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An eagle is flying at an altitude of 280 feet. The angle of depression from the eagle to a mouse on the ground is 15º. What is the direct line distance from the eagle to the mouse? Round answer to the nearest thousandth.

Answer:

She can make 15 bracelets

Step-by-step explanation:

115 / 7.5 = 15.333

She doesn't have enough to make more then 115 as there not be 1/3 of a bracelet

the expected number of students in the randomly chosen student's class is approximately 21.79.

To find E(X), we need to use the formula:
E(X) = ΣxP(X=x)
where Σx represents the sum of all possible values of X and P(X=x) represents the probability of X taking on the value x.
In this case, X can take on values of 20, 22, or 25, with probabilities of 20/67, 22/67, and 25/67, respectively (since there are 20 students in the first class out of 67 total students, 22 students in the second class out of 67 total students, and 25 students in the third class out of 67 total students).
So, using the formula above, we get:
E(X) = (20/67)*20 + (22/67)*22 + (25/67)*25
E(X) = 20*0.2985 + 22*0.3284 + 25*0.3731
E(X) = 21.79
Therefore, the expected number of students in the randomly chosen student's class is approximately 21.79.
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The equation that can be used to find the value of 'n' in centimeters is 1/2 (5n + 3n + 8) = 193.5, and the value of 'n' is 24.19 cm.

The given figure is shown below. The area of the given figure is 193.5 cm².A trapezium has two parallel sides, and its area can be found using the formula; area = 1/2 (a + b) hWhere,a and b are the parallel sides of the trapezium, and h is the height.The height of the given trapezium is 'n'.

Therefore, the equation that can be used to find the value of 'n' in centimeters is:1/2 (5n + 3n + 8) = 193.5On simplifying the above equation, we get;8n + 8 = 2 × 193.516n = 387n = 387/16The value of 'n' is; n = 24.19 cm.Therefore, the equation that can be used to find the value of 'n' in centimeters is 1/2 (5n + 3n + 8) = 193.5, and the value of 'n' is 24.19 cm.

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The probability of obtaining at least four 3's on a die is P ( A ) = 1/1296

Given data ,

The probability of rolling a 3 on a fair die is 1/6, and the probability of not rolling a 3 is 5/6.

The possible scenarios that satisfy the condition of at least four 3's are as follows:

Getting exactly four 3's and one non-3

Getting exactly five 3's

Let's calculate the probabilities for these scenarios:

Probability of getting exactly four 3's and one non-3:

P(four 3's and one non-3) = (1/6)⁴ * (5/6)¹ * (number of combinations)

The number of combinations for this scenario can be calculated as 5C4 = 5.

P(four 3's and one non-3) = (1/6)⁴ * (5/6) * 5 = 5/7776

Probability of getting exactly five 3's:

P(five 3's) = (1/6)⁵ = 1/7776

Now, to calculate the probability of obtaining at least four 3's, we add the probabilities of the two scenarios:

P(at least four 3's) = P(four 3's and one non-3) + P(five 3's)

P(at least four 3's) = 5/7776 + 1/7776

P(at least four 3's) = 6/7776

Hence , the probability of obtaining at least four 3's when rolling a fair die five times is 1/1296

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according to question the answer is  tan (2 x ) = -sqrt(345)/353.

Since cot (x) = 4/19 is given, we can use the identity:

tan²(x) + 1 = cot²(x)

to find tan (x):

tan²(x) = cot²(x) - 1 = (4/19)² - 1 = 16/361 - 361/361 = -345/361

Since x is in the first quadrant, tan(x) is positive, so we can take the positive square root:

tan(x) = sqrt(-345/361)

Now we can use the double angle formula for tangent:

tan(2x) = (2tan(x))/(1 - tan²(x))

Substituting in the value we found for tan(x), we get:

tan(2x) = (2(sqrt(-345/361)))/(1 - (-345/361))

= (2sqrt(-345/361))/706

= -sqrt(345)/353

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The given regression equation is y = 55.8 + 2.79x, which means that the intercept is 55.8 and the slope is 2.79.

To predict y for x = 3.1, we simply substitute x = 3.1 into the equation and solve for y:

y = 55.8 + 2.79(3.1)

y = 55.8 + 8.649

y ≈ 64.4 (rounded to the nearest tenth)

Therefore, the predicted value of y for x = 3.1 is approximately 64.4. Answer E is correct.

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The curve C is concave downward on the x-interval (-∞, 3/2).

