A researcher reports a significant treatment effect with t(15) - 2.56, p < .05. The study used a sample of n = 15 participants. True False

The estimated value of the slope is given by B. b1.

In a simple linear regression model with one predictor variable x, the slope coefficient is denoted as β1 in the population and estimated as b1 from the sample data. The slope represents the change in the response variable y for a unit increase in the predictor variable x. Therefore, b1 is the estimated value of the slope coefficient based on the sample data, and it can be used to make predictions for new values of x.

what is slope?

In mathematics and statistics, the slope is a measure of how steep a line is. It is also known as the gradient or the rate of change.

In the context of linear regression, the slope refers to the coefficient that measures the effect of an independent variable (often denoted as x) on a dependent variable (often denoted as y).

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Answer:

To determine the coordinates of the intercepts with the axes, we need to find the points where a graph intersects the x-axis (x-intercept) and the y-axis (y-intercept).

X-Intercept:

To find the x-intercept, we set y = 0 and solve for x. This means we are looking for the point(s) where the graph crosses the x-axis.

Y-Intercept:

To find the y-intercept, we set x = 0 and solve for y. This means we are looking for the point(s) where the graph crosses the y-axis.

Let's work through an example to illustrate this process:

Suppose we have an equation of a line: y = 2x + 3.

X-Intercept:

Setting y = 0:

0 = 2x + 3

2x = -3

x = -3/2

The x-intercept is (-3/2, 0).

Y-Intercept:

Setting x = 0:

y = 2(0) + 3

y = 3

The y-intercept is (0, 3).

Therefore, for the equation y = 2x + 3, the intercepts with the axes are (-3/2, 0) for the x-intercept and (0, 3) for the y-intercept.

We would predict an on-base percentage of approximately .290 for a player with a batting average of .206, assuming average values for walks, hit by pitch, and sacrifice flies.

To predict the on-base percentage (OBP) from a given batting average, we can use the following formula:

OBP = (Hits + Walks + Hit by Pitch) / (At Bats + Walks + Hit by Pitch + Sacrifice Flies)

Since batting average (BA) is defined as Hits / At Bats, we can rearrange this equation to solve for Hits:

Hits = BA * At Bats

Substituting this expression for Hits in the OBP formula, we get:

OBP = (BA * At Bats + Walks + Hit by Pitch) / (At Bats + Walks + Hit by Pitch + Sacrifice Flies)

Now we can plug in the given batting average of .206 and solve for OBP:

OBP = (.206 * At Bats + Walks + Hit by Pitch) / (At Bats + Walks + Hit by Pitch + Sacrifice Flies)

Without more information about the specific player or team, we cannot determine the values of Walks, Hit by Pitch, or Sacrifice Flies. However, we can make a prediction based solely on the batting average. Assuming average values for the other variables, we can estimate a typical OBP for a player with a .206 batting average.

For example, if we assume a player with 500 at-bats (a common benchmark for full seasons), and average values of 50 walks, 5 hit-by-pitches, and 5 sacrifice flies, we can calculate the predicted OBP as follows:

OBP = (.206 * 500 + 50 + 5) / (500 + 50 + 5 + 5)

= (103 + 50 + 5) / 560

= 0.29

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Juanita opens a new account in the bank and deposits 7,500. The bank pays 1.2% interest compounded annually on the account. Laura makes no additional deposits or withdrawals.

We are required to find the account balance at the end of 5 years  .Step 1: Calculate the compound interest earned for the first year. Interest for the first year will be: [tex]I = P × R × T= 7,500 × 1.2% × 1= 90[/tex]Step 2: Add the compound interest to the principal to find the new balance. Therefore, after the first year the balance will be 7,590. Step 3: Now, the balance of the account at the end of 5 years will be: Balance = [tex]P(1 + R/100)T= 7,500(1 + 1.2/100)5= 7,959.93.[/tex]Thus, 7,960.93 is the closest to the account balance at the end of 5 years. Therefore, option C is correct.

