Round 26,127 to the nearest hundred.

The second-degree Taylor polynomial for the function ()=63 at =0 is simply 63.

To find the Taylor polynomial 2() for the function ()=63 at =0, we need to use the formula for the nth-degree Taylor polynomial:
2() = f(0) + f'(0)() + (1/2!)f''(0)()^2 + (1/3!)f'''(0)()^3 + ... + (1/n!)f^(n)(0)()^n

Since we are only interested in the second-degree Taylor polynomial, we need to calculate f(0), f'(0), and f''(0):
f(0) = 63
f'(x) = 0 (the derivative of a constant function is always 0)
f''(x) = 0 (the second derivative of a constant function is always 0)

Substituting these values into the formula, we get:
2() = 63 + 0() + (1/2!)0()^2
2() = 63

Therefore, the second-degree Taylor polynomial for the function ()=63 at =0 is simply 63.

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Answer:

B, C

Step-by-step explanation:

A would round up to 5

B would round up to 4.9

C would round down to 4.9

D would round down to 5

E would round up to 5

Out of all these only B and C round to 4.9

Answer:

B and C

Step-by-step explanation:

A 4.95  --- this would round to 5.00.

B 4.87 - - - this would round to 4.9

C 4.93 - - - this would round to 4.9

D 5.04 - - - - this would round to 5.0

E 4.97 - - - this would round to 5.0

The amount of golden delicious apples than red delicious apples that Mr. Myers picked would be 14 1/8.

The extra amount of golden delicious apples that Mr. Myers picked in comparison to the red delicious apples that Mr. Myers picked would be gotten by subtracting the amount of golden delicious apples from red delicious apples as follows:

27 2/8 - 13 1/8

= 14 1/8

So, the amount with which the number of golden delicious apples that Mr. Myers got was greater than the red delicious apples is 14 1/8

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Complete Question:

Mr.Myers picked 13 1/8 bushels of red delicious apples and 27 2/8 bushels of golden delicious apples. How many bushels of golden delicious apples than of red delicious apples did he pick?

The statement "Decisions can never be made without the benefit of knowledge gained from sampling" is false.

Sampling refers to the process of selecting a subset of data from a larger population to make inferences about that population. While sampling can be useful in some decision-making contexts, it is not always necessary or appropriate.

In many decision-making situations, there may not be a well-defined population to sample from. For example, a business owner may need to decide whether to invest in a new product line based on market research and other available information, without necessarily having a representative sample of potential customers.

In other cases, the costs and logistics of sampling may make it impractical or impossible.

Additionally, some decision-making approaches, such as decision tree analysis, rely on modeling hypothetical scenarios and their potential outcomes without explicitly sampling from real-world data. While sampling can be a valuable tool in decision-making, it is not a requirement and decisions can still be made without it.

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Answer:

Step-by-step explanation:

True. But there is a very high chance of not happening

The probability that it won't rain given that your algorithm predicted a rainy day is approximately 9.1%. The probability that it will rain given that your algorithm predicted a dry day is approximately 0.01%.

When the algorithm predicts rain, it has a 90% accuracy rate, meaning that it correctly predicts rain 90% of the time. However, since the overall probability of rain in city X is only 1%, most of the algorithm's rainy predictions will be false positives. Using conditional probability, we can calculate the probability of no rain given a rainy prediction as follows: (0.01 * 0.1) / (0.01 * 0.1 + 0.99 * 0.9) ≈ 0.0091 or 9.1%.

Conversely, when the algorithm predicts a dry day, it has a 99% accuracy rate, meaning that it correctly predicts no rain 99% of the time. Since the overall probability of rain is 1%, the algorithm's dry predictions will mostly be true negatives. Using conditional probability again, we can calculate the probability of rain given a dry prediction as follows: (0.99 * 0.01) / (0.99 * 0.01 + 0.01 * 0.9) ≈ 0.0001 or 0.01%.

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If f and g be functions such that, f(0)=2, g(0)=3, f'(0)=-10, g'(0)=-3, then :

h'(0) = -36.

