Suppose that a category of world class runners are known to run a marathon (26 miles) in an average of 146 minutes with a standard deviation of 15 minutes. Consider 49 of the races. Let X = the average of the 49 races.Find the probability that the average of the sample will be between 143 and 147 minutes in these 49 marathons. (Round your answer to four decimal places.)Find the 60th percentile for the average of these 49 marathons. (Round your answer to two decimal places.)______ minFind the median of the average running times._____min

Answer:

0.718 = 71.8% probability that X is less than 38 minutes

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

[tex]f(x)=\mu e^{-\mu x}[/tex]

In which [tex]\mu=\frac{1}{m}[/tex] is the decay parameter.

The probability that x is lower or equal to a is given by:

[tex]P(X\leq x)=\int\limits^a_0f ({x)} \, dx[/tex]

Which has the following solution:

[tex]P(X\leq x)=1-e^{-\mu x}[/tex]

If X has an average value of 30 minutes

This means that [tex]m=30,\mu=\frac{1}{30}[/tex]

What is the probability that X is less than 38 minutes?

[tex]P(X\leq 38)=1-e^{-\frac{38}{30} }[/tex]

0.718 = 71.8% probability that X is less than 38 minutes

The scale factor of dilation of the shaded in gray to the shaded in black is 3

Given that

The preimage = shaded in gray

The image = shaded in black

From the graph, we have the following values on the image and the preimage

The preimage = shaded in gray = 4

The image = shaded in black = 12

The scale factor of dilation is then calculated as

Scale factor = shaded in black/shaded in gray

So, we have

Scale factor = 12/4

Evaluate

Scale factor = 3

Hence, the scale factor dilation is 3

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To highlight the line y = 0 on the graph in black/grey, draw a straight line passing through all points whose y-coordinate is 0.

What is graph?

In mathematics, a graph is a visual representation of a set of data, typically as a set of points or lines on a coordinate plane. Graphs are used to represent various types of data, such as numerical values, functions, relationships, and patterns.

Assuming that the graph is a coordinate plane with the x-axis and y-axis, do the following:

To highlight the point (9, 8) on the graph in red, locate the point (9, 8) on the coordinate plane and mark it with a red color.

To highlight the point (20, f(20)) on the graph in green, you need to know the value of f(20) first. Once you have that value, locate the point (20, f(20)) on the coordinate plane and mark it with a green color.

To highlight the line y = 5 on the graph in blue, draw a straight line passing through all points whose y-coordinate is 5. This line should be parallel to the x-axis and should be marked with a blue color.

Therefore To highlight the line y = 0 on the graph in black/grey, draw a straight line passing through all points whose y-coordinate is 0. This line should be parallel to the x-axis and should be marked with a black/grey color.

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From the figure 1. Two line segments are LA and EP. 2. Two rays are EC and AH. 3. Two lines are b and AP.

A ray is a segment of a line with a single endpoint and unlimited length in a single direction. A ray cannot be measured in terms of length.

The ends of a line segment are two. These endpoints are included, along with every point on the line that connects them. A segment's length can be measured, while a line's length cannot.

A line is a collection of points that extends in two opposing directions and is endlessly long and thin.

From the given figure we observe that,

1. Two line segments are LA and EP.

2. Two rays are EC and AH.

3. Two lines are b and AP.

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Answer:

father age 45 son age 0 this is answer

False. The second moment about mu for a binomial experiment is not given by the second derivative of [tex](p+qeAt)[/tex]with respect to t evaluated at t=0.

The binomial distribution is a discrete probability distribution that describes the number of successes in a fixed number of independent trials, each with the same probability of success. The probability of success in each trial is denoted by p, and the probability of failure is denoted by q=1-p.

The second moment about mu is a measure of the variability of the binomial distribution, and is given by the formula[tex]E[(X-mu)^2][/tex] , where X is the random variable, mu is the mean, and E is the expected value operator.