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The travel time T between home and office is expected to be between 20 and 40 minutes depending upon traffic. Based on experience, the average travel time is 30 minutes and the corresponding variance is 20 minutes.What is the expected value of the travel time?The expected value of the travel time is the average of the travel time between the home and office, which is given as 30 minutes.What is the standard deviation of the travel time?The standard deviation of the travel time is the square root of the variance which is given as follows:Variance = 20 minutesStandard deviation = √Variance= √20= 4.47 minutes.What is the probability of travel time being less than 25 minutes?Let X be the random variable for travel time between home and office.X ~ N(30, 20)We need to find P(X < 25).First, we find the z-score as follows:z = (x - μ) / σz = (25 - 30) / 4.47z = -1.12Using a standard normal distribution table, we can find the probability as:P(X < 25) = P(Z < -1.12) = 0.1314Therefore, the probability of travel time being less than 25 minutes is 0.1314.

a) The expected travel time is : 30 minutes.

b) The standard deviation of travel times is: 4.47 minutes

c) The probability that the travel time is less than 25 minutes is 0.1314.  

a) The expected travel time is simply the average travel time between home and office, given as 30 minutes.

b) The standard deviation of travel times is simply the square root of the variance and is expressed as:

Difference = 20 minutes

therefore:

standard deviation = √variance

standard deviation = √20

Standard deviation = 4.47 minutes.

c) Let X be the random variable for travel time between home and office. X to N(30,20)

I need to find P(X < 25).

First, find the Z-score from the following formula:

z = (x - μ)/σ

z = (25 - 30)/4.47

z = -1.12

The probabilities from the online p-values ​​in the Z-score calculator are:

P(X < 25) = P(Z < -1.12) = 0.1314

Therefore, the probability that the travel time is less than 25 minutes is 0.1314.  

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Complete question is:

The travel time T between home and office is expected to be between 20 and 40 minutes depending upon traffic. Based on experience, the average travel time is 30 minutes and the corresponding variance is 20 minutes.

What is the expected value of the travel time?

What is the standard deviation of the travel time?

What is the probability of travel time being less than 25 minutes?

The y-intercept of Jacob's graph representing the relationship between time and money is $60. If Jacob doesn't deposit any additional money into the account, he will have $79.49 in eight years, rounded to the nearest cent.

In the given equation, 60(1 + 0.04)x, the initial deposit of $60 is represented by the coefficient 60. The term (1 + 0.04) represents the factor by which the initial amount is multiplied each year, accounting for the 4% interest rate. The variable x represents the number of years.

The y-intercept of the graph represents the initial amount of money when x (the number of years) is 0. In this case, when Jacob hasn't invested for any years yet, his balance is the initial deposit of $60. Therefore, the y-intercept of Jacob's graph is $60.

To calculate the amount of money Jacob will have in eight years without any additional deposits, we can substitute x = 8 into the equation. The calculation would be 60(1 + 0.04)8. Evaluating this expression yields approximately $79.49. Rounding to the nearest cent, Jacob will have $79.49 in eight years without making any additional deposits.

In summary, the y-intercept of Jacob's graph is $60, and if he doesn't deposit any more money, he will have $79.49 in eight years.

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The line integral of the function f(x,y) = x²y³ -√x along the curve C, which is the arc of the curve y = √x from (0,0) to (4,2), has a value of -88/45.

To evaluate the line integral ∫C(x²y³ - √x) dy, where C is the arc of the curve y = √x from (0,0) to (4,2), we need to parameterize the curve and substitute the values into the integrand.

Let's parameterize the curve as x = t² and y = t, where t varies from 0 to 2. Then, dx/dt = 2t and dy/dt = 1.

Substituting these values into the integrand, we get:

(x²y³ - √x) dy = (t⁴t³ - t√t)dt

Integrating from t = 0 to t = 2, we get:

∫C(x²y³ - √x)dy = ∫0²(t⁷/2 - t³/²)dt

Evaluating this integral, we get:

Therefore, the value of the line integral is -88/45.

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The curve is concave downward on the interval (-∞, -1/5).

To determine the intervals where the curve y=∫(from 0 to x) (5t^2 + 2t + 7)dt is concave downward, we'll first find its second derivative. Since y is given as an integral, we can find the first derivative, y', by differentiating the integrand with respect to x:
y'(x) = 5x^2 + 2x + 7
Next, we'll find the second derivative, y''(x), by differentiating y'(x) with respect to x:
y''(x) = 10x + 2
Now, to find where the curve is concave downward, we need to determine where y''(x) is negative. To do this, we'll solve the inequality:
10x + 2 < 0
Subtract 2 from both sides:
10x < -2
Now, divide by 10:
x < -1/5
Therefore, the curve is concave downward on the interval (-∞, -1/5). In interval notation, this is written as:
Answer: (-∞, -1/5)

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The conditional probability that the sum is greater than 2 given that neither die is a 1 is 5/6.