To find h'(0), we can use the product rule for derivatives. The product rule states that if h(x) = f(x)g(x), then h'(x) = f'(x)g(x) + f(x)g'(x).

Applying this to our function h(x) = g(x)f(x), we get:

h'(x) = g'(x)f(x) + g(x)f'(x)

Now we can evaluate this expression at x = 0, since we are looking for h'(0). Plugging in the given values, we get:

h'(0) = g'(0)f(0) + g(0)f'(0)
      = (-3)(2) + (3)(-10)
      = -6 - 30
      = -36

Therefore, we can state that the value of h'(0) = -36.

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The singular point(s) of the differential equation are x = 4.

To find the singular points of the differential equation xy'' + y' + y(x + 2)/(x - 4) = 0, we need to find the values of x at which the coefficient of y'' or y' becomes infinite or undefined, since these are the points where the equation may behave differently.

The coefficient of y'' is x, which is never zero or undefined, so there are no singular points due to this term.

The coefficient of y' is 1, which is also never zero or undefined, so there are no singular points due to this term.

The coefficient of y is (x + 2)/(x - 4), which becomes infinite or undefined when x = 4, so 4 is a singular point of the differential equation.

Therefore, the singular point(s) of the differential equation are x = 4.

Note that this analysis does not consider any initial or boundary conditions, which may affect the behavior of the solution near the singular point(s).

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The first three nonzero terms in the Taylor polynomial approximation to $y(t)$ are $t + \frac{1}{3}t^2 + O(t^3)$.

Using these initial conditions, we can write the first few terms of the Taylor polynomial approximation as:

\begin{align*}

y(t) &\approx y(0) + y'(0)t + \frac{y''(0)}{2!}t^2 \

&= t + \frac{1}{2}y''(0)t^2 \

&= t + \frac{1}{2}\left(\frac{6\cos(0)}{9\cdot 0 + 9}\right)t^2 \

&= t + \frac{1}{3}t^2

\end{align*}

Therefore, the first three nonzero terms in the Taylor polynomial approximation to $y(t)$ are $t + \frac{1}{3}t^2 + O(t^3)$.

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Given that P is a function that gives the cost, in dollars, of mailing a letter from the United States to Mexico in 2018 based on the weight of the letter in ounces, w.In order to write a function, we must find the rate at which the cost changes with respect to the weight of the letter in ounces.

Let C be the cost of mailing a letter from the United States to Mexico in 2018 based on the weight of the letter in ounces, w.Let's assume that the cost C is directly proportional to the weight of the letter in ounces, w.Let k be the constant of proportionality, then we have C = kwwhere k is a constant of proportionality.Now, if the cost of mailing a letter with weight 2 ounces is $1.50, we can find k as follows:1.50 = k(2)⇒ k = 1.5/2= 0.75 Hence, the cost C of mailing a letter from the United States to Mexico in 2018 based on the weight of the letter in ounces, w is given by:C = 0.75w dollars. Answer: C = 0.75w

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The given argument "K ⊃ Q / Q ⊃ ∼ K // K ≡ Q" is valid.

To determine the validity of the argument "K ⊃ Q / Q ⊃ ∼ K // K ≡ Q," we construct an ordinary truth table. The argument consists of two premises and a conclusion. The symbol "⊃" represents the conditional implication, "∼" represents negation, and "≡" represents equivalence.

We assign truth values (T or F) to the atomic propositions K and Q and evaluate the truth values of the premises and the conclusion based on the given argument. By systematically filling out the truth table, we can examine all possible combinations of truth values for K and Q.

After constructing the truth table, we observe that in every row where the premises K ⊃ Q and Q ⊃ ∼ K are true, the conclusion K ≡ Q is also true. Therefore, the argument is valid.

