Example: Chebyshev's Theorem
What is the minimum percentage of the items in a data set which lie within 3 standard deviations of the mean?

The probability P(−1.60 ≤ Z ≤ 0) is 0.44500.

The probability P(−1.60 ≤ Z ≤ 0) can be found using a standard normal distribution table or calculator.

Using a standard normal distribution table, we can look up the area under the curve between z = −1.60 and z = 0, which is 0.44500. Therefore, the answer is 0.44500.

Alternatively, we can use a calculator that can calculate probabilities for a standard normal distribution. In this case, we would enter the following: P(−1.60 ≤ Z ≤ 0) = normdist(0, 1, 0, TRUE) − normdist(-1.60, 1, 0, TRUE), which also gives us 0.44500 as the answer.

Therefore, the probability P(−1.60 ≤ Z ≤ 0) is 0.44500.
To find the probability P(-1.60 ≤ Z ≤ 0), we'll use the standard normal distribution table or Z-table.

Step 1: Look up the Z-scores in the standard normal distribution table.
For Z = -1.60, the table value is 0.0548, which represents the probability P(Z ≤ -1.60).
For Z = 0, the table value is 0.5000, which represents the probability P(Z ≤ 0).

Step 2: Calculate the probability P(-1.60 ≤ Z ≤ 0).
Subtract the probability of Z ≤ -1.60 from the probability of Z ≤ 0.
P(-1.60 ≤ Z ≤ 0) = P(Z ≤ 0) - P(Z ≤ -1.60)
P(-1.60 ≤ Z ≤ 0) = 0.5000 - 0.0548

Step 3: Solve for the probability.
P(-1.60 ≤ Z ≤ 0) = 0.4452

Therefore, the probability P(-1.60 ≤ Z ≤ 0) is approximately 0.4450.

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We can expect that 40,000 of these light bulbs will last at least 1200 hours.

A defect-free bulb has a normal lifetime of 1200 hours with a standard deviation of 150 hours, so we know that the normal lifetime dissemination for these bulbs is 1200 hours with a standard deviation of 150 hours.

To decide the number of bulbs anticipated to final at the slightest 1200 hours, we ought to decide the rate of bulbs with normal life anticipation of at slightest 1200 hours. 

Using the standard normal distribution, we can find the area under the right curve at 1200 hours.

The Z-score for a bulb with a life expectancy of 1200 hours can be calculated as follows:

z = (1200 - 1200) / 150 = 0

Using the standard normal distribution table, we find that the area to the right of z=0 is 0.5. This means that 50% of the lamps should last at least 1200 hours.

For 80,000 bulbs produced, multiply that percentage by the total number of bulbs to find the number of bulbs expected to last at least 1200 hours.

number of bulbs = percentage × total number of bulbs

= 0.5 × 80,000

= 40,000

Therefore, with 40,000 of these bulbs, we can assume that they will last at least 1200 hours.  

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dA/dt = πa(db/dt) + πb(da/dt)

b is not changing at this time.

(a) To find dA/dt, we can use the product rule of differentiation:

A = πab
dA/dt = π(db/dt)a + πb(da/dt)
dA/dt = πa(db/dt) + πb(da/dt)  (since a and b can be interchanged)


(b) When a = 880 inches, da/dt = -2 inches/min (since a is decreasing by 2 inches per minute) and A is constant. We can use the formula for A and plug in the given values:

A = πab
π(880)(175) = constant
b = constant/(πa)
db/dt = (-πa constant')/(πa^2)  (using the quotient rule of differentiation)

Substituting the given values, we get:

db/dt = (-π(880)(175)(0))/(π(880)^2)
db/dt = 0 inches/min

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The MLE of θ into the formula for the MLE of 9(θ) is. [tex]\approx 0.607[/tex]

The Maximum Likelihood Estimator (MLE) of θ, we need to find the value of θ that maximizes the likelihood function.

The likelihood function is given by:

[tex]L(\theta|x1, x2, ..., xn) = \theta^n (x1 x2 ... xn)^(\theta-1)[/tex]

Taking the logarithm of the likelihood function, we get:

ln[tex]L(\theta|x1, x2, ..., xn) = n ln \theta + (\theta - 1) \Sigma ln xi[/tex]

To find the MLE, we differentiate the log-likelihood function with respect to θ, set the derivative equal to zero, and solve for θ:

[tex]d/d\theta (ln L(\theta|x1, x2, ..., xn)) = n/\theta + \Sigma ln xi = 0[/tex]

[tex]\theta = - n / \Sigma ln xi[/tex]

Since[tex]\theta > 0[/tex], we need to check that this value of θ actually maximizes the likelihood function.