To find the conditional probability, we need to first calculate the probability of the event "the sum of two fair 6-sided dice is greater than 2" and "neither die is a 1".

The probability of the sum being greater than 2 can be calculated by listing all the possible outcomes and counting the number of outcomes that satisfy the condition.

There are 36 possible outcomes, and the only outcomes that don't satisfy the condition are (1,1), so there are 35 outcomes that satisfy the condition.

Therefore, the probability of the sum being greater than 2 is 35/36.

The probability of neither die being a 1 can be calculated by considering the complementary event, which is the probability of at least one die being a 1.

The probability of one die being a 1 is 1/6, so the probability of at least one die being a 1 is 2/6 = 1/3 (since there are two dice).

Therefore, the probability of neither die being a 1 is 1 - 1/3 = 2/3.

Now, to find the conditional probability, we need to use Bayes' theorem:

P(sum > 2 | neither die is 1) = P(neither die is 1 | sum > 2) * P(sum > 2) / P(neither die is 1)

We have already calculated P(sum > 2) and P(neither die is 1), so we just need to find P(neither die is 1 | sum > 2).

To find P(neither die is 1 | sum > 2), we need to consider the outcomes that satisfy the condition "sum > 2".

There are 35 such outcomes, and of those, 10 have at least one 1 (namely, (1,2), (2,1), (1,3), (3,1), (1,4), (4,1), (1,5), (5,1), (1,6), and (6,1)). Therefore, the probability of neither die being a 1 given that the sum is greater than 2 is:
P(neither die is 1 | sum > 2) = (35 - 10) / 35 = 3/7

Plugging this and the previously calculated probabilities into Bayes' theorem, we get:
P(sum > 2 | neither die is 1) = (3/7) * (35/36) / (2/3) = 5/6

Therefore, the conditional probability that the sum is greater than 2 given that neither die is a 1 is 5/6.

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If the range on an X-bar-R chart suddenly and significantly increases, it indicates an increase in process variation. In this scenario, the mean (X-bar) may or may not be affected.

The mean represents the central tendency or average value of the process, while the range measures the dispersion or variation within the process.

If the mean remains stable and unaffected despite the increase in range, it suggests that the process average is still within control. However, if the range increase is accompanied by a significant shift in the mean, it indicates a potential shift in the process average.

To make a definitive determination, additional analysis and investigation are necessary to identify the underlying cause of the increased range and its impact on the process mean.

This could involve examining individual data points, performing hypothesis testing, or conducting further statistical analysis to assess the process stability and potential issues.

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After 19 hours, approximately 53.36 milligrams of the radioactive substance will remain.

To find out how many milligrams of the radioactive substance will remain after 19 hours, we need to use the exponential decay formula:  [tex]N(t) = N(0) (e)^{-λt}[/tex]
Where:
N(t) = amount of substance remaining at time t
N0 = initial amount of substance (160 mg)
e = base of natural logarithm (approximately 2.718)
λ = decay constant
t = time in hours

-First, we need to find the decay constant (λ). We know that after 12 hours, 80 mg of the substance remains:
[tex]80 = 160 e^{(-λ (12))}[/tex]
-Divide by 160: [tex]0.5 = e^{(-λ (12))}[/tex]
-Take the natural logarithm of both sides: [tex]ln(0.5) = 12 (-λ)[/tex]
-Now, find λ:  λ = [tex]λ = \frac{-ln(0.5)}{12}= 0.0578[/tex]

Next, we need to find the amount of substance remaining after 19 hours:
[tex]N(19) = 160 e^{(-0.0578)(19))}[/tex]
[tex]N(19) = 160 e^{(-1.0928)} = 160(0.3335)[/tex]
N(19) = 53.36 mg

So, after 19 hours, approximately 53.36 milligrams of the radioactive substance will remain.

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Answer:

115 persons in a population of 4,399 became ill with a disease of unknown etiology. The 115 cases occurred in 77 households.

Step-by-step explanation:

The answer of the given question based on the Triangle is the length of AC is approximately 12.81 centimeters (rounded to the nearest tenth of a centimeter).

We have,

The Pythagorean theorem is  mathematical principle that relates to three sides of right triangle. It states that in  right triangle, square of length of hypotenuse (side opposite the right angle) is equal to sum of the squares of the lengths of other two sides.

