If P(A) = 0.58, P(B) = 0.44, and P(A ? B) = 0.25, then P(A ? B) = a. 0.11. b. 0.77. c. 0.39. d. 1.02.

To identify the location(s) of max normal and shear stresses, we need to use the equations for normal and shear stresses in a beam.

The normal stress, σ, can be calculated using the formula σ = My/I, where M is the bending moment at a given point, y is the distance from the neutral axis, and I is the area moment of inertia. The maximum normal stress will occur at the point with the maximum bending moment.

The shear stress, τ, can be calculated using the formula τ = VQ/It, where V is the shear force at a given point, Q is the first moment of area of the portion of the beam to the left of the point, t is the thickness of the beam, and I is the area moment of inertia. The maximum shear stress will occur at the point with the maximum shear force.

To find the bending moment and shear force at a given point, we can use the equations for the elastic curve. The slope of the elastic curve, θ, is given by θ(x) = d/dx(w(x)), where w(x) is the deflection of the beam at a given point. The bending moment at a given point is then given by M(x) = EIθ(x), and the shear force is given by V(x) = d/dx(M(x)).

Using the given equation for the elastic curve, we can calculate the slope and curvature at any point. From there, we can find the bending moment and shear force at that point, and then calculate the normal and shear stresses. The location(s) of max normal and shear stresses will occur at the point(s) with the highest stress values.

It's important to note that the given equation for the elastic curve assumes a uniformly distributed load, so it may not accurately represent the actual loading condition. However, it can still be used to find the location(s) of max stresses as long as we assume that the actual loading condition is similar to a uniformly distributed load.

To determine the location of maximum normal and shear stresses in the simply supported beam with a deformed shape given by y(x) = -[T(x^4)]/[(L^4) sin(x/L)], we need to consider the beam's cross-section (b units wide and h units tall), Young's Modulus (E), and area moment of inertia (I).

Maximum normal stress occurs when the bending moment is maximum, and it can be calculated using the formula: σ = M(y)/I, where M is the bending moment and y is the distance from the neutral axis.

To find the maximum bending moment, first, find the first derivative of y(x) with respect to x and set it equal to zero. Solve for x to obtain the location of maximum bending moment. Then, use the obtained x value to find the corresponding maximum bending moment (M) and apply the normal stress formula.

For maximum shear stress, it occurs when the shear force is maximum. Shear force can be calculated using the formula: V = -dM/dx, where dM/dx is the derivative of the bending moment with respect to x.

To find the maximum shear force, first, find the second derivative of y(x) with respect to x and set it equal to zero. Solve for x to obtain the location of maximum shear force. Then, use the obtained x value to find the corresponding maximum shear force (V). Finally, apply the shear stress formula: τ = VQ/(Ib), where Q is the first moment of area and b is the beam's width.

By following these steps, you can identify the location(s) of maximum normal and shear stresses in the given beam.

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The final cost of the chair is $76.

Given that a chair has a sticker price of $100, with 20% discount, $20 off and $2 mail in rebate.

So, the price will be =

$100 - $20 - $100 × 0.20 - $2

= $80 - $2 - $2

= $80 - $4

= $76

Hence, the final cost of the chair is $76.

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Comparing a census of a large population to a sample drawn from it, we expect that the sample is usually a more practical method of obtaining the desired information.

This is because a census involves collecting data from every individual in the population, which can be time-consuming, expensive, and logistically challenging, especially for large populations. In contrast, a sample is a smaller, more manageable subset of the population, making it easier to gather and analyze data.

However, it's essential to note that the accuracy of the observations in the census is generally higher than in the sample, as the census covers the entire population, eliminating any sampling error. In comparison, a sample may be subject to various biases or inaccuracies, depending on the sampling technique used and the sample size.

To ensure that the sample accurately represents the population, it is crucial to select a sample that is both random and of an appropriate size. While the sample doesn't need to be a large fraction of the population, it should be sufficiently large to provide reliable estimates and minimize sampling error. Overall, sampling is a practical and efficient approach to obtaining information about a population when properly conducted, balancing the need for accuracy with resource constraints.

