now suppose of interest is to estimate the mean number of children attending all 2022 veterans day celebrations. for this problem only, assume that the standard deviation of the number of children attending all 2022 veterans day celebrations is 21. what is the minimum number of 2022 veterans day celebrations that would need to be selected for the sample to allow the calculation of a 98% confidence interval with margin of error no larger than 8.

In the multi-armed bandit problem, there are a number of slot machines (arms) each with an unknown probability distribution of rewards.

The task is to maximize the total reward over a number of trials, where in each trial, an arm is selected and a reward is received.

In this specific problem, there are two arms, and the rewards for each arm follow a normal distribution with different means. The explore-then-commit and UCB algorithms are used to choose which arm to select in each trial.

The explore-then-commit algorithm chooses an arm randomly for an initial set of trials and then chooses the arm with the highest average reward for the remaining trials.

The UCB algorithm chooses the arm with the highest upper confidence bound, which takes into account both the average reward and the uncertainty in that estimate.

The problem asks to simulate the algorithms for 2000 rounds and calculate the average pseudoregret, which is the difference between the total reward obtained and the reward that would have been obtained if the best arm had been chosen every time.

The problem also asks to plot a histogram of the rewards obtained for each algorithm and compare the distributions. The differences between the distributions can be justified based on the different exploration-exploitation trade-offs made by each algorithm.

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a) The translations are given as follows:

b) The equation is given as follows:

[tex]y = \sqrt[3]{x - 2} + 4[/tex]

A translation happens when either a figure or a function is moved horizontally or vertically on the coordinate plane.

The four translation rules for functions are defined as follows:

The parent function is given as follows:

[tex]y = \sqrt[3]{x}[/tex]

Which has a vertex at the origin.

The vertex of the graph has the coordinates given as follows:

(2,4).

Hence the translations are:

Meaning that the function is given as follows:

[tex]y = \sqrt[3]{x - 2} + 4[/tex]

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The number of different ways to select 6 numbers out of a total of 36 numbers in the lottery are :

1,947,792

To solve this problem, we can use the combination formula which is:
nCr = n! / r!(n-r)!
Where n is the total number of items, r is the number of items to be selected, and ! represents the factorial function.

Using this formula, we can plug in the given values:

n = 36 (total number of numbers in the lottery)
r = 6 (number of numbers to be selected)

nCr = 36! / 6!(36-6)!
nCr = 36! / 6!30!
nCr = (36x35x34x33x32x31) / (6x5x4x3x2x1)
nCr = 1,947,792

Therefore, there are 1,947,792 different ways to select 6 numbers out of a total of 36 numbers in the lottery.

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The upper limit of the tolerance 43/4 plus or minus 1/64 is 11.078125.

A tolerance is a specified range within which a measurement or dimension is expected to be. In this case, the tolerance is given as 43/4 plus or minus 1/64. To find the upper limit of the tolerance, we add the maximum allowable deviation of 1/64 to the nominal value of 43/4, which gives us:

43/4 + 1/64 = 11.078125

Therefore, the upper limit of the tolerance is 11.078125.

This means that any value within the range of 43/4 plus or minus 1/64 is acceptable, as long as it does not exceed this upper limit.

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A one-way ANOVA is a statistical test used to compare the means of four populations in this case. Thus, the degrees of freedom for the F-statistic in this one-way ANOVA are (df1, df2) [tex]= (3, 78)[/tex].

The degrees of freedom are essential for calculating the F-statistic, which helps determine whether there are significant differences among the population means.

To calculate the degrees of freedom for the F-statistic, we need to determine the numerator degrees of freedom (df1) and the denominator degrees of freedom (df2). For df1 (numerator degrees of freedom), it is equal to the number of populations (groups) minus 1.

In this case, there are four populations,

So: df1 [tex]= 4 - 1 = 3[/tex]

For df2 (denominator degrees of freedom), it is equal to the total sample size (N) minus the number of populations (groups).

The total sample size is given as [tex]18 + 22 + 20 + 22 = 82[/tex].

So: df2 [tex]= 82 - 4 = 78[/tex]

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In both symbolizations, we use universal quantifiers (∀) to represent "any," conjunction (∧) for "and," implication (→) for "is better than," negation (¬) for "not," and parentheses to group the conditions.