Since ABCD is a kite, we know that AC and BD are diagonals of the kite, and they intersect at right angles. Let E be the point where AC and BD intersect. Also, since DE = EB, we know that triangle EDB is Equilateral.

We can use Pythagorean theorem to find length of AC. Let's call length of AC "x". Then we have:

(AD)² + (CD)² = (AC)² (by Pythagorean theorem in triangle ACD)

Substituting the given values, we get:

(8)² + (10)² = (x)²

64 + 100 = x²

164 = x²

Taking square root of both sides, we will get:

x ≈ 12.81

Therefore, the length of AC is approximately 12.81 centimeters (rounded to the nearest tenth of a centimeter).

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Checking by differentiation,

y′ = -6sin(3t) + 12cos(3t)

y′′ = -18cos(3t) - 36sin(3t)

9y = y′ = -6sin(3t) + 12cos(3t)

y ′′ + 9y = 0

To verify that y=2cos3t+4sin3t is a solution to y ′′ +9y=0, we need to differentiate y twice and substitute the result into the differential equation.

First, we find the first derivative of y with respect to t:

y′ = -6sin(3t) + 12cos(3t)

Then, we take the second derivative of y with respect to t:

y′′ = -18cos(3t) - 36sin(3t)

Next, we substitute y′′ and y into the differential equation:

y′′ + 9y = (-18cos(3t) - 36sin(3t)) + 9(2cos(3t) + 4sin(3t))

Simplifying this expression, we get:

y′′ + 9y = -18cos(3t) - 36sin(3t) + 18cos(3t) + 36sin(3t)

y′′ + 9y = 0

Therefore, we have shown that y=2cos3t+4sin3t is a solution to y ′′ +9y=0, as the sum of the two terms reduces to 0 when substituted into the differential equation. This verifies that the function y satisfies the differential equation.

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Answer:

30 yards:6 inches = 5 yards per inch

(5 yards/inch)(2 inches) = 10 yards

The actual length of the east side is 10 yards.

The charge required to form 1.00 pound (454 g) of Al(s) from an Al³⁺ salt is 3) 1.62 x 10⁶ C.

To determine the charge required to form 1.00 pound (454 g) of Al(s) from Al³⁺ salt, we need to calculate the number of moles of Al and then convert it to coulombs using Faraday's constant.

Calculate the number of moles of Al:

Given mass of Al = 454 g

Molar mass of Al = 26.98 g/mol

Number of moles of Al = mass of Al / molar mass of Al

Number of moles of Al = 454 g / 26.98 g/mol ≈ 16.84 mol

Convert moles of Al to coulombs:

Given: 1 Faraday = 96,500 C

Charge (coulombs) = Number of moles of Al * Faraday's constant

Charge (coulombs) = 16.84 mol * 96,500 C/mol

Charge (coulombs) ≈ 1.62 x 10⁶ C

Therefore, the charge required to form 1.00 pound (454 g) of Al(s) from an Al³⁺ salt is approximately 1.62 x 10⁶ C (option 3).

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The probability of rolling a 1 on the first roll and a 6 on the second roll is 1/36.

Since each roll is independent of the other, the probability of the first roll being a 1 and the second roll being a 6 is the product of the probabilities of each event happening separately.

The probability of rolling a 1 on the first roll is 1/6, and the probability of rolling a 6 on the second roll is also 1/6. Therefore, the probability of both events occurring is:

1/6 × 1/6 = 1/36

So the probability of rolling a 1 on the first roll and a 6 on the second roll is 1/36.

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R = C (S/P)^k where C = ±C' is a constant of integration. This is the general solution to the differential equation.

To solve the differential equation dR/R = k S/P, we can separate the variables and integrate both sides with respect to their respective variables:

dR/R = k S/P

ln|R| = k ln|S/P| + C

where C is an arbitrary constant of integration. Exponentiating both sides, we get:

|R| = e^(k ln|S/P| + C)

|R| = e^(ln|S/P|^k) e^C

|R| = C' (S/P)^k

where C' = e^C is another arbitrary constant of integration. Since the absolute value of R is always positive, we can drop the absolute value signs and write:

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Answer:

Step-by-step explanation:

Step-by-step explanation:

Firstly, let's get the fractions out of mixed form.

2 1/5 = 11/5

(To do this, multiply 2 times 5 and add to the 1.)

1 5/6 = 11/6

(To do this, multiply 1 times 6 and add to the 5.)