To calculate the second moment about mu for a binomial distribution with parameters n and p, we can use the formula npq, where np is the mean and q=1-p. This formula can also be derived using the properties of variance, which state that [tex]Var(X)=E[X^2] - (E[X])^2.[/tex]

Therefore, the statement that the second moment about mu for a binomial experiment is given by the second derivative of [tex](p+qeAt)[/tex]with respect to t evaluated at t=0 is false. This statement does not relate to the binomial distribution or its properties, and is not a relevant formula for measuring the variability of a binomial experiment.

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Answer:

False.

Step-by-step explanation:

A concave polygon can never be a regular polygon as it can never be equiangular. Each side of a regular polygon must be the same length, and all interior angles must also be equal.

a) An automorphism of a group G is a bijective map φ:G→G that preserves the group structure. That is, φ(ab) = φ(a)φ(b) and φ(a⁻¹) = φ(a)⁻¹ for all a, b ∈ G.

b) The set-theoretic composition φψ of any two automorphisms φ, ψ is an automorphism, as it preserves the group structure and is bijective.

c) The set Aut(G) of all automorphisms of G, with the operation of composition of maps, is a group. This is because it satisfies the four group axioms: closure, associativity, identity, and inverses. Therefore, Aut(G) is a group under composition of maps.

An automorphism of a group G is a bijective map φ:G→G that preserves the group structure, meaning that for any elements a,b∈G, we have φ(ab) = φ(a)φ(b) and φ(a⁻¹) = φ(a)⁻¹. In other words, an automorphism is an isomorphism from G to itself.

To show that the set-theoretic composition φψ is an automorphism, we need to show that it satisfies the two conditions for being an automorphism. First, we have

(φψ)(ab) = φ(ψ(ab)) = φ(ψ(a)ψ(b)) = φ(ψ(a))φ(ψ(b)) = (φψ)(a)(φψ)(b)

using the fact that ψ and φ are automorphisms. Similarly,

(φψ)(a⁻¹) = φ(ψ(a⁻¹)) = φ(ψ(a))⁻¹ = (φψ)(a)⁻¹

using the fact that ψ and φ are automorphisms. Therefore, φψ is an automorphism.

To show that Aut(G) is a group, we need to show that it satisfies the four group axioms

Closure: If φ,ψ∈Aut(G), then φψ is also in Aut(G), as shown above.

Associativity: Composition of maps is associative, so (φψ)χ = φ(ψχ) for any automorphisms φ,ψ,χ of G.

Identity: The identity map id:G→G is an automorphism, since it clearly preserves the group structure and is bijective. It serves as the identity element in Aut(G), since φid = idφ = φ for any φ∈Aut(G).

Inverses: For any automorphism φ∈Aut(G), its inverse φ⁻¹ is also an automorphism, since it is bijective and preserves the group structure. Therefore, Aut(G) is closed under inverses.

Since Aut(G) satisfies all four group axioms, it is a group under composition of maps.

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Answer:

See below.

Step-by-step explanation:

We can solve this linear programming problem using the simplex method. We will start by converting the problem into standard form

Maximize z = 3x₁ + 5x₂ + 0s₁ + 0s₂

subject to

x₁ - 5x₂ + s₁ = 35

3x₁ - 4x₂ + s₂ = 21

x₁, x₂, s₁, s₂ ≥ 0

Next, we create the initial tableau

Basis x₁ x₂ s₁ s₂ RHS

s₁ 1 -5 1 0 35

s₂ 3 -4 0 1 21

z -3 -5 0 0 0

We can see that the initial basic variables are s₁ and s₂. We will use the simplex method to find the optimal solution.

Step 1: Choose the most negative coefficient in the bottom row as the pivot element. In this case, it is -5 in the x₂ column.

Basis x₁ x₂ s₁ s₂ RHS

s₁ 1 -5 1 0 35

s₂ 3 -4 0 1 21

z -3 -5 0 0 0

Step 2: Find the row in which the pivot element creates a positive quotient when each element in that row is divided by the pivot element. In this case, we need to find the minimum positive quotient of (35/5) and (21/4). The minimum is (21/4), so we use the second row as the pivot row.