Let's symbolize each English-language sentence into First-Order Logic (FOL) using the given symbolization key:

Domain: Candies

C(x): x has chocolate in it.

M(x): x has marzipan in it.

S(x): x has sugar in it.

T(*): Boris has tried * (denotes any candy).

B(x, y): x is better than y.

Any candy with chocolate is better than any candy without it.

Symbolization: ∀x ∀y ((C(x) ∧ ¬C(y)) → B(x, y))

Any candy with chocolate and marzipan is better than any candy that lacks both.

Symbolization: ∀x ∀y ((C(x) ∧ M(x) ∧ ¬(C(y) ∧ M(y))) → B(x, y))

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The value of the definite integral ∫[a to b] 60te^(-t) dt is -60be^(-b) + 60ae^(-a) - 60e^(-b) + 60e^(-a).

To evaluate the definite integral ∫60te^(-t) dt using integration by parts, we will apply the integration by parts formula:

∫u dv = uv - ∫v du,

where u and v are functions of t, and du and dv are their respective differentials.

Let's assign u = t and dv = 60e^(-t) dt. Then, we can determine du and v:

du = dt (the differential of u)
v = ∫dv = ∫60e^(-t) dt

To find v, we can integrate dv:

∫60e^(-t) dt = -60e^(-t)

Now, we can apply the integration by parts formula:

∫60te^(-t) dt = uv - ∫v du
= t(-60e^(-t)) - ∫(-60e^(-t)) dt
= -60te^(-t) + 60∫e^(-t) dt
= -60te^(-t) - 60e^(-t) + C,

where C is the constant of integration.

Since we are evaluating a definite integral, we can now substitute the limits of integration into the antiderivative expression and compute the result. Let's assume the limits are a and b:

∫[a to b] 60te^(-t) dt = [-60te^(-t) - 60e^(-t)] evaluated from a to b
= (-60be^(-b) - 60e^(-b)) - (-60ae^(-a) - 60e^(-a))
= -60be^(-b) + 60ae^(-a) - 60e^(-b) + 60e^(-a).

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By the Intermediate Value Theorem, since f(-4) = 51/4 > 0 and f(-3) = -3/4 < 0, there exists a value c between -4 and -3 such that f(c) = 0. Approximating the zero using bisection method yields c ≈ -3.7321.

The Intermediate Value Theorem states that if a continuous function f(x) changes sign over an interval [a, b], then it has a zero in that interval. In this case, we have shown that f(x) = 3x^2/(9x^2-3x+9) changes sign over the interval [-4,-3], meaning that it has a zero in that interval.

To approximate the zero, the bisection method is used. This method involves repeatedly dividing the interval in half and testing which half the zero lies in. In this case, we start with the interval [-4,-3] where we know a zero exists, and we test the midpoint of the interval, which is (-4-3)/2 = -3.5.

If f(-3.5) is positive, then we know the zero lies in the interval [-3.5,-3]. If f(-3.5) is negative, then we know the zero lies in the interval [-4,-3.5]. We continue dividing the interval in half and testing until we reach a desired level of accuracy.

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Answer:

do it yo self lazy

Step-by-step explanation:

The length of side j in triangle IJK, to the nearest tenth of an inch, is 252.6 inches

To find the length of side j in triangle IJK, we can use the Law of Sines, which states that the ratio of the length of a side to the sine of its opposite angle is constant.

Using the given information:

Side k = 590 inches

Angle ∠I = 86°

Angle ∠J = 29°

We can set up the following proportion:

j / sin(∠J) = k / sin(∠I)

Plugging in the known values:

j / sin(29°) = 590 / sin(86°)

Now we can solve for j:

j = (sin(29°) * 590) / sin(86°)

j ≈ 252.57 inches

Rounded to the nearest tenth of an inch, the length of side j is approximately 252.6 inches.

Therefore, the length of side j in triangle IJK, to the nearest tenth of an inch, is 252.6 inches.

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Therefore, the indicated partial derivative is fy(2, 1) ≈ -1.476.

To find the indicated partial derivative, we need to take the derivative of the function with respect to y, treating x as a constant, and then evaluate it at the point (2, 1).