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find the area of cuboid with measurements 18cm x 12cm x 10 cm

[tex]\sf \implies1032 \: c {m}^{2} [/tex]

Area of cuboid = 2 (length × breadth + breadth × height + length × height)

Given:-

[tex]\sf \implies Area \: \: of \: \: cuboid \: \: = 2 (l \times b) +(b \times h)+ (l \times h) \\ [/tex]

[tex]\sf \implies2(18 \times 12) + (12 \times 10) + (18 \times 10) \\ [/tex]

[tex]\sf \implies2(216) + (120) + (180) \\ [/tex]

[tex]\sf \implies2 \times 516 \\ [/tex]

[tex]\sf \implies1032 \: c {m}^{2} [/tex]

[tex] {\rule{200pt}{10pt}}[/tex]

To find the area of a cuboid, we need to determine the sum of the areas of all its faces.

The formula for the surface area of

a cuboid is:

Surface Area = 2(lw + lh + wh)

Where l, w, and h are the length, width, and height of the cuboid.

Plugging the given measurements into the formula, we get:

Surface Area = 2(18 × 12 + 18 × 10 + 12 × 10)

Surface Area = 2(216 + 180 + 120)

Surface Area = 2(516)

Surface Area = 1032 cm²

Therefore, the area of the given cuboid is 1032 cm².

a) The probability that X lies between 2 and 3.5 is 0.2.

b) The probability that Y lies between 18 and 22 is 0.1196.

a) Since X is a uniform continuous random variable on the interval (0, 5), the probability density function (pdf) of X is given by:

f(x) = 1/(b-a) = 1/(5-0) = 1/5 for 0 < x < 5

To find P(2 < x < 3.5), we integrate the pdf f(x) over the interval (2, 3.5):

P(2 < x < 3.5) = ∫[tex]2.5^{3.5[/tex] f(x) dx = ∫[tex]2.5^{3.5[/tex] (1/5) dx = (1/5) * [x][tex]2.5^{3.5[/tex] = (1/5) * (3.5 - 2.5) = 0.2

Therefore, the probability that X lies between 2 and 3.5 is 0.2.

b) If Y has an exponential distribution with mean 20, then the pdf of Y is given by:

f(y) = (1/20) * exp(-y/20) for y > 0

To find P(18 < Y < 22), we integrate the pdf f(y) over the interval (18, 22):

P(18 < Y < 22) = ∫[tex]18^{22[/tex] f(y) dy = ∫[tex]18^{22[/tex] [(1/20) * exp(-y/20)] dy

Using integration by substitution, let u = -y/20, then du = -dy/20:

= ∫[tex](-9/20)^{(-11/20)} exp(u)[/tex] du

= [tex]exp(u)^{(-11/20)[/tex]

= [exp(-11/20) - exp(-9/20)]

= 0.1196

Therefore, the probability that Y lies between 18 and 22 is 0.1196.

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The estimated volume of the solid is 99 cubic units (rounded to the nearest integer).

To use Simpson's Rule with n = 10, we need to divide the interval [0, 10] into 10 equal subintervals. The width of each subinterval is:

h = (10 - 0)/10 = 1

We can then use Simpson's Rule to approximate the volume of the solid:

V ≈ (1/3)[f(0) + 4f(1) + 2f(2) + 4f(3) + 2f(4) + 4f(5) + 2f(6) + 4f(7) + 2f(8) + 4f(9) + f(10)]

where f(x) = πy(x)²

We can use the given formula for y(x) to compute the values of f(x) for each subinterval:

f(0) = π(2/(1 + [tex]e^0[/tex]))² ≈ 3.1416

f(1) = π(2/(1 + [tex]e^-1[/tex]))² ≈ 2.6616

f(2) = π(2/(1 + [tex]e^-2[/tex]))² ≈ 2.4605

f(3) = π(2/(1 + [tex]e^-3[/tex]))² ≈ 2.4885

f(4) = π(2/(1 + [tex]e^-4[/tex]))² ≈ 2.6669

f(5) = π(2/(1 +[tex]e^-5[/tex]))² ≈ 2.9996

f(6) = π(2/(1 + [tex]e^-6[/tex]))² ≈ 3.4851

f(7) = π(2/(1 + [tex]e^-7[/tex]))² ≈ 4.1612

f(8) = π(2/(1 + [tex]e^-8[/tex])² ≈ 5.1216

f(9) = π(2/(1 + [tex]e^-9[/tex]))² ≈ 6.4069

f(10) = π(2/(1 + [tex]e^-10[/tex]))² ≈ 8.0779

Substituting these values into the formula for V and using a calculator, we get:

V ≈ 99

Therefore, the estimated volume of the solid is 99 cubic units (rounded to the nearest integer).