We can do this by taking the second derivative of the log-likelihood function with respect to θ:

[tex]d^2/d\theta^2 (ln L(\theta|x1, x2, ..., xn)) = -n/\theta^2 < 0[/tex]

Since the second derivative is negative, the value of θ that we obtained is a maximum.

The MLE of θ is:

[tex]\theta = - n / \Sigma ln xi[/tex]

The MLE of [tex]9(\theta) = e^{-1}/θ[/tex], we substitute the MLE of θ into the expression for 9(θ):

[tex]\^9 = e^{-1}/(\theta) = e^\Sigma ln xi / n[/tex]

Substituting the observed values into the formula for the MLE of θ, we have:

[tex]\^ \theta= - n / \Sigma ln xi[/tex][tex]\theta = - n / \Sigma ln xi[/tex]

[tex]= - 6 / (ln 0.41 + ln 0.52 + ln 0.94 + ln 0.83 + ln 0.84 + ln 0.60)[/tex]

[tex]\approx 2.112[/tex]

Substituting the MLE of θ into the formula for the MLE of 9(θ), we have:

[tex]\^9= e^{\Sigma ln xi / n}[/tex]

[tex]= e^{(-6/n \Sigma ln (1/xi))}[/tex]

[tex]\approx 0.607[/tex]

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The means diameter of ball bearings produced in this manufacturing process is 3.65 millimeters.

Since the diameter of ball bearings is uniformly distributed over the interval of 2.55 to 4.75 millimeters, we can use the formula for the mean of a continuous uniform distribution:

mean = (b + a) / 2

where a is the lower limit of the interval (2.55) and b is the upper limit of the interval (4.75).

Therefore, the mean diameter of ball bearings produced in this manufacturing process is:

mean = (4.75 + 2.55) / 2 = 3.65 millimeters.

Therefore, the mean diameter of ball bearings produced in this manufacturing process is 3.65 millimeters.

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lim(x→0) sin(x) = 0

To show that lim(x→0) sin(x) = 0, we will use the squeeze theorem, which states that if a function g(x) is bounded between two other functions f(x) and h(x) such that lim(x→a) f(x) = lim(x→a) h(x) = L, then lim(x→a) g(x) = L.

Here, f(x) = -x, g(x) = sin(x), and h(x) = x. The hint given is that -x ≤ sin(x) ≤ x for all x ≥ 0.

As x approaches 0, both f(x) and h(x) also approach 0:

lim(x→0) -x = 0 and lim(x→0) x = 0

Now, we apply the squeeze theorem. Since -x ≤ sin(x) ≤ x and both f(x) and h(x) have a limit of 0 as x approaches 0, then:

lim(x→0) sin(x) = 0

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A reflection across the line x = 1 for the quadrilateral is: L'(1, 1), K'(2, 0), M'(4, 0), J'(1, 2)

There are different types of transformation of objects namely:

Reflection

Rotation

Translation

Dilation

Now, the coordinates of the given quadrilateral are:

J(2, 2), K(-3, 0), L(1, 1), M(2, -4)

With a reflection across x = 1, we have the new coordinates as:

L'(1, 1), K'(2, 0), M'(4, 0), J'(1, 2)

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(a) The exponential equation to represent the amount of money in the account after t years is [tex]A(t) = 1600(1.0103)^{(4t)}[/tex].

(b) On solving the  exponential equation the amount of money that will be in the account after 7 years is $2,177.61.

What is an exponential function?

The formula for an exponential function is [tex]f(x) = a^x[/tex], where x is a variable and a is a constant that serves as the function's base and must be bigger than 0.

(a) The exponential function to represent the amount of money in the account after t years with quarterly compounding is -

[tex]A(t) = P(1 + \frac{r}{n})^{(nt)}[/tex]

where -

P = initial deposit = $1600

r = annual interest rate = 4.12%

n = number of compounding periods per year = 4 (since interest is compounded quarterly)

t = time in years

Substituting the given values, in the equation we get -

[tex]A(t) = 1600(1 + \frac{0.0412}{4})^{(4t)}[/tex]

Simplifying -

[tex]A(t) = 1600(1.0103)^{(4t)}[/tex]

Therefore, the equation is [tex]A(t) = 1600(1.0103)^{(4t)}[/tex].