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Potassium chloride (KCl) is a binary ionic compound consisting of potassium cations (K+) and chloride anions (Cl-).  a) KCl(s) → K+(aq) + Cl-(aq). b) CaBr2(s) → Ca2+(aq) + 2Br-(aq). c) Fe2(SO4)3(s) → 2Fe3+(aq) + 3SO42-(aq).

a) KCl(s) → K+(aq) + Cl-(aq)

Potassium chloride (KCl) is a binary ionic compound consisting of potassium cations (K+) and chloride anions (Cl-). When KCl is dissolved in water, it dissociates into its constituent ions, i.e., K+ and Cl-. This process is represented by the above chemical equation.

b) CaBr2(s) → Ca2+(aq) + 2Br-(aq)

Calcium bromide (CaBr2) is also a binary ionic compound consisting of calcium cations (Ca2+) and bromide anions (Br-). When CaBr2 is dissolved in water, it dissociates into its constituent ions, i.e., Ca2+ and 2Br-. This process is represented by the above chemical equation.

c) Fe2(SO4)3(s) → 2Fe3+(aq) + 3SO42-(aq)

Iron(III) sulfate (Fe2(SO4)3) is a complex ionic compound consisting of two iron cations (Fe3+) and three sulfate anions (SO42-). When Fe2(SO4)3 is dissolved in water, it dissociates into its constituent ions, i.e., 2Fe3+ and 3SO42-. This process is represented by the above chemical equation.

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Thus, the object travels a distance of 10π units along the given trajectory.

To find out how far an object travels along a given trajectory, we need to calculate the arc length of the curve. The formula for arc length is given by:

L = ∫_a^b √[dx/dt]^2 + [dy/dt]^2 dt

where L is the arc length, a and b are the start and end points of the curve, and dx/dt and dy/dt are the derivatives of x and y with respect to time t.

In this case, we have the trajectory r(t) = (10 cos 2t, 10 sin 2t) for 0 ≤ t ≤ π/2. Therefore, we can calculate the derivatives of x and y as follows:

dx/dt = -20 sin 2t
dy/dt = 20 cos 2t

Substituting these values into the formula for arc length, we get:

L = ∫_0^(π/2) √[(-20 sin 2t)^2 + (20 cos 2t)^2] dt
 = ∫_0^(π/2) √400 dt
 = ∫_0^(π/2) 20 dt
 = 20t |_0^(π/2)
 = 10π

Therefore, the object travels a distance of 10π units along the given trajectory.

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Sampling error refers to the discrepancy between sample characteristics and population characteristics. It can be diminished by increasing the sample size, using random sampling techniques, and improving response rates.

A) Sampling error refers to the difference between the characteristics of a sample and the characteristics of the population from which it was drawn.

In other words, sampling error refers to the degree to which the sample statistics deviate from the population parameters.

B) Sampling error can be diminished by increasing the sample size, using random sampling techniques to ensure that the sample is representative of the population, and minimizing sources of bias in the sampling process.

C) Convenience sampling, snowball sampling, and judgmental sampling are all methods of non-probability sampling, which means that they do not involve random selection of participants.

As a result, these methods are more likely to yield sampling error than probability sampling methods.

Convenience sampling involves selecting participants who are readily available, which may not be representative of the population of interest.

Snowball sampling involves using referrals from existing participants, which may create biases in the sample.

Judgmental sampling involves selecting participants based on the researcher's judgment of who is most relevant to the study, which may not be representative of the population of interest.

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Three different ways to write [tex]5^{11}[/tex] as the product of two powers are:

[tex]5^{1} * 5^{10} \\\\5^{5} * 5^{6} \\\\5^{3} * 5^{8}[/tex]

To write the powers in different ways that all translate to 5 raised to the power of 11, we need to first recall that the product of the same bases is gotten by summing up the bases.

In this case, 1 times 10 is 1 plus 10 which is 11. The same applies for 5 and 6 and 3 and 8. So, the above are three ways to rewrite the expression.

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the upper sum for f(x) = 17 - [tex]x^{2}[/tex] on the interval [3, 4] using four subintervals is approximately 6.46875.

To calculate the upper sum, we divide the interval [3, 4] into four subintervals of equal width. The width of each subinterval is (4 - 3) / 4 = 1/4.