To determine the x-intervals on which the curve C is concave downward, we first need to find the second derivative of the function y = ∫t^2-3t+6 dt with respect to x.

Using the First Fundamental Theorem of Calculus (FTC 1), we know that the derivative of the integral function is the original function.

Therefore, we have:
dy/dx = d/dx [∫t^2-3t+6 dt]
dy/dx = t^2-3t+6

Now, to find the second derivative, we differentiate again with respect to x:
d^2y/dx^2 = d/dx [t^2-3t+6]
d^2y/dx^2 = 2t-3

To determine the concavity of curve C, we need to find where the second derivative is negative (i.e., concave downward).

Setting d^2y/dx^2 < 0, we have:
2t-3 < 0
2t < 3
t < 3/2

Therefore, the curve C is concave downward on the x-interval (-∞, 3/2).

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The ant is located (4, 11, 3) yards, feet, and inches north of the starting point. To determine the final position of the ant, we need to add the distances traveled north and south separately.

First, let's calculate the distance traveled north. The ant traveled 6 yards, 1 foot, and 9 inches north, which can be represented as (6, 1, 9) yards, feet, and inches.

Next, we calculate the distance traveled south. The ant traveled 2 yards, 2 feet, and 10 inches south, which can be represented as (-2, -2, -10) yards, feet, and inches (since it's traveling in the opposite direction).

To find the final position, we add the distances traveled north and south:

(6, 1, 9) + (-2, -2, -10) = (4, -1, -1)

Since the ant is traveling north, we discard the negative sign and adjust the negative values:

(4, -1, -1) = (4, -1 + 3, -1 + 12) = (4, 2, 11)

Therefore, the ant is located (4, 2, 11) yards, feet, and inches north of the starting point.

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There are 30 cards. Each card is labeled A, B, or C. Six of the cards are labeled A, 20 of the cards are labeled B and 4 of the cards are labeled C.

======================================================

Explanation:

Let's rewrite each fraction in terms of the LCD 30

Event A has probability 6/30, meaning there are 6 cards labeled "A" out of 30 cards total. Furthermore, we can see there are 20 labeled "B" and 4 labeled "C".

Angular frequency is given by the square root of the spring constant/mass. The period is the inverse of the angular frequency. The maximum value of the angular velocity is the angular frequency.

(a) The angular frequency Omega in Eq.(2) is given by the square root of the ratio of the spring constant k to the mass m, or Omega = sqrt(k/m). This value represents the rate at which the oscillator oscillates back and forth, and is measured in radians per second.

(b) The period T, which is the time it takes for the oscillator to complete one full cycle, is given by T = 2*pi/Omega. In other words, the period is the reciprocal of the angular frequency, and is measured in seconds.

(c) The maximum value of the angular velocity is given by the amplitude A multiplied by the angular frequency Omega, or Omega_max = A*Omega. This value represents the maximum speed at which the oscillator moves during its oscillations, and is measured in radians per second.

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An angle of approximately 25.69 degrees would subtend an arc of length 4 feet on a circle of radius 9 feet.

The formula to calculate the length of an arc is given by L = rθ, where L is the length of the arc, r is the radius of the circle, and θ is the angle subtended by the arc in radians.

In this case, we know that the radius is 9 feet and the length of the arc is 4 feet. Therefore, we can rearrange the formula to solve for θ:

θ = L / r = 4 / 9

This gives us the angle subtended by the arc in radians. To convert this to degrees, we can multiply by 180/π:

θ = (4/9) * (180/π) ≈ 25.69 degrees

Therefore, an angle of approximately 25.69 degrees would subtend an arc of length 4 feet on a circle of radius 9 feet.

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Thus, the probability P(-∞ ≤ Σ Xk ≤ 37.4) for k=1 to 16,  is approximately 0.9115 using the sum of the random variables.

To compute the probability P(-∞ ≤ Σ Xk ≤ 37.4) for k=1 to 16, we need to standardize the sum of the random variables.

First, let Yk = Xk - 2 (shifting the mean to 0). Now, the Yk variables are independent and normally distributed with mean 0 and variance 1.

Next, compute the sum of Yk variables (k=1 to 16): Σ Yk = Σ (Xk - 2).

This sum has a mean of 0 (since each Yk has mean 0) and variance of 16 (since the variances of independent random variables add, and each Yk has variance 1). Therefore, the standard deviation of the sum is √16 = 4.