(b) To find the amount of money in the account after 7 years, we need to substitute t = 7 in the equation -

[tex]A(7) = 1600(1.0103)^{(4\times7)}[/tex]

A(7) = 1600(1.3610)

A(7) = $2,177.61 (rounded to the nearest cent)

Therefore, the amount of money in the account after 7 years, assuming no additional deposits or withdrawals, will be $2,177.61.

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Answer:

A: A = P(1 + r/n)^nt

B: $2131.72

Step-by-step explanation:

A = P(1 + r/n)^nt

A = 1600(1 + .0412/4)^(4)(7)

A = 1600(1 + .0103)^(28)

A = 1600(1.0103)^(28)

A = $2131.72

The total amount accrued, principal plus interest, on a principal of $1600 at a rate of 4.12% per year compounded 4 times a year over 7 years is $2131.72.

To determine if a sequence converges or diverges, we need to find its general term and analyze its behavior as n approaches infinity. The given sequence has the general term:
a(n) = (n + p)(n + 2p)(n^2 + p)


ii. To find the first five terms of the sequence, we will plug in n = 1, 2, 3, 4, and 5:

a(1) = (1 + p)(1 + 2p)(1 + p^2)
a(2) = (2 + p)(2 + 2p)(4 + p^2)
a(3) = (3 + p)(3 + 2p)(9 + p^2)
a(4) = (4 + p)(4 + 2p)(16 + p^2)
a(5) = (5 + p)(5 + 2p)(25 + p^2)

These are the first five terms of the sequence, but their exact values will depend on the value of p.

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The standard deviation for the number of wins is approximately 0.1567.

The probability of winning the lottery is [tex]$p = 1/9999$[/tex]. If a person plays the lottery [tex]$n$[/tex] times, the number of wins [tex]$X$[/tex] follows a binomial distribution with parameters [tex]$n$[/tex] and [tex]$p$[/tex]. The mean of [tex]$X$[/tex] is given by [tex]$\mu = np$[/tex], and the variance is given by [tex]$\sigma^2 = np(1-p)$[/tex]. Therefore, the standard deviation is[tex]$\sigma = \sqrt{np(1-p)}$[/tex].

In this case, the person plays the lottery 246 times. Thus, the expected number of wins is [tex]$\mu = np = 246 \times \frac{1}{9999} = 0.0246$[/tex], and the variance is [tex]$\sigma^2 = np(1-p) = 246 \times \frac{1}{9999} \times \frac{9998}{9999} = 0.0245$[/tex]. Therefore, the standard deviation is [tex]$\sigma = \sqrt{0.0245} \approx 0.1567$[/tex].

Thus, the standard deviation for the number of wins is approximately 0.1567.

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The solution of the initial value problem y′=3cosx+2 with y(3π/2)=8 is y(x) = 3sin(x) + 2x - 1/2π.

To solve the initial value problem y′=3cosx+2 with y(3π/2)=8, we need to find a function y(x) that satisfies the differential equation and the initial condition.

First, we find the antiderivative of 3cos(x) + 2, which is 3sin(x) + 2x + C, where C is a constant of integration. Then, we apply the initial condition y(3π/2) = 8 to determine the value of C.

y(3π/2) = 3sin(3π/2) + 2(3π/2) + C = -3/2π + 3π + C = 8

Solving for C, we get C = -1/2π. Thus, the solution to the initial value problem is:

y(x) = 3sin(x) + 2x - 1/2π

To verify that this solution satisfies the differential equation, we can take its derivative:

y′(x) = 3cos(x) + 2

Substituting this expression into the differential equation y′=3cosx+2, we see that y(x) is indeed a solution.

In summary, we solved the initial value problem y′=3cosx+2 with y(3π/2)=8 by finding the antiderivative of the given function, applying the initial condition to determine the constant of integration, and verifying that the resulting function satisfies the differential equation.

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There are 79,833,600 ways for the 8 girls and 5 boys to sit together if two boys wish to sit together.