Basis x₁ x₂ s₁ s₂ RHS

s₁ 4/5 0 1/5 1 28/5

x₂ -3/4 1 0 -1/4 -21/4

z 39/4 0 15/4 3/4 105

Step 3: Use row operations to create zeros in the x₂ column.

Basis x₁ x₂ s₁ s₂ RHS

s₁ 1 0 1/4 7/20 49/10

x₂ 0 1 3/16 -1/16 -21/16

z 0 0 39/4 21/4 525/4

The optimal solution is x₁ = 49/10, x₂ = 21/16, and z = 525/4.

Therefore, the maximum value of z is 525/4, which occurs when x₁ = 49/10 and x₂ = 21/16.

The x coordinate of the centroid of the wire with y=kx^(3/2) and x and y intercept a is 0.546a. The y coordinate is 8a/5.

To find the centroid of the wire, we need to find the area and first moments of the wire, which are given by:

Area, A = ∫y dx, where x ranges from -a to a

First moment with respect to x, Mx = ∫xy dx, where x ranges from -a to a

Then the x coordinate of the centroid is given by:

xc = Mx / A

We can start by finding the area:

A = ∫y dx = ∫kx^(3/2) dx = (2/5)kx^(5/2) + C

At x = a, y = 0, so C = - (2/5)ka^(5/2)

At x = -a, y = 0, so A = 2(2/5)ka^(5/2) = (4/5)ka^(5/2)

Now we need to find the first moment with respect to x:

Mx = ∫xy dx = ∫kx^(5/2) dx = (2/7)kx^(7/2) + C'

At x = a, y = 0, so C' = - (2/7)ka^(7/2)

At x = -a, y = 0, so Mx = 0

Therefore, the x coordinate of the centroid is:

xc = Mx / A = 0 / [(4/5)ka^(5/2)] = 0

This means that the centroid lies on the y-axis. To find its y coordinate, we can use the formula:

yc = ∫x dy / A = ∫x (dy/dx) dx / A

Using the equation y = kx^(3/2), we can find dy/dx:

dy/dx = (3/2)kx^(1/2)

Substituting this into the formula for yc and simplifying, we get:

yc = (4/5)ka^(5/2) / (5/8)ka^(5/2) = (8/5)a

Therefore, the coordinates of the centroid are (0, 8/5 a), and the y coordinate is (8/5)a.

To find the x coordinate of the centroid, we need to use the formula:

xc = (1/A) ∫x y dx

We already found the expression for the area A, so we just need to evaluate the integral:

xc = (1/A) ∫x y dx = (1/A) ∫x kx^(3/2) dx

Integrating this by substitution with u = x^(1/2), we get:

xc = (2/5a^(5/2)) ∫u^4 du = (2/5a^(5/2)) (u^5/5) + C

where C is a constant of integration.

At x = a, y = 0, so u = a^(1/2) and C = -(2/25)a^(5/2).

At x = -a, y = 0, so the contribution to the integral is zero.

Therefore, the x coordinate of the centroid is:

xc = (2/5a^(5/2)) (u^5/5) - (2/25a^(5/2)) = (2/25)a(5√2 - 1)

Plugging in a = 1, we get:

xc = 0.546a

So the x coordinate of the centroid is 0.546 times the x and y intercept value a.

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_____The given question is incomplete, the complete question is given below:

a homogeneous wire is bent into the shape shown of graph y = kx^(3/2), x and y intercept is 'a'. determine the x coordinate of its centroid by direct integration. express your answer in terms of a. Also find y- coordinate.

Answer: 1.85

Step-by-step explanation:

Answer: 14.5

Step-by-step explanation:

When there is a decimal point, you start counting from the left any number that is not zero. If the zero is at the end, then you count it.

For example, if the answer is 0.000145 then the number of significant figures is still three because you start counting from the first nonzero number from the left.