Next, let's get the common denominator. When making a common denominator, keep in mind you must multiply/divide both the numerator and denominator

Answer:

-----------------

Given the quadratic equation:

Solve it in the following steps:

So the solution is: {- 3/2; 1/3}

Answer: 45 degrees

Step-by-step explanation: Over 5 on the x-axis and whatever point is above it on the y-axis which would be 45

To find the equation of the tangent plane to surface S at point (7,5,4), we first need to find the partial derivatives of the parameterization function r(u,v).
∂r/∂u = <3, 15u^2, 0>
∂r/∂v = <2, 0, 2v>
Evaluating these partial derivatives at (7,5,4), we get
∂r/∂u (7,5) = <3, 1875, 0>
∂r/∂v (7,5) = <2, 0, 8>
Next, we can find the normal vector to the tangent plane by taking the cross product of these partial derivatives:
N = ∂r/∂u x ∂r/∂v = <-15000, 6, -5625>
The equation of the tangent plane can then be written as:
-15000(x-7) + 6(y-5) - 5625(z-4) = 0
To set up the double integral that represents the surface area of S, we can use the formula:
Surface area = ∫∫ ||∂r/∂u x ∂r/∂v|| dA
where dA = ||∂r/∂u x ∂r/∂v|| du dv
Plugging in our parameterization function and taking the cross product of the partial derivatives as before, we get:
||∂r/∂u x ∂r/∂v|| = sqrt(2250000u^2 + 4v^2 + 42187500u^4)
So the surface area of S can be found by integrating this expression over the given ranges of u and v:
∫∫ sqrt(2250000u^2 + 4v^2 + 42187500u^4) du dv, 0 ≤ u ≤ 8, 0 ≤ v ≤ 6.

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The value of x is 4.

Since, △ABC≅ △EDF

We know by the property of Congruence

AB = DF

CB = DE

AC = FE

and, <A = <F, <B = <D, <C = <E

So, <A = <F

3x + 3=  5x - 7

3x - 5x = -7 - 3

-2x = -8

x = 4

Thus, the value of x is 4.

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The absolute maximum value of h(x) over the interval [-3,3] is 4.

To find the absolute maximum value, we need to look at the critical points and the endpoints of the interval. Taking the derivative of h(x) and setting it equal to 0, we get 4x-1=0. Solving for x, we get x=1/4.

Plugging this value into h(x), we get h(1/4)=-15/8. However, this is not within the interval [-3,3], so we need to evaluate h(-3), h(3), and h(1/4). We find that h(-3)=10, h(3)=-16, and h(1/4)=-15/8.

Therefore, the absolute maximum value of h(x) over the interval [-3,3] is 4, which occurs at x=-1/2.

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The probability that less than 30 out of 100 people with Paroxysmal Nocturnal Hemoglobinuria (PNH) will survive is 0.000162.

In a sample of 100 PNH patients, the probability of an individual recovering from the disease is 0.40. We can calculate the probability of less than 30 survivors using the binomial probability formula. Let X represent the number of survivors, and using the formula, we find P(X < 30) = Σ P(X = k) for k = 0 to 29. This probability is calculated as 0.000162, indicating an extremely low likelihood.

In this case, the probability of an individual recovering from PNH is given as 0.40. We can apply the binomial probability formula to determine the likelihood of having less than 30 survivors out of the 100 patients. This involves summing up the individual probabilities of having 0, 1, 2,..., 29 survivors. After performing the calculations, we find that the probability of less than 30 survivors is 0.000162, or approximately 0.0162%.

This extremely low probability suggests that the chances of fewer than 30 individuals surviving out of the 100 PNH patients are quite slim. It highlights the severity and life-threatening nature of the disease, emphasizing the need for timely and effective medical interventions to improve patient outcomes.

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To find the curl of F, we need to compute the determinant of the following matrix:

| i    j    k   |

| ∂/∂x ∂/∂y ∂/∂z |

| 2xy  10y  7z  |

Expanding the determinant, we get:

i(7 - 0) - j(0 - 0) + k(0 - 20x)

= (7 - 20x)k

Therefore, the curl of F is (0, 0, 7 - 20x).

To check if there is a function f such that F = ∇f, we need to compute the partial derivatives of each component of F with respect to the corresponding variable. If these partial derivatives are equal, then there exists a scalar function f such that F = ∇f.