To solve this problem, we can think of the two boys who wish to sit together as a single unit. Therefore, we have 7 girls, 3 individual boys, and 1 unit of two boys.
The number of ways to arrange these 11 people is 11! (11 factorial), which equals 39,916,800. However, within the unit of two boys, there are 2! (2 factorial) ways to arrange them. Therefore, we need to multiply 11! by 2! to get the total number of ways:
11! x 2! = 79,833,600

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The amplitude of vibrations after a very long time is 0.

a) The equation of motion of a mass-spring system is given by

m x'' + kx = f(t)

where m is the mass, k is the spring constant and f(t) is the external force. Substituting the given values, we get

6x'' + 54x = 30e−4t cos 5t

The solution of this equation is given by

x(t) = A cos (ωt + θ)

where A is the amplitude, ω is the angular frequency and θ is the phase angle.

Substituting the given values, we get

x(t) = A cos (5t + θ)

At t = 0, x(0) = A cos θ

At t = π, x(π) = A cos (5π + θ)

Therefore, the position of the mass when t = π is given by

x(π) = A cos (5π + θ)

b) The amplitude of vibrations after a very long time is given by

A = x(0) = A cos θ

Therefore, the amplitude of vibrations after a very long time is 0.

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The open intervals on which f(x) is concave up are (-1/√6, 1/√6) and the open intervals on which f(x) is concave down are (-∞, -1/√6) and (1/√6, ∞). The x-coordinates of the inflection points are x = ±1/√6.

To determine where f(x) is concave up or down, we need to find the

second derivative of f(x) and examine its sign. The second derivative of

f(x) is:

[tex]f''(x) = -18x^2 + 3[/tex]

To find the intervals where f(x) is concave up, we need to solve the

inequality:

f''(x) > 0

[tex]-18x^2 + 3 > 0[/tex]

Solving this inequality, we get:

[tex]x^2 < 1/6[/tex]

-1/√6 < x < 1/√6

Therefore, f(x) is concave up on the interval (-1/√6, 1/√6).

To find the intervals where f(x) is concave down, we need to solve the inequality:

f''(x) < 0

[tex]-18x^2 + 3 < 0[/tex]

Solving this inequality, we get:

[tex]x^2 > 1/6[/tex]

x < -1/√6 or x > 1/√6

Therefore, f(x) is concave down on the intervals (-∞, -1/√6) and (1/√6, ∞).

To find the inflection points, we need to find the x-coordinates where the

concavity changes, i.e., where f''(x) = 0 or is undefined.

From [tex]f''(x) = -18x^2 + 3[/tex], we see that f''(x) is undefined at x = 0. At x = ±1/

√6, f''(x) changes sign from positive to negative or vice versa, so these

are the inflection points.

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Sarina used 51/10 or 5.1 cups of brown sugar for her 8.5 batches of sugar cookies.

To find out how many cups of brown sugar Sarina used for 8.5 batches of sugar cookies.

We need to multiply the number of batches (8.5) by the amount of brown sugar per batch (3/5 cups).

We are doing the step by step calculation,

1. Write down the given values: 8.5 batches and 3/5 cups of brown sugar per batch.

2. Multiply the number of batches (8.5) by the amount of brown sugar per batch (3/5 cups): 8.5 × (3/5).

To perform the multiplication: (8.5) × (3/5) = (17/2) × (3/5) = (17×3) / (2×5) = 51/10

Hence, Sarina used 51/10 or 5.1 cups of brown sugar for her 8.5 batches of sugar cookies.

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The plane’s altitude will be approximately 6,336 feet when it has flown over 7 miles of ground and climbed at a 9° angle.
What is meant by miles?

Miles are a unit of distance used in the United States and some other countries, equal to 5,280 feet or 1.609 kilometres. Miles are commonly used to measure distances between cities, countries, and other geographical locations.

What is meant by angle?

A geometric shape known as an angle is created by two rays or line segments that have a similar terminal (referred to as the vertex). Angles can be expressed as radians or degrees and are used to describe how lines and shapes are oriented, situated, and related to one another. Angles can be categorised according to their size and shape as acute, right, obtuse, straight, or reflex.

According to the given information

The problem can be solved using trigonometry 1. The altitude of the plane can be calculated using the tangent function as follows: tan(9°) = h/7 h = 7 tan(9°) ≈ 1.2 miles ≈ 6,336 feet

Therefore, the plane’s altitude will be approximately 6,336 feet when it has flown over 7 miles of ground and climbed at a 9° angle.

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The 99th percentile of the IQ distribution is an IQ score of approximately 111.65.

The 99th percentile of the IQ distribution, we need to find the IQ score that is greater than or equal to 99% of the scores in the distribution.