Next, we evaluate the function at the right endpoint of each subinterval and multiply it by the width of the subinterval. For this function, we need to find the maximum value within each subinterval. Since the function f(x) = 17 - [tex]x^{2}[/tex] is a downward-opening parabola, the maximum value within each subinterval occurs at the left endpoint.

Using four subintervals, the right endpoints are: 3 + (1/4), 3 + (2/4), 3 + (3/4), and 3 + (4/4), which are 3.25, 3.5, 3.75, and 4 respectively.

Evaluating the function at these right endpoints, we get: f(3.25) = 8.5625, f(3.5) = 10.75, f(3.75) = 13.5625, and f(4) = 13.

Finally, we calculate the upper sum by summing the products of each function value and the subinterval width: (1/4) × (8.5625 + 10.75 + 13.5625 + 13) = 6.46875.

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The y-intercept represents the area of the bed and room together when the length and width of the bed are both zero and the function is given by the relation A(x) = x² + 12x + 35

Given data ,

To find the standard form of the function A(x), we first expand the expression:

A(x) = (x + 5) (x + 7)

A(x) = x² + 7x + 5x + 35

A(x) = x² + 12x + 35

So the standard form of the function A(x) is:

A(x) = x² + 12x + 35

To find the y-intercept, we set x = 0 in the function:

A(0) = 0² + 12(0) + 35

A(0) = 35

So the y-intercept is 35. In the context of the problem, the y-intercept represents the area of the bed and room together when the length and width of the bed are both zero.

Hence , the function is solved

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The population mean, denoted by u, is 83, and the standard deviation of the population, denoted by sigma, is 14. The sample size, denoted by n, is 49.
Hi! I'd be happy to help you with your question. Based on the given parameters of the population and sample size, we need to determine µ (mean) and σ (standard deviation).

From the information provided, we have the following parameters:

1. Population mean (µ) = 83
2. Population standard deviation (σ) = 14
3. Sample size (n) = 49

Using these parameters, we can determine the mean and standard deviation for the sample. Since the population mean is given, the sample mean will also be 83.

To find the standard error (SE), which is the standard deviation for the sample, use the formula:

SE = σ / √n

Plugging in the values, we get:

SE = 14 / √49
SE = 14 / 7
SE = 2

So, the sample mean (µ) is 83, and the sample standard deviation (SE) is 2.

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the mass moment of inertia about the x-axis is ixx = (2/7)[tex]mr^{2}[/tex] and about the y-axis is iyy = (1/7)[tex]mr^{2}[/tex] + (2/3)[tex]mh^{2}[/tex]

To find the mass moment of inertia, we consider the solid of revolution as a collection of infinitesimally thin disks or cylinders stacked together along the axis of revolution. Each disk or cylinder has a mass element dm.

For the mass moment of inertia about the x-axis (ixx), we integrate the contribution of each mass element along the axis of revolution:

ixx = ∫ [tex]r^{2}[/tex] dm

Since the solid is homogeneous, dm = ρ dV, where ρ is the density and dV is the volume element. For a solid of revolution, dV = πr^2 dh, where h is the height of the solid.

Substituting the expressions and performing the integration, we get:

ixx = ∫ [tex]r^{2}[/tex] ρπr^2 dh

= ρπ ∫ [tex]r^{4}[/tex] dh

= [tex](1/5)\beta \pi r^{4}[/tex] h

Since the solid is homogeneous, the mass m = [tex]\beta \pi r^{2}[/tex] h. Substituting this in the equation above, we get:

ixx = (1/5)m [tex]r^{2}[/tex]

Similarly, for the mass moment of inertia about the y-axis (iyy), we integrate along the radius r:

iyy = ∫[tex]r^{2}[/tex]  dm

= ∫ [tex]r^{2}[/tex] [tex]\beta \pi r^{2}[/tex] dh

= ρπ ∫ [tex]r^{4}[/tex] dh

= (1/5)[tex]\beta \pi r^{4}[/tex] h

Since the height of the solid is h, substituting [tex]\beta \pi r^{2}[/tex] h = m, we get:

iyy = (1/5)m [tex]r^{2}[/tex] + [tex](2/3)mh^{2}[/tex]

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2) a. The average daily balance for the billing period, which ends on June 11. May has 31 days is $547.56.

b. $0.71 is the finance charge calculated on June 11. The monthly periodic rate is 1.3%.

c.  $548.27 is the Smith's new credit card balance on June 12.