If the answer is 14.50, then the number of significant figures is four because you start counting from the first nonzero number from the left.

14.53 is the answer to the equation but because you want to correct it to 3 significant figures, you round down because 3 is less than 5 and 14.5 ends up being the final answer.

Answer:

Step-by-step explanation:

u got this

If "f" is differentiable and f(1) < f(2), then there is a number "c", in the interval  (1, 2)  such that f'(c)>  0.

Applying the  Mean Value Theorem for derivatives, if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in the interval (a, b) such that:

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

In the scenario above, we have that f is differentiable, and that f(1) < f(2).

choosing a = 1 and b = 2.

Then applying the Mean Value Theorem, there exists at least one number c in the interval (1, 2) such that:

f'(c) = (f(2) - f(1)) / (2 - 1)

f'(c) = f(2) - f(1)

We have that f(1) < f(2), we have:

f(2) - f(1) > 0

We can conclude by saying that there exists a number c in the interval (1, 2) such that:

f'(c) = f(2) - f(1) > 0

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Answer:

height = 7 yards

Step-by-step explanation:

the area (A) of a trapezoid is calculated as

A = [tex]\frac{1}{2}[/tex] h (b₁ + b₂ )

where h is the perpendicular height between the 2 parallel bases

here b₁ = 12 , b₂ = 9 and A = 73.5 , then

[tex]\frac{1}{2}[/tex] h(12 + 9) = 73.5 ( multiply both sides by 2 to clear the fraction )

h(21) = 147

21h = 147 ( divide both sides by 21 )

h = 7 yards

Answer:

F) 12

Step-by-step explanation:

To answer, we use the perfect square formula:

(a + b)² = a² + 2ab + b²

(√3 + √3)² = (√3)² + 2(√3)(√3) + (√3)²

Simplify:

√3² = 3

2(√3)(√3) = 2 x (√3)² = 2 x 3 = 6

Plug in:

(√3 + √3)² = 3 + 6 + 3 = 12

Answer:

A large equilateral triangle pyramid stands in front of the city's cultural center. Each side of the base measures 40 feet and the slant height of each lateral side of the pyramid is 50 feet.

A painter can paint 100 square feet of the pyramid in 18 minutes.

How long does it take the painter to paint 75% of the pyramid?

Step-by-step explanation:

The total surface area of the pyramid can be calculated using the formula for the lateral surface area of a pyramid:

Lateral surface area = (1/2) × perimeter of base × slant height

Since the base is an equilateral triangle, the perimeter is 3 times the length of one side:

Perimeter of base = 3 × 40 feet = 120 feet

Lateral surface area = (1/2) × 120 feet × 50 feet = 3000 square feet

To paint 75% of the pyramid, the painter needs to paint:

0.75 × 3000 square feet = 2250 square feet

Since the painter can paint 100 square feet in 18 minutes, the time required to paint 2250 square feet can be calculated as:

2250 square feet ÷ 100 square feet per 18 minutes = 225 ÷ 10 × 18 minutes = 405 minutes

Therefore, the painter would need 405 minutes or 6 hours and 45 minutes to paint 75% of the pyramid.

The standard deviation that the student would have in order to be publicly recognized is given as 1.17

We would have to assume that the students score follows a normal distribution

This is given as

X ~ (μ, σ)

(μ, σ) are the mean and the standard deviation

1 - 12 percent =

0.88 = 88 percent

using the excel function given as NORMS.INV() we would find the standard deviations

=NORM.S.INV(0.88)

= 1.17498

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A student would have to score approximately 0.89 standard deviations above the mean to be in the top 12% and be publicly recognized.

we can use the empirical rule to estimate the number of standard deviations a student has to score above the mean to be in the top 12 percent, assuming it is a normal distribution

The empirical rule states that for a normal distribution:

we will use the complement rule since our aim is to find the number of standard deviations a student has to score above the mean to be in the top 12%.

The complement of being in the top 12% is being in the bottom 88%.