A standard normal distribution table, we can find the z-score corresponding to the 99th percentile, which is approximately 2.33.

The formula for standardizing a normal distribution to find the IQ score corresponding to this z-score:

[tex]z = (X - \mu) / \sigma[/tex]

z is the z-score, X is the IQ score we want to find, [tex]\mu[/tex]is the mean IQ of the distribution (100), and [tex]\sigma[/tex] is the standard deviation of the distribution (5).

Substituting the values we have:

2.33 = (X - 100) / 5

Multiplying both sides by 5:

11.65 = X - 100

Adding 100 to both sides:

X = 111.65

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The area of the region enclosed by y = ln(x) is e - 1.

The area of the region enclosed by y = ln(x), the x-axis, the y-axis, and y = 1.

(A) Using the method of horizontal slices (dx), we can integrate with respect to x:

The limits of integration are x = 1 (where the curves intersect) and x = e (where y = 1).

The height of the slice is y = 1 - ln(x)

Therefore, the area is given by:

A = ∫[1,e] (1 - ln(x)) dx

= x - x ln(x) |[1,e]

= e - e ln(e) - 1 + 1 ln(1)

= e - 1

Therefore, the area of the region is e - 1 square units.

(B) Using the method of vertical slices (dy), we can integrate with respect to y:

The limits of integration are y = 0 (where the curve intersects the x-axis) and y = 1.

The width of the slice is x = [tex]e^y[/tex]

Therefore, the area is given by:

A = ∫[0,1] [tex]e^y[/tex] dy

= [tex]e^y[/tex] |[0,1]

= e - 1

Therefore, the area of the region is e - 1 square units, which is the same as the result obtained using horizontal slices.

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Sarah's percentage improvement in her time is 18% between Year 8 and Year 11.

A percentage is a means to represent a percentage of 100 as a part of a whole. "%" is the symbol for percentage. For instance, if there are 25 female students in a class of 100, we may say that there are 25% of female students in the class because 25 is 25/100, or 0.25 when represented as a fraction of 100.

In a variety of areas, including finance, statistics, and daily life, percentages are used. They are frequently used to compare values that are stated in different units, such as weight or height, and to describe changes, such as percentage increases or decreases. Many professions require the ability to understand percentages, and it is frequently vital to be able to convert between percentages, fractions, and decimals.

The percentage improvement can be given by the formula:

percentage improvement = ((old time - new time) / old time) x 100%

Converting the time in one unit we have:

3 minutes 20 seconds = 3(60) + 20 = 200 sec

2 minutes 44 seconds = 2(60) + 44 = 164 sec

Substituting the values we have:

percentage improvement = (200 sec - 164 sec) / 200 sec x 100%

percentage improvement = 18%

Hence, Sarah improved her time by 18% between Year 8 and Year 11.

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The equation that describes the greatest horizontal length, H, in terms of the greatest vertical length, V is given as follows:

H = 8V/7.

The ratio between two amounts a and b is given as follows:

a to b.

Which is also the division of the two amounts.

The ratio of the horizontal length and the vertical length is 8:7, hence:

H/V = 8/7

Applying cross multiplication, the equation is given as follows:

H = 8V/7.

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we cannot determine the values of f(x) and g(x) at any other point, except for the given points f(2) = 8, g(2) = 0, f(4) = 10, and g(4) = 8.

To answer this question, we need to use the Mean Value Theorem (MVT) for differentiation. According to MVT, if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that:

f(b) - f(a) = f'(c) * (b - a)

We can apply this theorem to both functions f(x) and g(x) on the interval [2, 4]. Therefore, we have:

f(4) - f(2) = f'(c) * (4 - 2)
10 - 8 = f'(c) * 2
2 = f'(c)

g(4) - g(2) = g'(d) * (4 - 2)
8 - 0 = g'(d) * 2
4 = g'(d)

So, we know that f'(c) = 2 and g'(d) = 4. However, we do not know the exact values of c and d. We only know that they exist in the open interval (2, 4) for both functions.

Therefore, we cannot determine the values of f(x) and g(x) at any other point, except for the given points f(2) = 8, g(2) = 0, f(4) = 10, and g(4) = 8.

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After Simplifying the expression "8.a³.b¹.2", the equivalent Algebraic Expression will be 16a³b.

The "Algebraic-Expression" is defined as a "mathematical-phrase" that contains one or more variables, combined with constants and mathematical-operations such as addition, subtraction, multiplication, division, and exponentiation.