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To find the parametric equations for the path of a particle moving along the circle x^2 + (y - 3)^2 = 16 in the manner described, we can use the standard form for a circle equation and then convert it into parametric equations.

Given the equation of the circle: x^2 + (y - 3)^2 = 16, we can rewrite this in terms of parametric equations using a parameter, often denoted as t (for time).

Since it's a circle, we can use the following parametric equations for a circle with radius r and center (h, k):
x(t) = h + r*cos(t)
y(t) = k + r*sin(t)

From the given circle equation, we can determine the center (h, k) = (0, 3) and radius r = 4. Now, we can plug these values into the parametric equations:
x(t) = 0 + 4*cos(t) = 4*cos(t)
y(t) = 3 + 4*sin(t)

So, the parametric equations for the path of a particle moving along the circle in the manner described are:
x(t) = 4*cos(t)
y(t) = 3 + 4*sin(t)

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The probability that at least 7 of the 10 washing machines will have a lifetime beyond the warranty period is  0.321.

we know that the mean is 8 years and the variance is 16 years^2. Solving these equations for α and β, we get:

α = (Mean / Variance)² = (8 / 16)² = 1/4

β = Variance / Mean = 16 / 8 = 2

Therefore, the pdf of the lifetime of the new model of washing machine is:

f(x) = x^(α-1) e^(-x/β) / (β^α Γ(α))

where Γ(α) is the gamma function.

Substituting the values of α and β, we get:

f(x) = 4 x^(1/4-1) e^(-x/2) / Γ(1/4)

(b) Let X be the number of washing machines out of the 10 that have a lifetime beyond the warranty period.

P(X > 5) = 1 - P(X ≤ 5) = 1 - F(5)

F(x) = Γ(α, x/β) / Γ(α)

where Γ(α, x/β) is the upper incomplete gamma function.

F(x) = Γ(1/4, x/2) / Γ(1/4)

Therefore, the probability that at least 7 of the 10 washing machines will have a lifetime beyond the warranty period is:

P(X > 5) = 1 - F(5) = 1 - Γ(1/4, 5/2) / Γ(1/4) = 0.321

So the probability that at least 7 of the 10 washing machines will have a lifetime beyond the warranty period is  0.321.

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Answer:

Step-by-step explanation:

Note that the graph that represents the equation or expression given is option A. See same attached.

A system of linear inequalities is similar to a system of linear equations, except that it is made up of inequalities rather than equations. To represent scenarios with various restrictions, systems of linear inequalities are utilized.

A line graph is an effective tool for displaying a connection between two variables. The line graph depicts how the number of frog croaks per minute varies with temperature. Line graphs can help you to spot trends and correlations in data, allowing you to infer values that you did not measure.

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The convergent alternating series ∑n=1[infinity](−1)nn!=l alternates between positive and negative terms and has a sum of approximately 0.3679.

A series is a sum of infinitely many terms. For example, the series 1/2 + 1/4 + 1/8 + ... is the sum of infinitely many terms, where each term is half of the previous term.

The term "convergent." A series is said to be convergent if the sum of its terms approaches a finite value as the number of terms in the series approaches infinity.

For example, the series 1 + 1/2 + 1/3 + ... is a divergent series because the sum of its terms does not approach a finite value as the number of terms in the series approaches infinity.

However, the series 1/2 + 1/4 + 1/8 + ... is a convergent series because the sum of its terms approaches 1 as the number of terms in the series approaches infinity.

Now, let's talk about the specific convergent alternating series ∑n=1[infinity](−1)nn!=l. This series alternates between positive and negative terms, where each term is the ratio of n to the factorial of n.

This series is convergent because the sum of its terms approaches a finite value as the number of terms in the series approaches infinity. In fact, the sum of this series is approximately 0.3679, which is also known as the natural logarithmic of 2.

In summary, a series is a sum of infinitely many terms, and a series is said to be convergent if the sum of its terms approaches a finite value as the number of terms in the series approaches infinity.

The convergent alternating series ∑n=1[infinity](−1)nn!=l alternates between positive and negative terms and has a sum of approximately 0.3679.