∂F_x/∂y = 2x

∂F_y/∂x = 2x

Since these partial derivatives are not equal, there is no function f such that F = ∇f. Therefore, the answer is "no" (n).

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Any elementary row operation on A can be expressed as a product of replacement operations on A. This means that A can be row reduced to U using only replacement operations, which is the converse of what was proved in the text.

The LU factorization of a matrix A involves decomposing it into a lower triangular matrix L and an upper triangular matrix U, such that A = LU. This means that A can be written as the product of two triangular matrices, one of which is lower triangular and the other is upper triangular.

To show that A can be row reduced to U using only replacement operations, we need to prove that any elementary row operation performed on A can be expressed as a product of replacement operations on A.

First, consider the operation of multiplying a row of A by a scalar. This is a replacement operation, since it replaces one row of A with a multiple of itself.

Next, consider the operation of adding a multiple of one row of A to another row. This is also a replacement operation, since it replaces one row of A with a linear combination of itself and another row.

Finally, consider the operation of interchanging two rows of A. This can be expressed as a sequence of replacement operations: first, add one row to the other, then subtract the original row from the first row, and finally add the second row back to the first row.

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To determine whether the exponential function represents growth or decay, we need to examine the base of the exponent, which is 0.909 in this case.

If the base is greater than 1, it represents growth. If the base is between 0 and 1, it represents decay.

In this case, the base is 0.909, which is less than 1. Therefore, the exponential function represents decay.

To determine the percentage rate of decrease, we can calculate the percentage decrease per unit change in x. In this case, the base of the exponent represents the rate of decrease.

The percentage rate of decrease can be found by subtracting the base from 1 and multiplying by 100.

Percentage rate of decrease = (1 - 0.909) * 100 = 0.091 * 100 = 9.1%

Therefore, the exponential function represents decay with a percentage rate of decrease of 9.1%.

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Answer:

  -1 ≤ x ≤ 2

Step-by-step explanation:

You want the solution to |x +1| +|x -2| = 3.

We find it convenient to solve these absolute value equations using a graphing calculator. When we subtract 3 from both sides, we have ...

  |x +1| +|x -2| -3 = 0

The solutions will show on the graph as places where the expression has a value of 0, that is, the x-intercepts.

The left-side expression has a value of 0 for all values of x between -1 and +2, inclusive. That is, the solution is ...

  -1 ≤ x ≤ 2

The absolute value function is piecewise defined:

  |x| = x . . . . for x ≥ 0

  |x| = -x . . . . for x < 0

That is, the behavior of the function changes at x=0.

In the given equation the absolute value function arguments are zero at ...

  x +1 = 0   ⇒   x = -1

  x -2 = 0   ⇒   x = 2

These x-values divide the domain of the equation into three parts.

In this domain, both arguments are negative, so the equation is actually ...

  -(x +1) -(x -2) = 3

  -2x +1 = 3

  -2x = 2

  x = -1 . . . . . . not in the domain

In this domain, the argument (x+1) is positive, but the argument (x-2) is negative. That means the equation is ...

  (x +1) -(x -2) = 3

  1 +2 = 3

True for all x in this domain.

In this domain, both arguments are positive, so the equation is ...

  (x +1) +(x -2) = 3

  2x -1 = 3

  2x = 4

  x = 2 . . . . in the domain (this point was excluded from x < 2).

The solution is -1 ≤ x ≤ 2.

In order to determine how many x values will be listed in the first column of a frequency distribution table for a sample of n = 12 scores that ranges from a high of x = 7 to a low of x = 4, we need to first determine the range of the data.

The range is simply the difference between the highest and lowest scores in the sample, which in this case is 7 - 4 = 3.
Next, we need to determine the width of the intervals that will be used in the frequency distribution table. A common rule of thumb is to use intervals that are approximately equal to the square root of the sample size. For a sample size of 12, this would suggest using intervals that are approximately 3 wide (since the square root of 12 is 3.464).Based on this information, we can create intervals that range from 4-6, 7-9, etc. There will be 2 intervals (4-6 and 7-9), which means that there will be 2 x values listed in the first column of the frequency distribution table.Alternatively, we could use narrower intervals, such as 4-4.9, 5-5.9, 6-6.9, 7-7.9, 8-8.9, and 9-9.9. In this case, there would be 6 intervals and 6 x values listed in the first column of the frequency distribution table. However, the intervals would be quite narrow and may not provide a very useful summary of the data.

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