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The units/one's digit of 5^221 is 5.

To determine the units/one unit digit of a number raised to a power, we only need to consider the units/one's digit of the base. In this case, the unit digit of 5 is 5.

Now, we need to look for a pattern in the units digit of 5 raised to different powers.

5^1 = 5 (units digit is 5)
5^2 = 25 (units digit is 5)
5^3 = 125 (units digit is 5)
5^4 = 625 (units digit is 5)
5^5 = 3125 (units digit is 5)
. . .and so on.

We can see that the unit digit of 5 raised to any power is always 5. Therefore, the units/one's digit of 5^221 is 5, not 3.

So, the statement "the units/one's digit of 5^221 is 3" is disproved.

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Given Statement: Both fortunes does not favor the foolish and love is not eternal. Translation: [tex]~F ~L[/tex]

Given Statement: Love is eternal if and only if neither the glass is half full nor temperance is a virtue.

Translation: L ↔ [tex]~(G[/tex] ∨ T)

Given Statement: If love is eternal and temperance is a virtue, then either fortune favors the foolish or avarice is a vice.

Translation: (L ∧ T) → (F ∨ A)

Given Statement: Avarice is a vice, given that both temperance is not a virtue and the glass is not half full.

Translation: ([tex]¬T[/tex] ∧ [tex]¬G[/tex]) → A

Given Statement: It is not the case that both temperance is a virtue and either love is eternal or avarice is a vice.

Translation: [tex]¬(T[/tex] ∧ (L ∨ A))

Given Statement: If the glass is half full, then if fortune favors the foolish, then love's being eternal implies that space is the final frontier.

Translation: G → (F → (L → S))

Given Statement: Avarice's not being a vice is a necessary condition for temperance's not being a virtue.

Translation: ¬A → ¬T

Given Statement: It is not the case that both temperance's being a virtue implies that avarice is a vice and space's being the final frontier implies that fortune favors the foolish.

Translation: ¬(T → A ∧ S → F)

Given Statement: Fortune's favoring the foolish is a necessary condition for space's being the final frontier; moreover, love's being eternal and the glass's being half full is a sufficient condition for avarice's not being a vice.

Translation: (F → S) ∧ ((L ∧ G) → ¬A)

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Answer: 6000

Step-by-step explanation:

For the numbers you choose, there are 10, then 20, then 30 possible numbers to choose from.

To find the total amount of possible combinations with repetition, you just do 10x20x30 = 6000.

this is a special triangle so the undefined lenght is 12

so the answer is 12/20 = 0.6

If the economy starts to weaken and the FOMC determines that employment is falling short of maximum employment, an appropriate policy implementation would be to lower the interest on reserve balances rate, ON RRP offering rate, and discount rate.

This would make it cheaper for banks to borrow money and encourage them to lend more, which could stimulate economic activity and create more job opportunities. Option c, Lower the interest on reserve balances rate, ON RRP offering rate, and discount rate, is the best answer. The other options are not as effective in this scenario - raising rates would likely make it more expensive for businesses and consumers to borrow money, which could further slow down the economy, while using open market operations to decrease reserves could lead to a shortage of liquidity in the banking system.

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If this is a sample of 6 scores, then  SS using the definitional formula would equal 17.5.

To find the SS (sum of squares) using the definitional formula, you need to first calculate the mean of the scores. Here's

1. Calculate the mean (µ) using ∑x and the number of scores (n):
Mean (µ) = (∑x) / n
µ = 39 / 6
µ = 6.5

2. Use the computational formula for SS:
SS = ∑x² - ( (∑x)² / n )
SS = 271 - (39² / 6)
SS = 271 - (1521 / 6)
SS = 271 - 253.5

3. Calculate sample score SS:
SS = 17.5

So, the answer is 17.5.

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(a) The probability that all 4 missiles will fire is 0.2098 and (b) The probability that at most 2 will not fire is 0.0099.

Here we need to use the concept of combinations to get our required answer.

Here we have been given that 4 out of 14 missiles are defective.

Hence 10 missiles are working.

A sample of 4 missiles was chosen

Hence the simple can be chosen in ¹⁴C₄ ways

a)

The probability that all 4 missiles will fire is

¹⁰C₄/¹⁴C₄

= 0.2098

b)

The probability that at most 2 will not fire is

1 - the probability of including all three defective missiles

= 10³C₃/¹⁴C₄

= 0.0099

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(a) Probability all 4 fire: 0.3297. (b) Probability at most 2 not fire: 0.9591.