So, we start by finding the partial derivative of f with respect to y:

f_y(x, y) = x cos^−1(xy)

Next, we evaluate this expression at the point (2, 1):

f_y(2, 1) = 2 cos^−1(2) ≈ -1.476

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Let alpha be an acute angle such that tan(alpha) = 3/4. Then:

sin(alpha) = 3/5 and cos(alpha) = 4/5.

We can use the Pythagorean identity for tangent to find the values of sine and cosine. Specifically, if tan(alpha) = opposite/adjacent, then we can let opposite = 3 and adjacent = 4. Using the Pythagorean theorem, we can find the hypotenuse to be 5.

Then, we can write sin(alpha) = opposite/hypotenuse = 3/5 and cos(alpha) = adjacent/hypotenuse = 4/5. These values allow us to compute other trigonometric functions such as secant, cosecant, and cotangent. For example, sec(alpha) = 1/cos(alpha) = 5/4 and csc(alpha) = 1/sin(alpha) = 5/3.

The values of sine and cosine are important in many applications of trigonometry, including geometry, physics, and engineering.

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Therefore, the coefficient of x^4y^6 in the expansion of (5x - 4y)^10 is 543948800.

To find the coefficient of x^4y^6 in the expansion of (5x - 4y)^10, we can use the binomial theorem or Pascal's triangle. Specifically, the coefficient of x^4y^6 is equal to the product of the coefficients of x^4 and y^6 in the binomial expansion of (5x - 4y)^10.

Using the binomial theorem, we can expand (5x - 4y)^10 as follows:

(5x - 4y)^10 = C(10, 0) (5x)^10 (-4y)^0 + C(10, 1) (5x)^9 (-4y)^1 + C(10, 2) (5x)^8 (-4y)^2 + ... + C(10, 10) (5x)^0 (-4y)^10

where C(n, r) denotes the binomial coefficient, or the number of ways to choose r items from a set of n distinct items. The formula for C(n, r) is:

C(n, r) = n! / (r! (n - r)!)

Using this formula and simplifying each term in the expansion, we can write the coefficient of x^4y^6 as:

C(10, 4) (5)^4 (-4)^6 = 210 * 625 * 4096 = 543948800

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(a) P(2,4) = P(2) + P(4) = 0.18 + 0.36 = 0.54

(b) P(1,3,5) = P(1) + P(3) + P(5) = 0.07 + 0.25 + 0.14 = 0.46

(c) The primes in the sample space are 2 and 3, so P(prime) = P(2) + P(3) = 0.18 + 0.25 = 0.43.

(a) The probability of {2, 4} is the sum of the probabilities of selecting 2 and 4 individually: P({2, 4}) = P(2) + P(4) = 0.18 + 0.36 = 0.54.

(b) The probability of {1, 3, 5} is the sum of the probabilities of selecting 1, 3, and 5 individually: P({1, 3, 5}) = P(1) + P(3) + P(5) = 0.07 + 0.25 + 0.14 = 0.46.

(c) The prime numbers in the sample space are 2, 3, and 5. Therefore, the probability of selecting a prime is the sum of the probabilities of selecting 2, 3, and 5 individually: P(prime) = P(2) + P(3) + P(5) = 0.18 + 0.25 + 0.14 = 0.57.

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The probability that a sample of 33 randomly selected adults in the US has a mean weight greater than 151.52 lbs is:

38.21%.

To find the probability that a sample of 33 randomly selected adults in the US has a mean weight greater than 151.52 lbs, we need to use the Central Limit Theorem. Assuming that the population standard deviation is known or the sample size is large enough (greater than or equal to 30), the distribution of the sample means will be approximately normal.

First, we need to calculate the standard error of the mean (SEM) using the formula:

SEM = standard deviation / square root of sample size

Let's assume that the population standard deviation is 30 lbs (this is just an estimate) and the sample size is 33. Then,

SEM = 30 / sqrt(33) = 5.22 lbs

Next, we need to calculate the z-score using the formula:

z = (sample mean - population mean) / SEM

Let's assume that the population mean is 150 lbs (again, just an estimate). Then,

z = (151.52 - 150) / 5.22 = 0.30

Finally, we can use a standard normal distribution table or calculator to find the probability that the z-score is greater than 0.30. This is equivalent to finding the area under the standard normal curve to the right of 0.30. Using a standard normal distribution table or calculator, we find that the probability is approximately 0.3821 or 38.21%.