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The reparametrization of the curve in terms of arclength starting from the point (−5,0,−4) is given by r(t(s)) = (-5 + 3s/2)i + (s/2)j + (-4 + s/2)k.

To reparametrize the curve in terms of arclength, we need to find the arclength function s(t) and then solve for t(s) and substitute it into the original curve equation.

First, we find the velocity vector v(t) = 3i + 2j - 2k and the speed ||v(t)|| = sqrt(17).

Then, we integrate the speed function to get the arclength function: s(t) = integral from 0 to t of ||v(u)|| du = (sqrt(17)/2)t^2 + C, where C is a constant of integration that we determine using the initial condition s(0) = 0. Thus, C = 0.

Next, we solve for t(s) by inverting the arclength function: t(s) = sqrt(2s/sqrt(17)).

Finally, we substitute t(s) into the original curve equation to get the reparametrized curve in terms of arclength: r(t(s)) = (-5 + 3s/2)i + (s/2)j + (-4 + s/2)k.

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It is true that the probability distribution of a random variable is a set of probabilities that indicates the likelihood of each possible outcome of the variable.

The distribution can take different forms depending on the nature of the variable, but it always adds up to 1. In the example given, the random variable has four possible outcomes with probabilities of 0.2, 0.1, 0.4, and 0.3 respectively. This distribution can be used to calculate the expected value and variance of the variable, as well as to make predictions about future observations. Understanding probability distributions is a fundamental concept in statistics and data analysis.

It is true that the  probability distribution of a random variable represents a set of probabilities associated with each possible outcome. In your example, the random variable has a distribution of 0.2, 0.1, 0.4, and 0.3, which indicates the probability of each outcome occurring. These probabilities must add up to 1, reflecting the certainty that one of the outcomes will happen. A probability distribution helps us understand the likelihood of different outcomes and enables us to make predictions based on the given data.

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No, the given set is not a subspace of Pg because it is not closed under vector addition and scalar multiplication, and does not contain the zero vector of P.

No, the given set is not a subspace of P₆ because it does not meet the necessary conditions for being a subspace. The zero vector of P₆ is not in the set because zero is not a negative real number. The set is not closed under vector addition because the sum of two negative real numbers can result in a non-negative real number. Additionally, the set is not closed under scalar multiplication because the product of a scalar and a negative real number is not necessarily a negative real number.

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Answer: 4/5 i think sorry if i am wrong

Step-by-step explanation:

The probability that Max will get the job is 17/28 or approximately 0.61. This is calculated by subtracting the sum of the probabilities of the other two companies getting the job from 1.

To calculate the probability that Max's construction company will get the job, we first need to understand that the sum of probabilities for all three companies must equal 1. Let the probability of Max's company getting the job be represented by P(Max).

Since the probabilities of the two competing companies are 1/7 and 1/4, we can write the equation:

P(Max) + 1/7 + 1/4 = 1

To solve for P(Max), we first need to find a common denominator for the fractions. The least common denominator for 7 and 4 is 28. So, we can rewrite the equation as:

P(Max) + 4/28 + 7/28 = 1

Now, we can combine the fractions:

P(Max) + 11/28 = 1

To find P(Max), we subtract 11/28 from 1:

P(Max) = 1 - 11/28

Since 1 is equal to 28/28, the equation becomes:

P(Max) = 28/28 - 11/28

Now, we subtract the fractions:

P(Max) = 17/28

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The value of c guaranteed by the Mean Value Theorem for Integrals for the function f(x) = 3√x over the interval [4, 9] is approximately 6.1084.

By the Mean Value Theorem for Integrals, there exists at least one value c in the interval [4, 9] such that:

f(c) = (1 / (9 - 4)) * ∫[4,9] f(x) dx

where f(x) = 3√x.

To find the value(s) of c, we first need to evaluate the integral:

∫[4,9] 3√x dx = 2[9^(3/2) - 4^(3/2)]

Using a calculator, we get:

∫[4,9] 3√x dx ≈ 24.0416

For the function f(x) = 3√x on the interval [4,9], we have:

f(a) = f(4) = 3√4 = 6

f(b) = f(9) = 3√9 = 9

Substituting this and f(x) into the equation above, we get:

3√c = (1/5) * 24.0416

Therefore, by the mean value theorem for integrals, there exists at least one value c in (4,9) such that:

f(c) = (1/(9-4)) * ∫[4,9] f(x) dx

= (1/5) * 19.3070

= 3.8614

Simplifying, we get:

c = (24.0416 / 15) ≈ 6.1084

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We first use the value of 1/5 = sin 65° to find the value of sin 65°, which is approximately 0.1305. The value of w to the nearest degree is 40 degrees by using inverse sine function:

[tex]\frac{1}{5}[/tex] = [tex]sin 65°[/tex]

[tex]v = 15 sin 55°[/tex]

[tex]sin w = \frac{2}{1} V[/tex]

[tex]sin w = 15 sin 65° 21[/tex]

Then, we use the value of v = 15 sin 55° to find the value of sin 55°. Dividing both sides by 15 gives:

sin 55° = v/15

Using a calculator, we find that sin 55° is approximately 0.8192.

Next, we use the value of [tex]sin w = (2/1)V[/tex]and the value of [tex]v/15 = sin 55°[/tex] to solve for sin w:

[tex]sin w = (2/1)(v/15)[/tex][tex]= (2/15)v = (2/15)(15 sin 55°)[/tex][tex]= 2 sin 55°[/tex]

Using a calculator, we find that sin w is approximately 1.338. However, this is not possible, since the range of the sine function is between -1 and 1. This means that there is an error in the given information.

Assuming that the correct value for sin w is 0.866 (which is the value of sin 30°), we can solve for w using the inverse sine function:

[tex]w = sin^(-1)(0.866)\\ =40 degrees[/tex]

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The probability that 7 mice are required to find 2 that have contracted the disease is 0.0002837 or approximately 0.028%.

The probability of contracting the disease is 1/11 for each mouse inoculated. Therefore, the probability that 2 mice will contract the disease in a row is (1/11) x (1/11) = 1/121.

To find the probability that 7 mice are required, we need to use the concept of binomial distribution.

The probability of getting 2 successful outcomes (i.e., mice that contract the disease) in 7 trials (i.e., inoculations) can be calculated using the binomial formula: P(2 successes in 7 trials) = (7 choose 2) x (1/121)^2 x (120/121)^5 = 21 x 1/14641 x 2482515744/1305167425 = 21 x 0.0000069 x 1.9037 = 0.0002837 or approximately 0.028%.

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The beta distribution that describes this scenario has parameters alpha = 6.17 and beta = 22.83.

To find the beta parameters for this scenario, we need to use the mean and mode of the distribution. The mean percentage is 22%, so we can use this to find the value of alpha + beta in the beta distribution formula. The formula for the mean of the beta distribution is:

mean = alpha / (alpha + beta)

Rearranging this formula, we get:

alpha + beta = alpha / mean

Substituting in the values we have, we get:

alpha + beta = alpha / 0.22

Solving for alpha, we get:

alpha = 0.22 * alpha + 0.22 * beta

The most-likely value of the percentage is 18%, which is the mode of the distribution. To find the value of alpha and beta that corresponds to this mode, we can use the following formula:

alpha - 1 / (alpha + beta - 2) = mode - 1

Substituting in the values we have, we get:

alpha - 1 / (alpha + beta - 2) = 18 - 1

Solving for alpha, we get:

alpha = 18 * (alpha + beta - 2)

Now we can substitute this value of alpha into our earlier equation to solve for beta:

alpha + beta = alpha / 0.22
(18 * (alpha + beta - 2)) + beta = alpha / 0.22

Solving for beta, we get:

beta = (alpha / 0.22) - alpha - 18

Substituting in our value of alpha, we get:

beta = (18 / 0.22) * (alpha + beta - 2) - alpha - 18

Simplifying this equation, we get:

beta = (78.26 * alpha) + (78.26 * beta) - 177.39

Finally, we can solve for alpha and beta simultaneously using these two equations:

alpha = 0.22 * alpha + 0.22 * beta
beta = (78.26 * alpha) + (78.26 * beta) - 177.39

Solving these equations, we get:

alpha = 6.17
beta = 22.83

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a. f(x) = ln(1-x) over [0, 2]:

This function is continuous on the closed interval [0, 2] since ln(1-x) is defined for x in the interval [0, 1]. Therefore, by the extreme value theorem, f(x) must have both an absolute maximum and an absolute minimum within the interval [0, 2].

b. g(x) = ln(1+x) over [1, 2]:

Similar to the previous explanation, this function is also continuous on the closed interval [1, 2] since ln(1+x) is defined for x in the interval [-1, ∞). Therefore, by the extreme value theorem, g(x) must have both an absolute maximum and an absolute minimum within the interval [1, 2].

c. h(x) = ln(x-1) over [1.4]:

This function is continuous on the closed interval [1.4, ∞) since ln(x-1) is defined for x > 1. Therefore, by the extreme value theorem, h(x) must have both an absolute maximum and an absolute minimum within the interval [1.4, ∞).

d. k(x) = x² over [1, 4]:

This function is also continuous on the closed interval [1, 4] as it is a polynomial function. Therefore, by the extreme value theorem, k(x) must have both an absolute maximum and an absolute minimum within the interval [1, 4].

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a. the monthly sales data for a popular ice cream parlor located on the beach.

b. the monthly electricity consumption data for a residential area.

a. An example of a time series with trend, cycles, and random fluctuations but not seasonal components would be the monthly sales data for a popular ice cream parlor located on the beach.

The trend would be an overall increase in sales as the summer months approach. Cycles would be the weekly fluctuations in sales, with higher sales on weekends and lower sales during weekdays. Random fluctuations would be the unpredictable changes in sales due to various factors such as weather, events, or competition.

b. An example of a time series with seasonal components and random fluctuations, but not trend or cycles, would be the monthly electricity consumption data for a residential area.

The seasonal component would be the regular patterns of higher electricity consumption during the summer months and lower consumption during the winter months. Random fluctuations would be the unpredictable changes in consumption due to various factors such as changes in weather, individual behavior, or appliance use.

There would be no trend as the overall consumption level would remain relatively stable over time, and no cycles as there would be no regular weekly or monthly patterns of consumption.

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The sample mean of 220 would not be considered very unusual or unexpected.

To find the probability of a sample mean being greater than 220, we need to use the formula for the standard error of the mean:

SE = σ/√n

SE = 5.5/√65

SE ≈ 0.68

Then, we can use the z-score formula to find the probability:

z = (X - μ) / SE

z = (220 - 219) / 0.68

z ≈ 1.47

Using a standard normal distribution table, we find that the probability of a z-score being greater than 1.47 is approximately 0.0708. This means that the probability of a sample mean being greater than 220 is about 0.0708 or 7.08%.

To determine whether the given sample mean would be considered unusual, we need to compare it to the population mean and consider the variability of the population. In this case, the sample mean of 220 is only slightly higher than the population mean of 219, and the standard deviation of the population is relatively small at 5.5.

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The sum of the given telescoping series is -1258/507. As n approaches infinity, the terms in the series approach zero, and so the limit of the partial sums is the value of the series.

To find the sum of the given telescoping series, we can use partial fraction decomposition. First, we can write:

[tex](4n-3)(4n+1) = [(4n-3) - (4n+1)] + (4n+1) = -4 + (4n+1)[/tex]

Therefore, we can rewrite the given series as:

[tex]\sum\limits_{n=1}^{\infty} [1/(4n-3) - 1/(4n+1)][/tex]

Now, we can see that each term in the series cancels out all the terms except for the first and the last one. Hence, we get:

[tex][1/(4(1)-3) - 1/(4(1)+1)] + [1/(4(2)-3) - 1/(4(2)+1)] + ...[/tex]

= -3/1 + 1/5 - 3/9 + 1/13 - 3/17 + ...

To find the sum of this alternating series, we can use the alternating series test, which tells us that the sum is equal to the limit of the partial sums, which alternate in sign and decrease in absolute value.

Evaluating the partial sums, we get:

s1 = -3/1 = -3

s2 = -3 + 1/5 = -14/5

s3 = -14/5 - 1/9 = -131/45

s4 = -131/45 + 1/13 = -1258/507

As n approaches infinity, the terms in the series approach zero, and so the limit of the partial sums is the value of the series. Therefore, the sum of the given telescoping series is -1258/507.

In summary, we can find the sum of the given telescoping series by first rewriting it as a series of differences between two terms and then using partial fraction decomposition. The resulting series is an alternating series, and we can use the alternating series test to find the sum.

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Complete Question:

Find the sum of the following telescoping series.

[tex]\sum\limits_{n=1}^{\infty} \frac{4}{(4n-3)(4n+1)}[/tex]

Step-by-step explanation:

(PEMDAS).