(a) To track down the chance that all of the 4 rockets will fire, we actually need to do not forget the quantity of methods we can select 4 operating rockets out of the eleven that are not unsuitable, isolated by the all out range of approaches we will select four rockets out of the whole component.

The amount of ways of selecting four operating rockets out of the eleven that aren't improper is given by way of the combination recipe:

C(eleven,four) = 330

The absolute quantity of methods of selecting four rockets out of the entire part is given via:

C(14,4) = 1001

Subsequently, the chance that each one of the four rockets will hearth is:

P(all four hearth) = C(11,four)/C(14,4) = 330/1001 ≈ zero.3297

(adjusted to 4 decimal spots)

(b) To find the likelihood that at most 2 rockets might not fire, we need to remember every one of the potential conditions in which 0, 1, or 2 poor rockets are selected, and afterward song down the likelihood of every case and upload them up.

Case 1: No defective rockets are selected

The amount of ways of selecting four working rockets out of the eleven that are not faulty is given through the mixture recipe:

C(11,four) = 330

Subsequently, the chance of choosing no defective rockets is:

P(0 inadequate) = C(eleven,4)/C(14,four) ≈ 0.3297

Case 2: One defective rocket is chosen

The quantity of ways of selecting three operating rockets out of the eleven that are not poor is given with the aid of the combo equation:

C(eleven,three) = 165

The amount of ways of selecting 1 damaged rocket out of the three that are available is given through the mix equation:

C(three,1) = 3

Subsequently, the all out number of approaches of selecting 1 insufficient and three running rockets is:

C(eleven,three) × C(3,1) = 495

Thusly, the probability of choosing one deficient rocket and three running rockets is:

P(1 deficient) = C(eleven,three) × C(three,1)/C(14,4) ≈ 0.4655

Case three: Two poor rockets are selected

The amount of approaches of choosing 2 poor rockets out of the three which might be handy is given through the combination recipe:

C(3,2) = 3

The amount of ways of choosing 2 running rockets out of the 11 that aren't faulty is given via the mix recipe:

C(11,2) = 55

In this way, the all out number of methods of selecting 2 broken rockets and 2 working rockets is:

C(3,2) × C(eleven,2) = one hundred sixty five

Subsequently, the chance of choosing two incorrect rockets and two operating rockets is:

P(2 insufficient) = C(3,2) × C(11,2)/C(14,four) ≈ 0.1638

Subsequently, the likelihood that at maximum 2 rockets might not fireplace is:

P(at maximum 2 don't fire) = P(zero imperfect) + P(1 blemished) + P(2 deficient) ≈ 0.9591

(adjusted to 4 decimal spots).

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There are 151200 distinguishable permutations for the word CONFERENCE

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

CONFERENCE

In the above word, we have

Letters = 10

Repeated C = 2

Repeated N = 2

Repeated E = 3

Using the above as a guide, we have the following:

The number of distinguishable permutations for the word is

Number = Letters!/Repeated letters!

This means that

Number = 10!/(2! * 2! * 3!)

Evaluate

Number = 151200

Hence, there are 151200 distinguishable permutations

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Vinny's new grade in math, if the playing time reduces to 6 hours per week, would be 85.33.

If we wish to illustrate how Vinny's Math Models grade relates to the amount of time she spends playing computer games, we may employ the formula that defines an inverse proportion:

k = G x T

k = 64 x 8 = 512

For 6 hours playing:

512 = G x 6

G = 512 / 6

= 85. 33

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The theoretical probability of the spinner not landing on blue is 3/4

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

Colors = 4 i.e. purple, yellow, blue and red

Blue = 1

Not blue = 3

So, we have

Theoretical probability = Not Blue /Colors

Substitute the known values in the above equation, so, we have the following representation

Theoretical probability = 3/4

Hence, theoretical probability is 3/4

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The solution is, the probability that exactly two of the dice show an even number is 0.3125.

We will use the Binomial Probability formula to find the answer

The formula is given by ⁿCₓ (p)ˣ (1-p)ⁿ⁻ˣ

Where:

n, the number of trials = 5

x, the number of success that we aim = 2

p, the probability of success = 0.5 ⇒ there are six out of twelve numbers on the dice that are even

Substitute these values into the formula, we have

P(2) = ⁵C₂ (0.5)² (1-0.5)⁵⁻²

P(2) = ⁵C₂ (0.5)² (0.5)³

P(2) = 0.3125

The solution is, the probability that exactly two of the dice show an even number is 0.3125.