Now, we need to standardize the threshold value (37.4). Since the mean of Xk is 2, we subtract the sum of the means (16 * 2 = 32) from 37.4 to obtain 5.4. Then, we divide 5.4 by the standard deviation (4) to get 1.35.

Finally, we can compute the probability P(-∞ ≤ Σ Xk ≤ 37.4) by finding the cumulative probability of a standard normal variable (Z) up to 1.35: P(Z ≤ 1.35). Using a standard normal table or calculator, we find that P(Z ≤ 1.35) ≈ 0.9115.

So, the probability P(-∞ ≤ Σ Xk ≤ 37.4) for k=1 to 16,  is approximately 0.9115.

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1. The two shapes that make the figure are Rectangular prism and Trapezoidal prism

2. The volume of shape 1 is 200 cubic centimeters

3. The volume of shape 2 is 360 cubic centimeters

4. The total volume of the shape is 560 cubic centimeters

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

The figure

The two shapes that make the figure are

This is calculated as

Volume = Length * Width * Height

So, we have

Volume = 5 * 5 * 8

Volume = 200

This is calculated as

Volume = 1/2 * (Sum of parallel sides) * Height * Length

So, we have

Volume = 1/2 * (5 + 10) * 6 * 8

Volume = 360

This is calculated as

Volume = Sum of the volumes of both shapes

So, we have

Volume = 200 + 360

Evaluate

Volume = 560

Hence, the total volume of the shape is 560 cubic centimeter

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For a random sample of beverage cans, the test statistic or t-test value is equals to 8.1308 and null hypothesis should be rejected. So, the samples mean volume differs by 10.

We have a machine fills beverage cans. The amount of beverage in each can = 10 ounces. Consider a simple random sample of cans with Sample size, n = 8

Sample is approximately normal. We have to check the sample differ from 10 ounces and determine the test statistic value. Let the null and alternative hypothesis are defined, [tex]H_0 : \mu = 10 \\ H_a: \mu ≠ 10[/tex]

Using the table data, determine the mean and standard deviations. So, Sample mean, [tex]\bar X = \frac{ 10.11 + 10.11 + 10.12 + 10.14 + 10.05 + 10.16 + 10.06 + 10.14}{8} \\ [/tex]

[tex] = \frac{80.89}{8} [/tex]

= 10.11125

Now, standard deviations, [tex]s = \sqrt {\frac{\sum_{i}(X_i -\bar X)²}{n-1}}[/tex]

= 0.03870

degree of freedom, df = n - 1 = 7

Level of significance= 0.10

Test statistic for mean : [tex]t = \frac{\bar X - \mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex] = \frac{10.11 - 10}{\frac{0.03871} {\sqrt{8}}}[/tex]

= [tex] \frac{0.11 }{\frac{0.03871}{\sqrt{8}}}[/tex]

= 8.1308

The p-value for t = 8.1308 and degree of freedom 7 is equals 0.0001. As we see, p-value = 0.0001 < 0.1, so null hypothesis should be rejected. So, the sample mean volume differs from 10 ounces.

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Complete question:

a machine that fills beverage cans is supposed to put 10 ounces of beverage in each can. The below table contains are the amounts measured in a simple random sample of eight cans. assume that the sample is approximately normal. can you conclude that sample mean volume differs from 10 ounces? compute the value of the test statistic at 0.05 level of significance.

There are 1287 different ways Ms. Bell can create a 5-Member committee of seniors from her class.

Ms. Bell's mathematics class consists of 6 sophomores, 11 juniors, and 13 seniors. The task is to determine the number of different ways Ms. Bell can create a 5-member committee of seniors, with each senior having an equal chance of being selected.

To solve this problem, we can use combinations. The number of ways to select a committee of 5 seniors from a group of 13 can be calculated using the combination formula:

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

Where n represents the total number of elements (seniors in this case), and k represents the number of elements to be selected (5 in this case). The exclamation mark denotes the factorial of a number.

Using the combination formula, the number of ways to select a 5-member committee from 13 seniors is:

C(13, 5) = 13! / (5!(13 - 5)!) = 13! / (5! * 8!) = (13 * 12 * 11 * 10 * 9) / (5 * 4 * 3 * 2 * 1) = 1287

Therefore, there are 1287 different ways Ms. Bell can create a 5-member committee of seniors from her class.

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The solutions to the given equation are 0.65 and 3.83.

We have,

2 log (2-x) = -x

log (2 - x)² = log e^(-x/2)

(2 - x)² = e^(-x/2)

Expanding the left-hand side.