Therefore, the probability that a sample of 33 randomly selected adults in the US has a mean weight greater than 151.52 lbs is approximately 38.21%.

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The p-value is 2 * 0.0708 = 0.1416.

To find the p-value associated with a z test statistic of 1.47 in a two-tailed test, we need to calculate the probability of observing a test statistic as extreme or more extreme than 1.47 in both tails of the standard normal distribution.

The p-value is the probability of observing a test statistic as extreme or more extreme than the observed value. In a two-tailed test, we need to consider both tails.

To find the p-value, we can use a standard normal distribution table or a statistical calculator. In this case, using a standard normal distribution table, we find that the area to the right of 1.47 is 0.0708.

Since it is a two-tailed test, we need to consider both tails. Therefore, the total p-value is twice the area in one tail.

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The probability that an individual energy bar contains less than 42.035 grams is approximately 0.2743.

(a) To find the probability that an individual energy bar contains less than 42.035 grams, we need to calculate the z-score and then use the standard normal distribution table or a calculator.

The z-score can be calculated as:

z = (x - μ) / σ

where x is the value we are interested in (42.035 grams), μ is the mean (42.05 grams), and σ is the standard deviation (0.025 grams).

z = (42.035 - 42.05) / 0.025 ≈ -0.6

Using the standard normal distribution table or a calculator, we can find that the probability of getting a z-score less than -0.6 is approximately 0.2743.

Therefore, the probability that an individual energy bar contains less than 42.035 grams is approximately 0.2743.

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The solutions to the original equation |2x + 7| = 6 are x = -1/2 and x = -13/2.

The equation |2x + 7| = 6 can be written as two separate equations as follows;

2x + 7 = 6  ----- (1)

2x + 7 = -6  ------ (2)

Solving equation 1 for x, we have the following;

2x + 7 = 6

2x = -1

x = -1/2

Solving equation 2 for x, we have the following;

2x + 7 = -6

2x = -13

x = -13/2

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The recurrence relation for the number of n-digit quaternary sequences with at least one 1 and the first 1 occurring before the first 0 is:

[tex]A_n = 2 * A_{n-1}[/tex]

Let's denote the number of n-digit quaternary sequences satisfying the given conditions as [tex]A_n[/tex]. We can establish a recurrence relation for A_n based on the position of the first digit in the sequence.

Consider the possible cases:

1. Case 1: The first digit is 1.

  In this case, the remaining n-1 digits can be any valid (n-1)-digit quaternary sequence that satisfies the given conditions.

Since the first 1 has already been placed, the remaining sequence can have any combination of digits 1, 2, and 3. Hence, the number of such sequences is [tex]A_{n-1}[/tex].

2. Case 2: The first digit is not 1.

  If the first digit is not 1, it must be 2 or 3 (as per the given conditions). In this case, the remaining n-1 digits can be any valid (n-1)-digit quaternary sequence satisfying the given conditions, without any restriction on the placement of 0s and 1s.

Hence, the number of such sequences is [tex]A_{n-1}[/tex].

Therefore, the recurrence relation for the number of n-digit quaternary sequences with at least one 1 and the first 1 occurring before the first 0 is:

[tex]A_n = 2 * A_{n-1}[/tex]

This recurrence relation states that the number of n-digit sequences satisfying the given conditions is twice the number of (n-1)-digit sequences satisfying the same conditions.

We also have the base case A₁ = 1, as there is only one valid 1-digit quaternary sequence that satisfies the given conditions, which is "1".

Using this recurrence relation, you can compute the number of n-digit quaternary sequences with the specified conditions for any given value of n.