50/(20-y)-1=4

50/(20-y)=5

Next, we can cross-multiply to get rid of the fraction:

50 = 5(20-y)

Simplifying further, we get:

50 = 100 - 5y

-50 = -5y

10 = y

Therefore, the solution to the equation is y=10.

Answer: y=10

Answer:

90

Step-by-step explanation: i used

Answer: 90

Step-by-step explanation:

hope this helps comment if you have any questions

Answer: 3 6 11

Step-by-step explanation:

Answer: 3:6:11

Step-by-step explanation: did the question and got this

Formula of  significance of  test as a function and value produces significance level is given by  F(x)  = 1 - [tex]e^{(-x /\mu)}[/tex], for x ≥ 0 with p-value  = [tex]e^{(-x /\mu)}[/tex], for x ≥ 0 and  x = μ log(1/α)respectively.

The significance of the test can be calculated using the following formula,

p-value = P(X > x),

where X is the duration of the observed call

And x is the cutoff value for distinguishing between voice and data calls.

Since X is an exponential random variable with expected value μ,

The probability density function of X is equal to,

f(x) = (1/μ) × [tex]e^{(-x /\mu)}[/tex], for x ≥ 0

The cumulative distribution function CDF of X is,

F(x) = P(X ≤ x)

      = [tex]\int_{0}^{x}[/tex] f(t) dt

      = 1 - [tex]e^{(-x /\mu)}[/tex] for x ≥ 0

The significance of the test is,

p-value = P(X > x)

             = 1 - F(x)

             = [tex]e^{(-x /\mu)}[/tex], for x ≥ 0

The value of x that produces a significance level α.

Since the exponential distribution is a continuous distribution,

Use the inverse of the CDF to find x.

Let F⁻¹ be the inverse of the CDF of X.

Then,

P(X > F⁻¹(1 - α)) = α

Substituting F(x) = 1 - [tex]e^{(-x /\mu)}[/tex], we get,

P(X > μ log(1/α)) = α

The value of x that produces a significance level α is,

x = μ log(1/α)

Therefore, the formula of  significance of the test as a function F(x)  = 1 - [tex]e^{(-x /\mu)}[/tex], for x ≥ 0 with p-value  = [tex]e^{(-x /\mu)}[/tex], for x ≥ 0 .

The value produces significance level is x = μ log(1/α).

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For the given vectors u = -i + 2j + k and v = -2i - 5j - 8k, the solution is:

u = (-1, 2, 1)

v = (-2, -5, -8)

u · v, = -16

Write the vectors in component form:
u = (-1, 2, 1)
v = (-2, -5, -8)

Compute the dot product (u · v) using the formula:
u · v = (u1 * v1) + (u2 * v2) + (u3 * v3)

Substitute the components of u and v into the formula:
u · v = (-1 * -2) + (2 * -5) + (1 * -8)
u · v = 2 - 10 - 8
u · v = -16

So, the given vectors are:
u = -i + 2j + k or (-1, 2, 1)
v = -2i - 5j - 8k or (-2, -5, -8)
and their dot product, u · v, is -16.

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The coordinates of B for the line segment AB where P divides it in the ratio 2: 1 is equal to (6 ,1).

Ratio that divides line segment AB is equal to,

AP : PB = 2 : 1

⇒ ( m : n ) = 2 : 1

Coordinates of point P (x ,y ) = ( 3 ,0 )

Coordinates of point A(x₁ , y₁ ) = ( -3, -2 )

Let us use the ratio of distances formula to find the coordinates of point B.

If point P divides the line segment AB in the ratio 2:1,

AP/PB = 2/1

Let the coordinates of point B be (x₂, y₂).

Use the midpoint formula to find the coordinates of the midpoint of the line segment AB.

which is also the coordinates of point P.

[ (mx₂ + nx₁ ) / (m + n) , (my₂ + ny₁ ) / (m + n) ] = ( x , y )

Substitute the values we have,

⇒ [ (2x₂ + (1)(-3) ) / (2 + 1) , (2y₂ + (1)(-2) ) / (2 + 1)  ] = ( 3 , 0 )

Equate the corresponding values we get,

⇒ 2x₂ -3 / 3 = 3 and 2y₂ -2 / 3 = 0

⇒x₂ = 6 and y₂ = 1

Therefore, the coordinates of point B for the line segment AB are (6 ,1).

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