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complete question:

Ben rolls 5 fair 12-sided dice. the 12 faces of each die are numbered from 1 to 12. what is the probability that exactly two of the dice show an even number?

To diagonalize C, we first need to find its eigenvalues and eigenvectors. The characteristic equation for C is det(C -

                                λI)  =  0, which gives us (1 - λ)(7 - λ)  -  6  =                                

                                   0. Solving for λ, we get λ1  =  1 and λ2  =                                    

               7. To find the eigenvector corresponding to λ1, we solve the system of equations (C -                

                   λ1I)x  =  0, which gives us the equation  - x1  +  2x2  =  0. Choosing x2  =                    

                                           1, we get the eigenvector v1  =                                          

                            [2,1]. Similarly, for λ2 we get the eigenvector v2  =  [1, -                            

                              1]. We can then diagonalize C by forming the matrix P  =                              

                         [v1, v2] and the diagonal matrix D  =  [λ1 0; 0 λ2]. We have C  =                        

                      -                                    -                                                        

                   PDP 1. To compute A5, we first compute C 1 as [7  - 2;  - 3 1] / 4. Then, A  =                    

                         -         -   -                                  5       5       5                          

                      CDC 1  =  PDP 1DC 1P. We have D  =  [1 0; 0 7], so D   =  [1  0; 0 7 ]  =                      

                                           5       5 -                                                              

                    [1 0; 0 16807]. Thus, A   =  PD P 1  =  [2 1; 1  - 1][1 0; 0 16807][1 / 3  -                    

                              1 / 3; 1 / 3 2 / 3]  =  [11203 11202; 16804 16805] / 9.                                

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To find a root-finding method with a third order of convergence, consider using the "Halley's method." Halley's method is an iterative numerical technique used for finding roots of a function. It has a third-order convergence, meaning the number of correct digits approximately triples with each iteration, resulting in a faster convergence rate compared to methods with lower orders of convergence.

Here's a step-by-step explanation of Halley's method:

1. Choose an initial guess x_0 for the root of the function f(x).

2. Calculate the first and second derivatives of the function f(x), denoted as f'(x) and f''(x), respectively.

3. Update the guess using the formula:
  x_(n+1) = x_n - (2 * f(x_n) * f'(x_n)) / (2 * (f'(x_n))^2 - f(x_n) * f''(x_n))

4. Check for convergence by comparing the difference between consecutive guesses (x_(n+1) - x_n) to a predefined tolerance level.

5. If the convergence criterion is not met, repeat steps 3 and 4 until convergence is achieved or a maximum number of iterations is reached.

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The probability of one success in five trials in the binomial experiment with a success probability of 5 % is 20. 4 %.

The formula for calculating the likelihood of one success in a binomial probability with a 5% chance of success is:

P ( X = 1) = (5 choose 1) x ( 0.05 ) x  (0.95 ) ⁴

Solving for this success would give :

= ( 0.05 ) x  ( 0. 95 ) ⁴

= 0.05 x 0.8145

= 0.040725

Then we multiply both sides to get :

P(X = 1) = 5 x 0.040725

= 0.203625

= 20. 4 %

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The equation of the line of best fit is y = 0.883x + 17.95.

We have,

To find the line of best fit, we want to find the equation of the line that comes closest to passing through all the points in the scatterplot.

One way to do this is to use linear regression analysis.

Using a calculator or statistical software,

We can find that the equation of the line of best fit for this data is:

y = 0.883x + 17.95

Thus,

The equation of the line of best fit is y = 0.883x + 17.95.

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Answer:

The tip is $17.00, and the cost of the meal is $62.50. To find the percentage that the tip represents of the cost of the meal, we need to divide the tip by the total cost and multiply by 100:

Percentage tip = (tip / total cost) x 100%

Percentage tip = (17.00 / (62.50 + 17.00)) x 100%

Percentage tip = (17.00 / 79.50) x 100%

Percentage tip = 0.214 x 100%

Percentage tip = 21.4%

Rounding to the nearest tenth of a percent, the tip represents 21.4% of the cost of the meal.

The tip is approximately 21.4% of the meal cost.

To find the percentage of the cost that is the tip, we need to first calculate the actual amount of the tip and then express it as a percentage of the meal cost.