From the empirical rule, we have that 68% of the data falls within one standard deviation of the mean.

Therefore, the remaining 32% (100% - 68%) falls outside one standard deviation of the mean.

Since we want to find the number of standard deviations a student has to score above the mean to be in the bottom 88%, we can assume that the remaining 32% is split evenly between the two tails of the distribution.      

Applying the z-score formula:

z = (x - μ) / σ

The z-score for a cumulative area of 0.44 is approximately -0.89  found by looking up the z-score corresponding to the cumulative area of 0.44 (half of 0.88) in a standard normal distribution table.

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To earn an A grade, a student needs to score at least 82.8 , calculated using the inverse normal cumulative distribution function with a mean of 70, a standard deviation of 10, and a 10th percentile of 0.10.

Given that the grades are normally distributed with a mean of 70 and a standard deviation of 10.

We need to find the score which is at the 10th percentile of the distribution.

Using the standard normal distribution table, we can find the z-score that corresponds to the 10th percentile.

From the table, we can see that the z-score is approximately -1.28.

Using the formula for standardizing a normal distribution:

z = (x - μ) / σ

where z is the z-score, x is the raw score, μ is the mean, and σ is the standard deviation.

Substituting the given values, we have:

-1.28 = (x - 70) / 10

Solving for x, we get:

x = (-1.28 * 10) + 70

x = 82.8

Therefore, a student would need to earn a score of approximately 82.8 to receive an A grade.

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Answer:

[tex]14 {y}^{2} \sqrt{ {x}^{3} {z}^{9} }[/tex]

Step-by-step explanation:

[tex] \sqrt{196 {x}^{3} {y}^{4}{z}^{9} }= \sqrt{196} \times \sqrt{ {x}^{3} } \times \sqrt{ {y}^{4} } \times \sqrt{ {z}^{9} } \\ \sqrt{196} = 14 \\ \sqrt{ {x}^{3} } = {x}^{ \frac{3}{2} } \\ \sqrt{ {y}^{4} } = {y}^{2} \\ \sqrt{ {z}^{9}} = {z}^{ \frac{9}{2} }[/tex]

A fractional exponent is not necessarily simpler so just take out the 1st and 3rd parts of the term which simplify nicely:

[tex] \sqrt{196 {x}^{3} {y}^{4}{z}^{9} } = 14 {y}^{2} \sqrt{ {x}^{3} {z}^{9} } [/tex]

Answer:

Yes

Step-by-step explanation:

Yes because 1+14=15 hours and that is more than two

The equation of the line passing through A and B is y = (4/5)x - (2/5).

A straight line's general equation is y = mx + c, where m is the gradient and y = c is the value at which the line intersects the y-axis. The y-axis intercept is denoted by the number c. A straight line with gradient m and intercept c on the y-axis has the equation y = mx + c.

The point-slope form of a linear equation can be used to find the equation of the line passing through points A and B:

y - y1 = m(x - x1) (x - x1)

where m denotes the slope of the line, (x1, y1) denotes the coordinates of point A or B, and (x, y) denotes the coordinates of any other point on the line.

To calculate the slope, we can use points A (3, 2) and B (8, 6).

m = (y2 - y1) / (x2 - x1) = (6 - 2) / (8 - 3)\s= 4 / 5

So the equation for the line connecting A and B is:

y - 2 = (4/5)(x - 3) (x - 3)

This equation can be simplified by multiplying both sides by 5:

5y - 10 = 4x - 12

Then we can rearrange it to form the slope-intercept equation, y = mx + b:

5y = 4x - 2

y = (4/5)x - (2/5)

As a result, the equation for the line connecting A and B is y = (4/5)x - (2/5).

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In the graph, Student B and C are both 10 years old and student C has a shoe size of 5. The coordinates of D are (12,6)

In graph theory, a graph is a framework that consists of a collection of objects, some of which are paired together to form "related" objects. The objects are represented by mathematical abstractions known as vertices (also known as nodes or points), and each set of connected vertices is known as an edge (also called link or line). A graph is typically shown diagrammatically as a collection of dots or circles representing the centres and lines or curves representing the edges.