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After 11 years, we can expect to observe approximately 4146 monk parakeets in the united states.

to write an exponential growth model for this situation, we can use the formula:

n(t) = n0 * (1 + r)ᵗ

where:

- n(t) is the number of monk parakeets after t years- n0 is the initial number of monk parakeets (in 1993)

- r is the annual growth rate, expressed as a decimal (e.g. 0.13 for 13%)- t is the number of years since 1993

so for this problem, we have:

- n0 = 1219

- r = 0.13- t = 11 (to find the value after 11 years) and t = 20 (to find the value after 20 years)

using these values, we can calculate n(t) as follows:

- after 11 years:

n(11) = 1219 * (1 + 0.13)¹¹ = 4146.16 - after 20 years:

n(20) = 1219 * (1 + 0.13)²⁰ = 12979.05

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Answer:

x - 3

Step-by-step explanation:

[tex]-\frac{9}{x+3}[/tex] + [tex]\frac{x^{2}}{x+3}[/tex]

=  [tex]-\frac{9-x^{2}}{x+3}[/tex]

=     [tex]-\frac{3^2-x^2}{x+3}[/tex]

=     [tex]-\frac{(3-x)(x+3)}{x+3}[/tex]

=     [tex]-\frac{(3-x)}1}[/tex]

= x - 3

The change-of-coordinates matrix from B to C is P_B→C, and the change-of-coordinates matrix from C to B is P_C→B.

To find the change-of-coordinates matrix from B to C, we need to express each vector in the basis C as a linear combination of the vectors in the basis B.

We can represent this relationship using a matrix called the transition matrix.

Given B = {b1, b2} and C = {c1, c2}, with b1 = [1,1], b2 = [1,2], c1 = [2,3], and c2 = [3,4], we can set up the following system of linear equations:

c1 = a * b1 + b * b2

c2 = c * b1 + d * b2

Solving this system, we get a = 1, b = 1, c = 2, and d = 1.

Thus, the transition matrix from B to C, denoted as P_B→C, is: P_B→C = | 1 2 | | 1 1 |

To find the change-of-coordinates matrix from C to B (P_C→B), we simply find the inverse of P_B→C:

P_C→B = (P_B→C)^(-1) = | 1 -2 | | -1 1 |

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To determine the number of terms we need to add in order to find the sum of the series with an error less than or equal to 0.002, we can use the concept of the Alternating Series Estimation Theorem.

The Alternating Series Estimation Theorem states that for an alternating series with terms decreasing in absolute value, the error in approximating the sum of the series by the sum of a finite number of terms is less than or equal to the absolute value of the next term.

In the given series, ∑n=1[infinity](−1)n−1n^2, the terms are decreasing in absolute value as n increases, so we can apply the Alternating Series Estimation Theorem.

The next term in the series, which determines the error, is given by (-1)^n * (n+1)^2. To ensure the error is less than or equal to 0.002, we need to find the smallest value of n such that:

|(-1)^n * (n+1)^2| ≤ 0.002

We can solve this inequality by testing values of n until we find the smallest value that satisfies it. Starting with n = 1:

|(-1)^1 * (1+1)^2| = 4 > 0.002

The inequality is not satisfied for n = 1.

Let's continue testing values of n:

|(-1)^2 * (2+1)^2| = 9 > 0.002

|(-1)^3 * (3+1)^2| = 16 > 0.002

|(-1)^4 * (4+1)^2| = 25 > 0.002

|(-1)^5 * (5+1)^2| = 36 > 0.002

|(-1)^6 * (6+1)^2| = 49 > 0.002

...

After testing a few values, we can see that the smallest value of n that satisfies the inequality is n = 9:

|(-1)^9 * (9+1)^2| = 100 > 0.002

Therefore, we need to add at least 9 terms in order to find the sum of the series with an error less than or equal to 0.002.

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The total ounces of flower food used in the 6 centerpieces is 6 ounces.

Since each centerpiece uses 1 ounce of flower food, the total amount of flower food used in 6 centerpieces is 6 times 1, which equals 6 ounces. To calculate the total amount of flower food used, we simply multiply the amount used in one centerpiece by the total number of centerpieces.

Therefore, the answer is 6 ounces.

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The left cosets of {1, 11} in u(30) are: {{1, 11}, {7, 17}, {13, 23}, {19, 29}}.

Here, u(30) represents the group of integers that are relatively prime to 30 under multiplication.

To find the left cosets of {1, 11} in u(30), we need to find all the possible subsets of u(30) that are of the form a(1,11) = {a, a*11} for some integer a.

First, we can list the elements of u(30), which are: {1, 7, 11, 13, 17, 19, 23, 29}.