The amount of the tip is $17.00, and the cost of the meal is $62.50, so the total amount paid is:

$62.50 + $17.00 = $79.50

To find the percentage of the cost that is the tip, we can use the formula:

(tip amount / total amount) x 100%

Plugging in the values we have:

($17.00 / $79.50) x 100% ≈ 21.4%

Rounding to the nearest tenth of a percent, we get:

21.4%

Therefore, the tip is approximately 21.4% of the meal cost.

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The probability that a small car was involved in the fatal accident is 0.483, the probability that a medium car was involved is 0.493, and the probability that a large car was involved is 0.024.

We can use Bayes' theorem to calculate the probabilities of small, medium, and large cars being involved in the fatal accident given that a fatal accident occurred. Let S, M, and L denote the events that a small, medium, and large car was involved, respectively, and F be the event that a fatal accident occurred. Then, we have:

P(S|F) = P(F|S) * P(S) / P(F)

P(M|F) = P(F|M) * P(M) / P(F)

P(L|F) = P(F|L) * P(L) / P(F)

where:

P(F|S) = 0.133 (the probability of a fatal accident given a small car is involved)

P(F|M) = 0.071 (the probability of a fatal accident given a medium car is involved)

P(F|L) = 0.045 (the probability of a fatal accident given a large car is involved)

P(S) = 0.28 (the probability of a small car on the road)

P(M) = 0.53 (the probability of a medium car on the road)

P(L) = 0.19 (the probability of a large car on the road)

P(F) = P(F|S) * P(S) + P(F|M) * P(M) + P(F|L) * P(L) (the total probability of a fatal accident)

We can calculate P(F) using the law of total probability:

P(F) = P(F|S) * P(S) + P(F|M) * P(M) + P(F|L) * P(L)

= 0.133 * 0.28 + 0.071 * 0.53 + 0.045 * 0.19

= 0.07694

Then, we can calculate the probabilities of small, medium, and large cars being involved:

P(S|F) = 0.133 * 0.28 / 0.07694 = 0.483

P(M|F) = 0.071 * 0.53 / 0.07694 = 0.493

P(L|F) = 0.045 * 0.19 / 0.07694 = 0.024

Therefore, the probability that a small car was involved in the fatal accident is 0.483, the probability that a medium car was involved is 0.493, and the probability that a large car was involved is 0.024.

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When measuring time, a part of a whole is not a whole when using units smaller than the whole unit of time.

For example, if we measure time in hours, then a part of an hour, such as 30 minutes, is not a whole. Similarly, if we measure time in minutes, then a part of a minute, such as 30 seconds, is not a whole.

In such cases, we need to convert the part into a fraction or decimal of the whole unit of time. For instance, 30 minutes is half of an hour, and 30 seconds is half of a minute.

It is important to keep track of the units of time being used and make appropriate conversions when necessary to ensure accurate and meaningful measurements.

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The length of the base of the triangle is √(2)/6, which can also be expressed as (√(2))/6.

In this case, we know the area (1/12 square centimeters), but we don't know the height or the base. However, we can use the fact that the area is equal to 1/2 times the base times the height to set up an equation:

1/12 = 1/2 x base x height

Now we need to solve for the base. We can do this by isolating the base on one side of the equation:

1/12 = 1/2 x base x height

1/6 = base x height

At this point, we need to make an assumption about the triangle.

We can use the Pythagorean theorem to solve for the length of h:

h² + (base/2)² = (base)²/4

Simplifying this equation, we get:

h² = (base)²/4 - (base)²/4

h² = (base)²/2

h = √((base)²/2)

h = base/√(2)

Now we can substitute this expression for h into our equation for the area:

1/6 = base x height

1/6 = base x (base/√(2))

Simplifying this equation, we get:

1/6 = (base²)/√(2)

Multiplying both sides by √(2), we get:

√(2)/12 = base²

Taking the square root of both sides, we get:

base = √(√(2)/12)

Simplifying this expression, we get:

base = √(2)/6

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The radius of the circle represented by the equation x² + y² - 4x - 8y - 16 = 0 is given as follows:

6 units.

The equation of a circle of center [tex](x_0, y_0)[/tex] and radius r is given by:

[tex](x - x_0)^2 + (y - y_0)^2 = r^2[/tex]

The equation for this problem is given as follows:

x² + y² - 4x - 8y - 16 = 0.

