression, we get:

[tex]$ \frac{1}{x} = \frac{5}{6}$[/tex]

Multiplying both sides by 6x, we get:

[tex]$6 = 5x$[/tex]

Dividing both sides by 5, we get:

[tex]$x = \frac{6}{5} hours$[/tex]

Therefore, it will take Jill and John [tex]$\frac{6}{5}$[/tex] hours or 1 hour and 12 minutes (rounded to the nearest minute) working together to complete the job.

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The probability that all 4 players have an ace is approximately 0.00000369 or about 0.00037%.

To find the probability that all 4 players have an ace, we can use the principle of multiplication, which states that the probability of two independent events occurring together is the product of their individual probabilities.

There are 4 aces in the deck of 52 cards. The probability that the first player gets an ace is 4/52. After the first ace has been dealt, there are 51 cards remaining, including 3 aces. Therefore, the probability that the second player gets an ace is 3/51. Similarly, the probability that the third player gets an ace is 2/50, and the probability that the fourth player gets an ace is 1/49.

Using the principle of multiplication, we can multiply these probabilities to find the probability that all four players get an ace:

P(all 4 players have an ace) = (4/52) x (3/51) x (2/50) x (1/49) = 0.00000369

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Answer:

I'd say probably 6, as 3 + 3 = 6, and 3 is the middle number.

**EDIT**

It's a 7, because there are 6 ways of rolling this value, making it the most probable.

Instead of performing repeated two-mean comparisons, we conduct an analysis of variance to compare means because doing so reduces the likelihood of Type I errors. Here option A is the correct answer.

ANOVA is preferred over multiple two-mean comparisons because it reduces the Type I error probability and offers several benefits over conducting multiple t-tests.

When conducting multiple two-mean comparisons, the likelihood of making a Type I error (i.e., rejecting a true null hypothesis) increases with each comparison made. This is known as the multiple comparison problem, and it can lead to an inflated false positive rate.

In contrast, ANOVA uses a single test to compare multiple means simultaneously, reducing the likelihood of making a Type I error. ANOVA achieves this by partitioning the total variation in the data into different sources of variation and estimating the amount of variation due to random chance.

Additionally, ANOVA offers several benefits over multiple t-tests. One advantage is that it provides a statistical significance test for the overall difference among the groups, not just for pairwise comparisons. ANOVA also allows for the testing of interactions between factors that influence the mean differences. Furthermore, ANOVA calculations can be more efficient than conducting multiple t-tests, especially when the number of groups is large.

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Complete question:

The reason why we perform an analysis of variance for comparing means rather than conducting multiple two-mean comparisons is

A. Because multiple two-mean comparisons increase the Type I error probability.

B. Simply because ANOVA has simpler calculations but otherwise offers no benefit over the multiple t-tests.

C. Power is compromised when using t-tests.

Based on the obtained chi-square value of 10.78 and the critical chi-square value of 3.841, we can conclude that there is a significant association between the two categorical variables being tested, and the null hypothesis is rejected.

A chi-square test is a statistical method used to determine if there is a significant association between two categorical variables.

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The function f(x) is continuous for all x ≠ 0 but is not continuous at x = 0.

The function f: R → R is defined by f(x) = sin(1/x) if x ≠ 0 and f(x) = 0 if x = 0. To determine whether f is continuous, we need to examine its behavior at x = 0 and for all x ≠ 0.

For x ≠ 0, the function f(x) = sin(1/x) is a composition of two continuous functions, sin(u) and u = 1/x. Since the composition of continuous functions is continuous, f(x) is continuous for x ≠ 0.

At x = 0, we need to check the limit of f(x) as x approaches 0. Let's examine the limit from both the left and right sides:

lim(x→0-) sin(1/x) and lim(x→0+) sin(1/x)

As x approaches 0 from either side, 1/x becomes increasingly large in magnitude, causing the sine function to oscillate rapidly between -1 and 1. There is no unique value that the function approaches. Therefore, the limit does not exist:

lim(x→0) sin(1/x) does not exist.

Since the limit does not exist at x = 0, the function f(x) is not continuous at x = 0.

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The expected value per share, taking into account the possibility of financial distress, is 7.92.

If the probability of financial distress is 10%, we need to adjust our valuation to account for the possibility of that scenario occurring.

Let's denote the probability of no financial distress as p and the probability of financial distress as (1-p). Then, the expected value of equity per share can be calculated as:

Expected value = p × 9.50 + (1-p) × 2.0

We know that (1-p) = 0.10, so we can substitute that in:

Expected value = p × 9.50 + 0.10  2.0

Solving for p, we get:

p = (Expected value - 0.10 × 2.0) / 9.50

Substituting the given values, we get:

p = (9.50 - 0.10 × 2.0) / 9.50 = 0.789

This means that the probability of no financial distress is 0.789, and the probability of financial distress is 0.211.

Now, we can calculate the value per share under the scenario of financial distress:

Value per share (distress) = 2.0

And the value per share under the scenario of no financial distress:

Value per share (no distress) = 9.50

So, the expected value per share is:

Expected value per share = p × Value per share (no distress) + (1-p) × Value per share (distress)

Expected value per share = 0.789 × 9.50 + 0.211 × 2.0 = 7.50 + 0.42 = 7.92

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When Solver is unable to find a solution to a problem, it can be an indication of various issues. One possible cause is that the problem may be too complex for Solver to solve within the given constraints.

In such cases, it may be necessary to adjust the problem parameters or seek alternative solutions. Another possible cause of Solver's inability to find a solution could be due to incorrect input data. This can lead to inconsistent or contradictory constraints, making it impossible for Solver to arrive at a feasible solution. Lastly, Solver may fail to find a solution due to numerical errors or limitations in the algorithm used. These issues can arise when dealing with large datasets or highly non-linear problems. In any case, when Solver is unable to find a solution, it is important to carefully examine the problem and its parameters, and consider alternative approaches. Sometimes, a small adjustment to the input data or constraints can make all the difference in arriving at a successful solution.  In such cases, Solver may struggle to find an accurate or optimal solution. Ill-conditioned problems typically involve numerical instability or poor scaling, while infeasible problems occur when the given constraints cannot be satisfied simultaneously. It is crucial to analyze and refine the problem to enable Solver to find a viable solution effectively. When there is a problem with Solver being unable to find a solution, it often indicates the presence of a(n) ill-conditioned or infeasible problem.

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When the spinner is spun, the P ( vowel, then P ) would be B. one-ninth.

To find the probability of P ( vowel, then P ), on the spinner given, the formula would be :

= P ( vowel ) x  P ( P )

We have six sides on the spinner which gives us:

P ( vowel ) = 4 / 6

P ( P ) = 1 / 6

The probability of P ( vowel, then P ) is :

= 4 / 6 x 1 / 6

= 4 / 36

= 1 / 9

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The total number of outcomes where all three dice show different numbers is 120.

To determine the number of outcomes in which all three dice show different numbers, you can use the multiplication principle. First, choose a number on the red die (6 options). Then, choose a different number on the blue die (5 options, since it cannot be the same as the red die). Finally, choose a different number on the green die (4 options, since it cannot be the same as the red or blue die). Multiply the options together: 6 x 5 x 4 = 120. Therefore, there are 120 possible outcomes in which the three dice all show different numbers.

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