Both directed and undirected lines are possible. For instance, if the edges between two individuals are handshakes, then the graph is undirected because any individual A can only shake hands with an individual B if B also holds hands with A. The graph is directed, however, if an edge from person A to person B indicates that A owes money to B because borrowing money is not always returned.

In the given graph,

Student B and C are both 10 years old and student C has a shoe size of 5.

The coordinates of D are (12,6)

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Therefore, each portion will be (1-x)/4 pounds.

Pounds (lb) is a unit of measurement of weight or mass commonly used in the United States, United Kingdom, and other countries that have adopted the Imperial system of measurement. One pound is equal to 0.453592 kilograms (kg). The symbol for pound is "lb", which comes from the Latin word libra. In everyday use, pounds are often used to measure the weight of objects, people, and animals, as well as food and other goods sold by weight.

Given by the question.

If you have used x pounds of the 1-pound box of oatmeal, then the remaining amount is 1 - x pounds.

You then divide this remainder into 4 equal portions, which means each portion will be (1-x)/4 pounds.

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Answer:

$31,142

Step-by-step explanation:

To convert a percentage into a decimal, you move the decimal two places to the left. 1354% converted into a decimal is 13.54.

$23,000 * 13.54 = $31,142

So, the probability of Freddie getting 4 base hits in a row is approximately 0.0088, or 0.88%.

Probability is a measure of the likelihood or chance of an event occurring. It is expressed as a number between 0 and 1, where 0 represents an impossible event and 1 represents a certain event.

by the question.

Assuming that each at-bat is independent of the others, the probability of Freddie getting a base hit in one at-bat is 0.305.

To find the probability that he gets 4 base hits in a row, we can use the multiplication rule for independent events. This rule states that the probability of two or more independent events occurring together is the product of their individual probabilities.

Therefore, the probability of Freddie getting 4 base hits in a row is:

0.305 x 0.305 x 0.305 x 0.305 = 0.0088 (rounded to four decimal places)

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Step-by-step explanation:

Refer to pic..........

Answer:

Let's assume that the rate of interest is r (in decimals), and the time period is also r years. Then we can use the simple interest formula:

I = P * r * t

where I is the interest paid, P is the principal amount (the loan amount in this case), r is the rate of interest per year, and t is the time period in years.

Substituting the given values, we get:

324 = 900 * r * r

Simplifying, we get:

r² = 324/900

r² = 0.36

Taking the square root of both sides, we get:

r = ±0.6

Since the rate of interest cannot be negative, we can take r = 0.6. Therefore, the rate of interest is 0.6 or 60% per year.

By using rational zeros theorem, we find that there are no real zeros of the polynomial function f(x) = x^3 - x^2 - 37x - 35, so we cannot factor f(x) over the real numbers.

To find the real zeros of the polynomial function f(x) = x^3 - x^2 - 37x - 35, we can use the rational zeros theorem, which states that any rational zeros of the function must have the form p/q, where p is a factor of the constant term (-35) and q is a factor of the leading coefficient (1).

The possible rational zeros of f are therefore ±1, ±5, ±7, ±35. We can then test each of these values using synthetic division or long division to see if they are zeros of the function. After testing all of the possible rational zeros, we find that none of them are actually zeros of the function.

Therefore, we can conclude that there are no real zeros of the function f(x) = x^3 - x^2 - 37x - 35.

However, we could factor it into linear and quadratic factors with complex coefficients using the complex zeros of f(x). But since the problem only asks for factoring over the real numbers, we can conclude that the factored form of f(x) is:

f(x) = x^3 - x^2 - 37x - 35 (cannot be factored over the real numbers)

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Answer:

680 ways

Step-by-step explanation:

C(17, 3) gives 17! / (14! 3!), or (17*16*15)/6 = 680 ways to select the 3 individuals.

Hope this helped!