We have to equivalent Algebraic Expression of 8a³b¹2.

⇒ 8.a³.b¹.2 = 8 × a³ × b¹ × 2,

Since multiplication is commutative, we rearrange terms without changing value of expression;

⇒ 8 × a³ × b¹ × 2 = 8 × 2 × a³ × b¹,

⇒ 8.a³.b¹.2 = 8 × 2 × a³ × b¹ = 16a³b,

Therefore, the simplified algebraic expression is 16a³b.

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The given question is incomplete, the complete question is

Simplifying and write the equivalent Algebraic Expression of 8.a³.b¹.2.

To find the critical value for a 98% confidence interval, we need to find the z-score that corresponds to the level of confidence. Since we are using a two-tailed test, we need to split the alpha level (2% or 0.02) into two equal parts (1% or 0.01 on each tail) and find the corresponding z-scores.

Using a standard normal distribution table or calculator, we can find that the z-score for a one-tailed area of 0.01 is approximately 2.33. Therefore, the z-score for a two-tailed area of 0.02 is approximately 2.33. So the critical value for a 98% confidence interval is 2.33.

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Since g is twice-differentiable and the limit of g as x approaches infinity is 4, we know that g'(x) approaches 0 as x approaches infinity (otherwise, the limit of g would not exist).

Using L'Hopital's rule, we can take the derivative of both the numerator and denominator of the expression 1/infinity, which gives us:
lim as x->infinity g'(x) / 1 = lim as x->infinity g''(x) / 0
Since g''(x) is the derivative of g'(x), we can apply the same logic and use L'Hopital's rule again:
lim as x->infinity g''(x) / 0 = lim as x->infinity g'''(x) / 0
We can continue applying L'Hopital's rule until we reach a finite limit. Since g is twice-differentiable, we know that g'''(x) exists, but we don't know what its limit is as x approaches infinity. However, we do know that g'(x) approaches 0 as x approaches infinity, so we can conclude that: lim as x->infinity g'(x) / 1 = 0
Therefore, 1/infinity multiplied by 0 is equal to 0.
In summary: 1/infinity times the limit of g'(x) as x approaches infinity is equal to 0.

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We should sample at least 61 farming regions to estimate the mean price of watermelon crops with a margin of error of ​$0.30 per 100 pounds and 90% confidence.

To estimate the mean price of farmers' watermelon crops with a margin of error of ​$0.30 per 100 pounds and 90% confidence, we need to use the formula:
The margin of error = (Z-value) x (standard deviation / square root of sample size)
Here, we want the margin of error to be ​$0.30 per 100 pounds, which is our desired precision level. The Z-value for 90% confidence is 1.645. We know that the standard deviation of watermelon prices is ​$1.99 per 100 pounds, as per prior studies.
Plugging these values into the formula, we get:
0.30 = 1.645 x (1.99 /[tex]\sqrt{ (n)}[/tex])
Solving for n, we get:
n = [tex](1.645 * 1.99 / 0.30)^2[/tex] = 60.19
Therefore, we should sample at least 61 farming regions to estimate the mean price of watermelon crops with a margin of error of ​$0.30 per 100 pounds and 90% confidence.

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There is a 30% chance that the stock price will be more than $27. Since the closing price of the stock is uniformly distributed between $20 and $30, we can assume that each value within that range has an equal chance of occurring. Therefore, the probability of the stock price being more than $27 is the same as the probability of the stock price falling between $27 and $30.

To get this probability, we can calculate the proportion of the total range that falls within the $27 to $30 range. This can be done by finding the length of the $27 to $30 range (which is $3), and dividing it by the length of the entire range ($30 - $20 = $10).
So the probability of the stock price being more than $27 is: $3 / $10 = 0.3, or 30%
Therefore, there is a 30% chance that the stock price will be more than $27.

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Answer:

Set your calculator to degree mode.

The figure is not shown--please sketch it to confirm my answer.

Let h = horizontal distance.

cos(6°) = h/3

h = 3cos(6°) = 2.983 km

A is the correct answer.