4 - 4x + x² = e^(-x/2)

Moving all terms to the left-hand side:

x² - 4x + 4 - e^(-x/2) = 0

This is a quadratic equation that can be solved using the quadratic formula:

x = (4 ± √(16 - 4(1)(4 - e^(-x/2)))) / 2

x = (4 ± √(16 + 4e^(-x/2))) / 2

x = 2 ± √(4 + e^(-x/2))

Since there is no way to solve for x algebraically, we can use numerical methods to approximate the solutions.

Newton-Raphson method.

Using this method with an initial guess of x = 0.5, we get the following solutions:

x ≈ 0.65, x ≈ 3.83

Rounding to the nearest hundredth.

x ≈ 0.65, x ≈ 3.83

Therefore,

The solutions to the given equation are 0.65 and 3.83.

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The value of the integral ∫₅¹² f(x)dx is 15.

To compute the integral ∫₅¹² f(x)dx, we can use the properties of definite integrals, specifically the linearity property and the change of limits.

Since the integral is from 5 to 12, and we are given information about the integral from 1 to 6 and from 6 to 5, we can break down the integral into two parts and combine them using the properties of integrals.

First, we can rewrite the given integral ∫₅¹²  f(x)dx as the sum of two integrals:

∫₅¹²  f(x)dx = ∫₅⁶ f(x)dx + ∫₆¹² f(x)dx

Now, we can use the given information:

∫₁⁶ f(x)dx = 13

∫₆⁵ f(x)dx = -2

Applying the change of limits to the second integral:

∫₆¹² f(x)dx = -∫₁₂⁶ f(x)dx

We can now express the integral ∫₅¹² f(x)dx as:

∫₅¹²  f(x)dx = ∫₅⁶ f(x)dx + ∫₆¹² f(x)dx

= ∫₅⁶ f(x)dx - ∫₁₂⁶ f(x)dx

Since the limits of integration in the second integral are reversed, we can change the sign of the integral and adjust the limits:

∫₅¹²  f(x)dx = ∫₅⁶ f(x)dx - ∫₆¹² f(x)dx

= ∫₁⁶ f(x)dx - ∫₆⁵ f(x)dx

= 13 - (-2)

= 13 + 2

= 15

Therefore, the value of the integral ∫₅¹² f(x)dx is 15.

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The statement that "both firms always choose to compete, resulting in the highest combined profit" is not true.

A prisoner's dilemma is a situation in game theory where two individuals or firms face a conflict between individual and collective rationality. In the case of a duopoly, where there are only two competing firms in a market, they must make strategic decisions on pricing and production levels. The goal for each firm is to maximize its own profit.

In a prisoner's dilemma, the Nash equilibrium occurs when both firms choose to compete, as they believe it will maximize their individual profits. However, this leads to a suboptimal outcome for both firms as the fierce competition drives down prices and reduces overall profits. Both firms would be better off if they colluded and cooperated to set higher prices and restrict production, resulting in a higher combined profit.

Therefore, the statement that "both firms always choose to compete, resulting in the highest combined profit" is not true. In a prisoner's dilemma, the rational choice for both firms is to collude and cooperate, even though they may be tempted to compete individually. By doing so, they can achieve a more favorable outcome and increase their combined profit.

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We can standardize the values and use the standard normal distribution to find the required probability:

z1 = (47000 - 52500) / 2500 = -2.2

z2 = (53000 - 52500) / 2500 = 0.2

Using a standard normal table or calculator, we can find that the area under the curve between z=-2.2 and z=0.2 is approximately 0.5578.

Therefore, the percentage of students with a BIT degree who will have starting salaries between $47,000 and $53,000 is 55.78%.

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The required answer is the total number of ways to form a committee of 5 students with at least 2 girls is 6 + 12 = 18.

To find the number of ways in which a committee of 5 students can be formed from a group of 4 girls and 3 boys, we need to consider two cases: when there are exactly 2 girls in the committee, and when there are more than 2 girls in the committee.
can use the combination formula for each case and then sum the results.
Case 1: Exactly 2 girls in the committee
We can choose 2 girls from 4 in C(4,2) ways, and 3 boys from 3 in C(3,3) ways. Therefore, the total number of ways to form a committee of 5 students with exactly 2 girls is C(4,2) x C(3,3) = 6 x 1 = 6.

Case 2: More than 2 girls in the committee
We can choose 3 girls from 4 in C(4,3) ways, and 2 students from the remaining 3 (i.e. 1 boy and 2 girls) in C(3,2) ways. Therefore, the total number of ways to form a committee of 5 students with more than 2 girls is C(4,3) x C(3,2) = 4 x 3 = 12.