To obtain the radius of the circle, we must complete the squares from the equation, as follows:

x² - 4x + y² - 8y = 16

(x - 2)² + (y - 4)² = 16 + 2² + 4²

(x - 2)² + (y - 4)² = 36.

Hence the center and the radius of the circle are given as follows:

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The correct statement is c. When marginal utility is positive, an increase in the quantity consumed will increase total utility.

This is because as long as the marginal utility of each additional unit consumed is positive, the total utility will continue to increase with each additional unit consumed. However, when marginal utility starts to decrease, consuming additional units will result in diminishing returns and eventually lead to a decrease in total utility. The statement in option a is incorrect because an increase in the quantity consumed can still increase total utility if the marginal utility is positive. The statement in option b is also incorrect because if the marginal utility is positive, consuming more will increase total utility, not decrease it. Option d is also incorrect because when marginal utility is increasing, it means that the additional units consumed are providing more utility than the previous ones, so decreasing the quantity consumed will result in a decrease in total utility.

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What most likely happened is that : Sandy's experimental probability was closest to the theoretical probability in the experiment with 3,000 flips.

During a coin toss trial, the probability of heads or tails is theoretically 50% for any outcome. Nevertheless, experimental probabilities exhibit convergence with theoretic probability over time as trials increase.

In Sandy's scenario, it follows that an experiment with more flips - precisely, 3,000 - would have a substantially higher chance of exhitibing experimental outcomes closest in percentage to the theoretical fraction of fifty-fifty proportionality than those conducted involving fewer combinations such as with only merely 80 and 800 flippages per iteration.

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To prove that the three points of intersection of the three circles with PA, PB, and PC as diameters are collinear, we'll use the following terms: circle, chord, diameter, intersection, and collinear.

Let X, Y, and Z be the points of intersection of circles with diameters PA, PB, and PC respectively. To prove that X, Y, and Z are collinear, we need to show that they lie on a straight line.

Consider triangles PAX, PBY, and PCZ. Since the diameters PA, PB, and PC are subtended by angles AXB, BYC, and CZA at the circumference of circle O, we have:

∠AXB = ∠BYC = ∠CZA = 90° (by the property of angles in a semicircle)

Now, let's consider the sum of the angles in quadrilateral ABYC:

∠AXC + ∠AXB + ∠BZC + ∠BYC = 360°

Since ∠AXB = ∠BYC = ∠CZA = 90°, we get:

∠AXC + 90° + ∠BZC + 90° = 360°

Simplifying, we have:

∠AXC + ∠BZC = 180°

This means that points X, Y, and Z are collinear, as the sum of angles ∠AXC and ∠BZC in a straight line is 180°. Therefore, we have proven that the three points of intersection of the three circles with PA, PB, and PC as diameters are collinear.

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Answer:

The density graph for all possible temperatures from 60 degrees to 160 degrees can be used to find the probability of a temperature falling within a certain range.

Option (A) is incorrect because it includes temperatures that are outside the range of the graph.

Option (B) is incorrect because it includes temperatures that are outside the range of the graph.

Option (C) is incorrect because it includes temperatures that are outside the range of the graph.

Option (D) is the only option that falls within the range of the graph. Therefore, the density graph can be used to find the probability of a temperature from 90 degrees to 120 degrees, which is option (D).

Step-by-step explanation:

The expression is equivalent to 4 − 3(x + 2) + 2(3 − 2x) will be 4 − 7x. Then the correct option is A.

Given that:

Expression, 4 − 3(x + 2) + 2(3 − 2x)

The equivalent is the expression that is in different forms but is equal to the same value.

Simplify the expression, then we have

4 − 3(x + 2) + 2(3 − 2x)

4 − 3x − 6 + 6 − 4x

4 − 7x

Thus, the correct option is A.

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The missing options are given below.

4 − 7x

4 + 7x

7 − 4x

7 + 4x

The best data set that represents the histogram is (4, 5, 7, 8, 12, 13, 15, 18, 19, 21, 24, 25, 26, 28, 29, 30, 32, 33, 35, 42, option C is correct.

From the histogram, we can see that there were 3 data points in the interval 1-10, 3 data points in the interval 11-20, 7 data points in the interval 21-30, 4 data points in the interval 31-40, and 3 data points in the interval 41-50.