The principal value of integral  ∫ x sin x/ X^4 + 4 dx can be evaluated as PV ∫ x sin x/ X^4 + 4 dx = (1/4) [2(π/2) - π] = π/4

To evaluate the principal value of the integral ∫ x sin x/ X^4 + 4 dx, we can use the substitution u = x^2, du = 2x dx. Then, we have:
∫ x sin x/ X^4 + 4 dx = (1/2) ∫ sin(u)/ (u^2 + 4) du
Next, we can use partial fractions to simplify the integrand:
sin(u)/ (u^2 + 4) = A/(u + 2) + B/(u - 2)
Multiplying both sides by (u + 2)(u - 2) and setting u = -2 and u = 2, we get:
A = -1/4, B = 1/4
Therefore, we have:
(1/2) ∫ sin(u)/ (u^2 + 4) du = (1/2)(-1/4) ∫ sin(u)/ (u + 2) du + (1/2)(1/4) ∫ sin(u)/ (u - 2) du
Using integration by parts on each integral, we get:
(1/2)(-1/4) ∫ sin(u)/ (u + 2) du = (-1/8) cos(u) - (1/8) ∫ cos(u)/ (u + 2) du
(1/2)(1/4) ∫ sin(u)/ (u - 2) du = (1/8) cos(u) + (1/8) ∫ cos(u)/ (u - 2) du
Substituting back u = x^2, we have:
∫ x sin x/ X^4 + 4 dx = (-1/8) cos(x^2)/(x^2 + 2) - (1/8) ∫ cos(x^2)/ (x^2 + 2) dx + (1/8) cos(x^2)/(x^2 - 2) + (1/8) ∫ cos(x^2)/ (x^2 - 2) dx
Note that since the integrand has poles at x = ±√2, we need to take the principal value of the integral. This means we split the integral into two parts, from -∞ to -ε and from ε to +∞, take the limit ε → 0, and add the two limits together. However, since the integrand is even, we can just compute the integral from 0 to +∞ and multiply by 2:
PV ∫ x sin x/ X^4 + 4 dx = 2 lim ε→0 ∫ ε^2 to ∞ [(-1/8) cos(x^2)/(x^2 + 2) + (1/8) cos(x^2)/(x^2 - 2)] dx
Using integration by parts on each integral, we get:
2 lim ε→0 [(1/8) sin(ε^2)/(ε^2 + 2) + (1/8) sin(ε^2)/(ε^2 - 2) + ∫ ε^2 to ∞ [(-1/4x) sin(x^2)/(x^2 + 2) + (1/4x) sin(x^2)/(x^2 - 2)] dx]
The first two terms tend to 0 as ε → 0. To evaluate the integral, we can use the substitution u = x^2 + 2 and u = x^2 - 2, respectively. Then, we have:
PV ∫ x sin x/ X^4 + 4 dx = ∫ 0 to ∞ [(-1/4(u - 2)) sin(u)/ u + (1/4(u + 2)) sin(u)/ u] du
= (1/4) ∫ 0 to ∞ [(2/u - 1/(u - 2)) sin(u)] du
Using the fact that sin(u)/u approaches 0 as u approaches infinity, we can apply the Dirichlet test to show that the integral converges. Therefore, we can evaluate it as:
PV ∫ x sin x/ X^4 + 4 dx = (1/4) [2(π/2) - π] = π/4

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Given the above problem on circles theorem, m∡AB = 140° This is resolved using the angle at the center theorem.

The Angle at the Center Theorem states that the measure of an angle formed by two intersecting chords in a circle is equal to half the sum of the measures of the arcs intercepted by the angle.

In other words, if two chords intersect inside a circle, and an angle is formed at the center of the circle by these chords, then the measure of that angle is equal to half the sum of the measures of the arcs intercepted by the angle.

Thus, since ∡CB is the arc formed by the angle at the center,

m∡AB = 360° - 120°-100°

m∡AB = 140°

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The 99% confidence interval is - 3.455 < μ < 6.205 for the population mean μ.

Assuming that the population is normally distributed for,

x = 88, S = 15 and n = 64

Thus, sample mean, x' = x/n = 88/ 64 = 1.375

The z- score of 99% confidence interval is 2.576.

Therefore the confidence interval of the population mean, say μ, is,

μ = x' ± [tex]z_{\alpha /2}[/tex] ( S /√n )

⇒ μ = 1.375 ± 2.576 ( 15 / √64 )

(where, [tex]z_{\alpha /2}[/tex] represents the z- score at the 99% confidence interval)

⇒ μ = 1.375 ± 2.576 ( 1.875)

⇒ μ = 1.375 ± 4.83

⇒ - 3.455 < μ < 6.205

Thus at 99% confidence interval of the population mean, μ is - 3.455 < μ < 6.205.

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