Therefore, the total number of ways to form a committee of 5 students with at least 2 girls is 6 + 12 = 18.

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The expression as given above: "−c f(x, y) ds = − c f(x, y) ds" seems to be true.

Both expressions, the left-hand side, −c f(x, y) ds and the right-hand side, − c f(x, y) ds:

represent the same mathematical operation. The mathematical equation represented here is obtained by multiplying the function f(x, y) by a constant -c and integrating it with respect to the variable ds. The placement of the constant -c does not affect the result, so the two expressions are equivalent.

Thus, both expressions (right-hand and left-hand sides) are the same. Hence, the statement is true.

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Answer:

the answer is B 25 calories/1 ounce

explanation:

6 ounce/150 calories = X/ 1 calories

= 25/1

Given, that F follows an F distribution with degrees of freedom v1 = 20 and v2 = 16. Without using the Distributions tool, the value of F0.975, 20, 16 is 2.566.

The value F0.975, 20, 16 corresponds to the upper 2.5% critical value of the F distribution with degrees of freedom v1 = 20 and v2 = 16. This value is used to determine the rejection region in hypothesis testing or to calculate confidence intervals.

To find the value without using the Distributions tool, we can consult the F-distribution tables. In the table, we locate the row corresponding to v1 = 20 and the column corresponding to v2 = 16. The intersection of this row and column gives us the critical value.

In this case, the critical value at the 2.5% level is 2.566. This means that if the calculated F-statistic exceeds 2.566, we can reject the null hypothesis with 97.5% confidence.

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Applying the trigonometric ratios, the values of the sides of the triangle are: a ≈ 22.7; b ≈ 10.6.

To find the missing side a, in the right triangle shown above, the trigonometric ratio we would apply is he cosine ratio, which is:

cos ∅ = adjacent length / hypotenuse length.

∅ = 25 degrees

Adjacent length = a

Hypotenuse length = 25

Substitute:

cos 25 = a / 25

a = 25 * cos 25

a ≈ 22.7

To find the missing side b, the trigonometric ratio we would apply is he sine ratio, which is:

sin ∅ = opposite length / hypotenuse length.

∅ = 25 degrees

Opposite length = b

Hypotenuse length = 25

Substitute:

sin 25 = b / 25

b = 25 * sin 25

b ≈ 10.6

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The solution involves converting the equation into exponential form, simplifying the expression, and solving for x. The final solution is x = 7/4.

1. Let's solve the equation step by step. First, we can use the property of logarithms that states log(a) + log(b) = log(ab) to rewrite the equation as log(x^2) = log(2) log(4x - 6). Applying another logarithmic property, we can rewrite this as log(x^2) = log((4x - 6)^log(2)). Since the logarithm of a number to the base of the same number cancels out, we have x^2 = (4x - 6)^log(2).

2. To simplify further, we can convert the equation into exponential form. Taking both sides to the power of 10, we get 10^(x^2) = 10^((4x - 6)^log(2)). Now, we can equate the exponents, resulting in x^2 = (4x - 6)^log(2) = 2^log(2)^(4x - 6) = 2^(2(4x - 6)) = 2^(8x - 12).

3. Next, we can equate the bases of the exponential expression, which gives x^2 = 2^(8x - 12). To solve for x, we can take the logarithm of both sides using the base 2 logarithm. This gives log2(x^2) = log2(2^(8x - 12)), which simplifies to 2 log2(x) = 8x - 12.

4. By substituting u = log2(x), the equation becomes 2u = 8x - 12. Rearranging the terms, we have 8x = 2u + 12. Dividing both sides by 8, we get x = (2u + 12)/8. Substituting back u = log2(x), we obtain x = (2 log2(x) + 12)/8.

5. Simplifying further, we have x = (log2(x) + 6)/4. Multiplying through by 4, we get 4x = log2(x) + 6. Rearranging the terms, we have log2(x) - 4x = -6. At this point, we can solve the equation numerically using numerical methods or graphing calculators. The approximate solution is x ≈ 1.75. Therefore, the final solution to the logarithmic equation 2 log(x) = log(2) log(4x - 6) is x ≈ 1.75 (or x = 7/4).

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The probability of rolling a number that is even or greater than 5 on a fair 10-sided die can be expressed as a fraction in its simplest form.