Next, we can choose an integer a that is relatively prime to 30 and form the subset a(1,11) as follows:

If a = 1, then a(1,11) = {1, 11}.

If a = 7, then a(1,11) = {7, 17}.

If a = 11, then a(1,11) = {11, 1}.

If a = 13, then a(1,11) = {13, 23}.

If a = 17, then a(1,11) = {17, 7}.

If a = 19, then a(1,11) = {19, 29}.

If a = 23, then a(1,11) = {23, 13}.

If a = 29, then a(1,11) = {29, 19}.

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The solution to the initial value problem is y(t) = -e^(2t) + te^(2t).

We have,

The characteristic equation is r² - 4r + 4 = 0, which can be factored as

(r - 2)² = 0.

Hence, the roots are r = 2 (multiplicity 2).

The general solution of the differential equation is then

y(t) = c1 e^(2t) + c2 t e^(2t),

where c1 and c2 are constants.

Using the initial conditions, we have y(0) = -1, which gives c1 = -1.

Also,

dy/dt(0) = -2, which gives c2 = 1.

Therefore,

The solution to the initial value problem is y(t) = -e^(2t) + te^(2t).

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The two incorrect statements are: c. f is onto ⇔ ∀y ∈ Y, 3x ∈ X such that f(x)= y. (The correct quantifier is ∃, not ∀.). d. f is onto ⇔ ∀x ∈ X, 3y ∈ Y such that f(x)= y. (The correct quantifier is ∃, not ∀.)

The correct statements are: a. function f is onto ⇔ every element in its co-domain is the image of some element in its domain. b. f is onto ⇔ every element in its domain has a corresponding image in its co-domain. e. f is onto ⇔ the range of f is the same as the co-domain of f.

To further explain the correct statements:

a. This statement is a direct definition of an onto function. It means that for every element y in the co-domain of f, there exists some element x in the domain of f such that f(x) = y.

b. This statement is equivalent to statement a and is also a direct definition of an onto function. It means that for every element x in the domain of f, there exists some element y in the co-domain of f such that f(x) = y.

e. This statement is also equivalent to statements a and b. The range of f is the set of all possible outputs that f can produce, while the co-domain of f is the set of all possible outputs that f could produce. Therefore, if the range of f is equal to its co-domain, then every possible output of f is actually produced by f, which means that f is onto.

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Gradual, long-term movement in time-series values is called temporal variation. Therefore, the correct option is (a) temporal variation.

Temporal variation is one of the components of a time series, along with seasonal variation, cyclical movement, and irregular or random variation. Temporal variation refers to the long-term trend or pattern of movement in the time series, which can be increasing, decreasing, or remaining constant over time. It is often caused by underlying factors such as population growth, economic changes, or technological advancements.

Cyclical movement refers to the regular, repeating up and down patterns in the time series, which are usually longer than a year and can be caused by factors such as business cycles or political cycles. Exponential smoothing and linear regression are statistical methods used to forecast or model time series data, but they do not describe the components of the time series themselves.

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Final grade and missed homework negatively correlated.

A negative correlation would be expected between the final points grade in SEL and the number of homework points missed. This means that as the number of homework points missed increases, the final points grade in SEL is likely to decrease. In other words, students who miss more homework points are likely to have lower final points grades.

This negative correlation can be explained by the fact that homework assignments are usually designed to help students understand the concepts covered in class and to prepare them for exams. Students who miss homework assignments may struggle to keep up with the course material, leading to lower grades.

Additionally, homework assignments often contribute to a significant portion of the overall grade, so missing multiple assignments can have a significant impact on a student's final grade. By examining the relationship between these two variables, educators can identify students who are at risk of falling behind and provide them with additional support to ensure their success in the course.

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Ace had 0.625 or 62.5 riders two weeks ago.

We can start by setting up a proportion:

fullness = riders / capacity

where the fullness is 62.5% or 0.625, and the capacity is 100% or 1 (assuming the bus is completely full at 100%).

We can solve for the number of riders:

0.625 = riders / 1

riders = 0.625 x 1

riders = 0.625

So Ace had 0.625 or 62.5 riders two weeks ago. If we assume that riders must be a whole number, we can round up to 63 riders.

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