Therefore, the best data set that represents the histogram is the one that has 3 data points in the range 1-10, 3 data points in the range 11-20, 7 data points in the range 21-30, 4 data points in the range 31-40, and 3 data points in the range 41-50.

4, 5, 7, 8, 10, 12, 13, 15, 18, 21, 23, 28, 32, 34, 36, 40, 41, 41, 42, 42 does not fit this pattern since it has more than 3 data points in some of the intervals.

4, 7, 10, 13, 14, 19, 22, 24, 26, 27, 29, 31, 33, 35, 36, 38, 40, 42, 42, 42 also does not fit the pattern since it has more than 3 data points in some of the intervals.

4, 5, 7, 8, 12, 13, 15, 18, 19, 21, 24, 25, 26, 28, 29, 30, 32, 33, 35, 42 fits the pattern and has 3 data points in each interval. This is the correct answer.

4, 6, 9, 12, 16, 18, 21, 24, 25, 26, 28, 29, 30, 32, 35, 36, 38, 41, 41, 42 does not fit the pattern since it has more than 3 data points in some of the intervals.

Therefore, the best data set that represents the histogram is  (4, 5, 7, 8, 12, 13, 15, 18, 19, 21, 24, 25, 26, 28, 29, 30, 32, 33, 35, 42

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The appropriate inference procedure based on the statistical study aim and the sample size, is the option;

t confidence interval for a difference in means

The sample size is the number of elements in the sample.

The details of the data are;

The aim of the study = To seek the difference in the mean fuel economy (measured in miles per gallon) for vehicles under two treatment

1) Driving with under inflated tyres

2) Driving with properly inflated tyres

The number of cars in the sample = 12 (6 for each test)

The appropriate inference procedure, for the above data and test aim, therefore is the test for the confidence interval for the difference in means, and the sample size of less than 30, indicates that the is the student t confidence interval, the correct option is therefore;

t confidence interval for a difference in means

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Answer:

130.63

Step-by-step explanation:

The formula for a circle is:

Area = πr²

Let's plug in our values.

Area = 3.14(6.45)²

= 3.14(41.6025)

= 130.63185

Now we round to the nearest hundredth to get 130.63.

Hope this helps and good luck on your homework!

Answer:

130.62

Step-by-step explanation:

The formula for the area of a circle is [tex]\pi r^{2}[/tex].

We use 3.14 for pi and 6.45 for r:

3.14 · 6.45²

3.14 · 41.6

130.624, rounded to the nearest hundredth 130.62

1. The modal number of pairs of shoes owned by the children will be; 3.

2. The median number of pairs of shoes owned by the children  will be;3.

3. The Mean is 3.

1. The modal number of pairs of shoes owned by the children would be 3.

2. The median number of pairs of shoes owned by the children are;

= 14/2 th term

= 7 th term

= 3

3. The Mean would be

= (1 x 2+ 2 x 3+ 3 x 5+ 4 x 2 + 5 x 1+ 6x 1)/ (2 +3 +5 +2 + 1 +1)

= 42/14

= 3

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The state lottery commission should have $513,446.50 in the bank initially to guarantee the payments.

To determine how much money the state lottery commission should have in the bank initially to guarantee the payments, we will calculate the present value of the annuity.

Given:
- 20 installments of $50,000 per year
- First payment made immediately
- n = 19 payments at the end of each year
- Interest rate = 4% per year compounded yearly

Step 1: Calculate the present value of the annuity.
PV = PMT * [(1 - (1 + r)^(-n)) / r]
where:
PV = present value of the annuity
PMT = periodic payment amount ($50,000)
r = interest rate per period (4% per year or 0.04 as a decimal)
n = number of periods (19 years)

Step 2: Plug in the given values and solve for PV.
PV = $50,000 * [(1 - (1 + 0.04)^(-19)) / 0.04]
PV ≈ $50,000 * [1 - (1.04)^(-19)] / 0.04
PV ≈ $50,000 * [1 - 0.629243] / 0.04
PV ≈ $50,000 * [0.370757] / 0.04
PV ≈ $50,000 * 9.26893
PV ≈ $463,446.50

Step 3: Add the first payment to the present value.
Since the first payment is made immediately, the commission should have the present value of the remaining 19 payments plus the first payment of $50,000 in the bank initially.

Initial amount = PV + first payment
Initial amount = $463,446.50 + $50,000
Initial amount = $513,446.50

The state lottery commission should have $513,446.50 in the bank initially to guarantee the payments.

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