3) $83.50 money was saved by making the payment earlier in the billing cycle.

a. It does matter when you make your payment because the finance charge is based on the balance at the end of the billing period.

b.  It also matters when you make your purchases because the daily balance is calculated based on the charges and payments up to and including each day.

2)

a. To find the average daily balance, we need to first calculate the balance for each day of the billing period. The balance for each day is the sum of charges and payments up to and including that day. We can calculate the balances as follows:

May 12: $378.50

May 13: $378.50 + $129.79 = $508.29

May 14-31: $508.29

June 1: $508.29 + $135.85 = $644.14

June 2-7: $644.14

June 8: $644.14 + $37.63 = $681.77

June 9: $681.77 - $50.00 = $631.77

June 10-11: $631.77

Next, we add up the daily balances and divide by the number of days in the billing period:

Average daily balance = (31 x $508.29 + 6 x $644.14 + 2 x $681.77) / 39

                                      = $21,328.99 / 39

                                      = $547.56

b. To calculate the finance charge, we first need to calculate the daily periodic rate, which is the monthly periodic rate divided by the number of days in a month:

Daily periodic rate = 1.3% / 30

                              = 0.04333%

Next, we multiply the average daily balance by the daily periodic rate and by the number of days in the billing period:

Finance charge = $547.56 x 0.0004333 x 30

                          = $0.71

c. The Smith's new credit card balance on June 12 is the sum of the average daily balance and the finance charge:

New balance = $547.56 + $0.71

                       = $548.27

3) The payment was made on June 9, which is 3 days before the end of the billing period. If the payment had been made on June 11, the balance would have been $631.77 instead of $548.27. This means that the payment saved the Smiths $83.50 in finance charges.

a) It does matter when you make your payment because the finance charge is based on the balance at the end of the billing period. If you make a payment earlier in the billing cycle, your balance will be lower at the end of the period and you will pay less in finance charges.

b) It also matters when you make your purchases because the daily balance is calculated based on the charges and payments up to and including each day. If you make a large purchase early in the billing cycle, your average daily balance will be higher and you will pay more in finance charges.

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The area of the parallelogram is 21 in².

Area is the region bounded by a plan shape.

To calculate the area of the parallelogram, we use the formula below

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

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The number of votes for each candidate would be:

Corrine Brown = 66,617

Bill Randall = 52,920

To determine the number of votes for each candidate, we will make some equations with the values given.

Equation 1 = CB + BR = 119,537

(BR + 13,696) + BR = 119,537

2BR + 13,696 = 119,537

Collect like terms

2BR = 119,537 - 13,696

2BR = 105841

Divide both sides by 2

BR = 52,920

This means that Corrine Brown received 52,920 +  13,696 =  66,617

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Complete Question:

In a recent election corrine brown received 13,696 more votes than bill Randall. If the total number of votes was 119,537, find the number of votes for each candidate

Given, Lincoln invested $2,800 in an account paying an interest rate of 5 3/8 % compounded continuously. Lily invested $2,800 in an account paying an interest rate of 5 7/8 % compounded quarterly.

After 15 years, we need to calculate how much more money would Lily have in her account than Lincoln, to the nearest dollar. Calculation of Lincoln's investment Continuous compounding formula is A = Pe^rt Where, A is the amount after time t, P is the principal amount, r is the annual interest rate, and e is the base of the natural logarithm.