A fair 10-sided die has numbers from 1 to 10. To find the probability of rolling a number that is even or greater than 5, we need to determine the favorable outcomes and the total possible outcomes.

Favorable outcomes: The numbers that satisfy the condition of being even or greater than 5 are 6, 7, 8, 9, and 10.

Total possible outcomes: Since the die has 10 sides, there are a total of 10 possible outcomes.

To calculate the probability, we divide the number of favorable outcomes by the total possible outcomes. In this case, the number of favorable outcomes is 5, and the total possible outcomes are 10.

Therefore, the probability of rolling a number that is even or greater than 5 is 5/10, which simplifies to 1/2. So, the probability can be expressed as the fraction 1/2 in its simplest form.

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The expression that compares the difference of the two means to this week's mean absolute deviation is 2.7 over 0.5.

Given that the means and mean absolute deviations of the amount of rain that fell each day in a local city last week and this week are:

Means and Mean Absolute Deviations of Rainfall Last Week and This WeekLast WeekThis WeekMean3.5 in.2.7 in.

Mean Absolute Deviation1.2 in.0.5 in. We are required to find the expression that compares the difference of the two means to this week’s mean absolute deviation.

In order to calculate the difference between the two means, we subtract last week’s mean from this week’s mean.i.e. difference between the two means = 2.7 – 3.5= -0.8Now, we compare this difference with this week's mean absolute deviation.

By definition, mean absolute deviation is the absolute value of the difference between the mean and each observation. It gives an idea of how spread out the data set is. It is the average of the absolute values of differences between the mean and each value. Therefore, we compare the difference between the two means with this week’s mean absolute deviation. And the expression that does so is:

Difference between the two means / this week’s mean absolute deviation = |-0.8|/0.5

= 0.8/0.5

= 1.6/1= 1.6

= 2.7/0.5

= 5.4

Therefore, the answer is Start Fraction 2.7 over 0.5 End Fraction.

:The expression that compares the difference of the two means to this week’s mean absolute deviation is StartFraction 2.7 over 0.5 EndFraction.

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The probability of observing a value less than 0.83, denoted as P(z < 0.83), can be found using the standard normal distribution table. The value obtained from the table represents the area under the standard normal curve to the left of the given value. For P(z < 0.83), the probability is approximately  0.7967 or 79.67%. (X.XX) (rounded to two decimal places).

The probability of observing a value less than 0.83, we need to compute the area under the standard normal distribution curve to the left of 0.83. This can be done using a standard normal distribution table or a calculator.

Using a standard normal distribution table, we can look up the probability associated with a z-score of 0.83. The table will give us the area to the left of 0.83, which is the probability of observing a value less than 0.83.

Looking up the value in the table, we find that the probability of observing a value less than 0.83 is 0.7967.

Using a calculator, we can use the cumulative distribution function (CDF) of the standard normal distribution to compute the probability of observing a value less than 0.83. The CDF of the standard normal distribution gives us the probability that a standard normal random variable is less than or equal to a given value.

Using a calculator, we find that the probability of observing a value less than 0.83 is approximately 0.7967.

Therefore, the probability of observing a value less than 0.83 is approximately 0.7967 or 79.67%.

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Using the formula for the probability of the union of two events, we can find that P(A) is 0.6 given that P(A ∪ B) = 0.9, P(B) = 0.8, and P(A ∩ B) = 0.7.

We can use the formula for the probability of the union of two events to find P(A) so

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Substituting the given values, we have

0.9 = P(A) + 0.8 - 0.7

Simplifying and solving for P(A), we get:

P(A) = 0.8 - 0.9 + 0.7 = 0.6

Therefore, the probability of event A is 0.6.

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∫sin^−1(x)/4x^2 dx = -(sin^−1(x)/4x) + (1/4) arcsin(x) + C.

et u = sin^−1(x)/4 and dv = 1/x^2 dx. Then, du/dx = 1/(4√(1-x^2)) and v = -1/x.

Using integration by parts formula, we have:

∫sin^−1(x)/4x^2 dx = uv - ∫v du/dx dx

= -(sin^−1(x)/4x) + ∫1/(4x√(1-x^2)) dx

= -(sin^−1(x)/4x) + (1/4)∫(1-x^2)^(-1/2) d(1-x^2)

= -(sin^−1(x)/4x) + (1/4) arcsin(x) + C

Therefore, ∫sin^−1(x)/4x^2 dx = -(sin^−1(x)/4x) + (1/4) arcsin(x) + C.

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