Lincoln invested $2,800 in an account paying an interest rate of 5 3/8 % compounded continuously .i.e. r = 5.375% = 0.05375 and P = $2,800Thus, A = Pe^rtA = $2,800 e^(0.05375 × 15)A = $2,800 e^0.80625A = $2,800 × 2.24088A = $6,292.44Step 2: Calculation of Lily's investmentThe formula to calculate the amount in an account with quarterly compounding is A = P (1 + r/n)^(nt)Where, A is the amount after time t, P is the principal amount, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is the time. Lily invested $2,800 in an account paying an interest rate of 5 7/8 % compounded quarterly.i.e. r = 5.875% = 0.05875, n = 4, P = $2,800Thus, A = P (1 + r/n)^(nt)A = $2,800 (1 + 0.05875/4)^(4 × 15)A = $2,800 (1.0146875)^60A = $2,800 × 1.96494A = $7,425.16Step 3: Calculation of the difference in the amount After 15 years, Lily has $7,425.16 and Lincoln has $6,292.44Thus, the difference in the amount would be $7,425.16 - $6,292.44 = $1,132.72Therefore, the amount of money that Lily would have in her account than Lincoln, to the nearest dollar, is $1,133.

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"The given statement is false."Any function that satisfies the condition of periodicity can have a transition from period 2 straight pi to an arbitrary period p equals 2 l. It does not have to be a trigonometric function.

"False". The transition from period 2 straight pi to an arbitrary period p equals 2 l can be achieved by any function that satisfies the condition f(x + p) = f(x) for all x. Such a function is said to be periodic with period p.

Trigonometric functions such as sine and cosine are examples of periodic functions with period 2π, but there are many other functions that can be periodic with different periods.

For instance, the function f(x) = x^2 is a periodic function with period 2, since f(x + 2) = (x + 2)^2 = x^2 + 4x + 4 = x^2 + 4(x + 1) = f(x) + 4. This means that the function repeats every 2 units. Similarly, the function f(x) = sin(πx) is a periodic function with period 2, since f(x + 2) = sin(π(x + 2)) = sin(πx + 2π) = sin(πx) = f(x).

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True. The transition from period 2 straight pi to an arbitrary period p equals 2 l is only possible if f is a trigonometric function.

True, the transition from a period of 2π to an arbitrary period P = 2L is only possible if f is a trigonometric function.
1. Trigonometric functions, such as sine and cosine, have a standard period of 2π.
2. In order to transition from the standard period to an arbitrary period P, we need to adjust the function by a factor.
3. The arbitrary period P can be represented as P = 2L, where L is a constant value.
4. For a trigonometric function f(x) with the standard period 2π, we can create a new function g(x) with period P by using the following transformation: g(x) = f(kx), where k = (2π)/P.
5. As a result, the new function g(x) will have the desired arbitrary period P = 2L.

This is because trigonometric functions are periodic and can have arbitrary periods, whereas non-trigonometric functions may not exhibit periodicity at all or may have a specific period that cannot be easily modified.
Thus, the statement is true.

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c = pi/2, and the value of c > 1 such that the average value of f(x) on the interval [2, 20] is equal to c is c = pi/2.

The average value of a function f(x) on the interval [a, b] is given by:

Avg = 1/(b-a) * ∫[a, b] f(x) dx

We want to find a value of c > 1 such that the average value of the function [tex]f(x) = (9pi/x^2)cos(pi/x)[/tex] on the interval [2, 20] is equal to c.

First, we find the integral of f(x) on the interval [2, 20]:

[tex]∫[2, 20] (9pi/x^2)cos(pi/x) dx[/tex]

We can use u-substitution with u = pi/x, which gives us:

-9pi * ∫[pi/20, pi/2] cos(u) du

Evaluating this integral gives us:

[tex]-9pi * sin(u) |_pi/20^pi/2 = 9pi[/tex]

Therefore, the average value of f(x) on the interval [2, 20] is:

[tex]Avg = 1/(20-2) * ∫[2, 20] (9pi/x^2)cos(pi/x) dx[/tex]

= 1/18 * 9pi

= pi/2

Now we set c = pi/2 and solve for x:

Avg = c

[tex]pi/2 = 1/(20-2) * ∫[2, 20] (9pi/x^2)cos(pi/x) dx[/tex]

pi/2 = 1/18 * 9pi

pi/2 = pi/2

Therefore, c = pi/2, and the value of c > 1 such that the average value of f(x) on the interval [2, 20] is equal to c is c = pi/2.

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The probability that the bit will fail before 10 hours of usage is:

P(X < 10) = F(10) = 1 - e^(-(10/50)^2) ≈ 0.3935

The Weibull distribution is given by the probability density function:

f(x) = (a/B) * (x/B)^(a-1) * e^(-(x/B)^a)

where a and B are the shape and scale parameters, respectively.

In this case, a = 2 and B = 50. We want to find the probability that the bit will fail before 10 hours of usage, i.e., P(X < 10), where X is the random variable representing the length of life of the drill bit.

Using the cumulative distribution function (CDF) of the Weibull distribution, we have:

F(x) = 1 - e^(-(x/B)^a)

Substituting the values of a and B, we get:

F(x) = 1 - e^(-(x/50)^2)

So the answer is approximately 0.4.

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The cubic function that has a local maximum value of 4 at 1 and a local minimum value of –1,184 at 7 is:

[tex]f(x) = (-28/15)x^3 + (59/15)x^2 - 23x - 149/3[/tex]

We can start by writing the cubic function in the general form:

[tex]f(x) = ax^3 + bx^2 + cx + d[/tex]

To find the coefficients of the function, we can use the given information about the local maximum and minimum values.

First, we know that the function has a local maximum value of 4 at x = 1. This means that the derivative of the function is equal to zero at x = 1, and the second derivative is negative at that point. So, we have:

f'(1) = 0

f''(1) < 0

Taking the derivative of the function, we get:

[tex]f'(x) = 3ax^2 + 2bx + c[/tex]

Since f'(1) = 0, we have:

3a + 2b + c = 0 (Equation 1)

Taking the second derivative of the function, we get:

f''(x) = 6ax + 2b

Since f''(1) < 0, we have:

6a + 2b < 0 (Equation 2)

Next, we know that the function has a local minimum value of -1,184 at x = 7. This means that the derivative of the function is equal to zero at x = 7, and the second derivative is positive at that point. So, we have:

f'(7) = 0

f''(7) > 0

Using the same process as before, we can get two more equations:

21a + 14b + c = 0 (Equation 3)

42a + 2b > 0 (Equation 4)

Now we have four equations (Equations 1-4) with four unknowns (a, b, c, d), which we can solve simultaneously to get the values of the coefficients.

To solve the equations, we can eliminate c and d by subtracting Equation 3 from Equation 1 and Equation 4 from Equation 2. This gives us:

a = -28/15

b = 59/15

Substituting these values into Equation 1, we can solve for c:

c = -23

Finally, we can substitute all the values into the general form of the function to get:

[tex]f(x) = (-28/15)x^3 + (59/15)x^2 - 23x + d[/tex]

To find the value of d, we can use the fact that the function has a local maximum value of 4 at x = 1. Substituting x = 1 and y = 4 into the function, we get:

4 = (-28/15) + (59/15) - 23 + d

Solving for d, we get:

d = -149/3

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The standard deviation of x can be  0.725.

The table describing the probability distribution of x is as follows

x P(X=x)

0 10/120

1 48/120

2 42/120

3 20/120

To find the probabilities, we can use the hypergeometric distribution formula:

P(X=x) = (C(4,x) * C(6,3-x)) / C(10,3)

where C(n,r) represents the number of combinations of n things taken r at a time.

The mean of x can be found using the formula:

E(X) = Σ(x * P(X=x))

= 0*(10/120) + 1*(48/120) + 2*(42/120) + 3*(20/120)

= 1.4

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A segment that connects two points on a circle is called a chord, which makes the option B correct.

In the context of circles, a chord refers to a line segment that connects two points on the circumference of the circle. It can also be defined as the longest possible segment that can be drawn between two points on a circle. Every chord in a circle creates two arcs, one on each side of the chord.

Note that diameter is a special type of chord that passes through the center of the circle. It is the longest possible chord in a circle, and it divides the circle into two congruent semicircles.

Therefore, a segment that connects two points on a circle is called a chord.

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