let x ∼ beta(α, β) and let y ∼ beta(α β, γ). if x and y are independent random variables find the joint pdf of u and v where u

The answer of the given question based on the Scale factor is , Part (a) the dimensions of the print are 3.51 feet by 6 feet. , Part (b) the area of the print is 21.06  feet².

Dimension refers to the measurement or size of an object in terms of length, width, and height. It can also refer to the number of parameters or variables that are required to describe a system or phenomenon.

Part A:

To find the dimensions of the print in feet, we need to first find the dimensions of the original painting in feet, and then multiply them by the scale factor of 3.

The width of the painting in feet is 14 inches divided by 12 inches/foot, which is:

14/12 = 1.17 feet

The length of the painting in feet is 24 inches divided by 12 inches/foot, which is:

24/12 = 2 feet

Multiplying these dimensions by the scale factor of 3, we get:

Width of print = 1.17 feet x 3 = 3.51 feet

Length of print = 2 feet x 3 = 6 feet

Therefore, the dimensions of the print are 3.51 feet by 6 feet.

Part B:

To find the area of the print in square feet, we need to multiply the width and length of the print together. Using the dimensions we found in Part A, we get:

Area of print = width x length

Area of print = 3.51 feet x 6 feet

Area of print = 21.06  feet²

Therefore, the area of the print is 21.06 feet².

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Answer:

The second one down

Step-by-step explanation:

The 26 ft  ladder is the hypotenuse of the right triangle

24 feet is the vertical leg

One bottle of liquid plant food is enough for at least 12 weeks, and potentially a bit more.

William uses 40 milliliters of liquid plant food per week, so to find out how much plant food he needs for 12 weeks, we can simply multiply the weekly usage by the number of weeks:

Amount of plant food needed for 12 weeks = 40 milliliters/week x 12 weeks = 480 milliliters

So, William needs 480 milliliters of liquid plant food for 12 weeks.

Since the bottle of liquid plant food, William purchased contains 0.5 liters or 500 milliliters of liquid plant food, we can see that one bottle is enough for 12 weeks since 480 milliliters is less than the total amount of liquid plant food in the bottle.

In fact, we can calculate how many weeks one bottle of liquid plant food will last William by dividing the total amount of liquid plant food in the bottle by the amount used per week:

Time to use up the liquid plant food = (Total amount of liquid plant food) / (Amount used per week) = 500 ml / 40 ml/week ≈ 12.5 weeks

So, we can say that one bottle of liquid plant food is enough for at least 12 weeks, and potentially a bit more.

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The correct question should be:

William bought a 0.5 -liter bottle of liquid plant food. He uses 40 milliliters each week? How much plant food does William need for 12 weeks? Is one bottle enough for 12 weeks?

Answer:

Step-by-step explanation:

If it is exponential dividing any term by the preceding term will be a constant, the common ratio or rate. 50/25=100/50...=2 The initial value, a, (at x=0) is 25 so the general form is

y=ar^x we have a=25 and r=2 so

y=25(2^x)

Answer:

C A person with an annual income of at least \$50{,}000$50,000dollar sign, 50, comma, 000 is more likely to be 454545 years old or above

A person in the 44 and under age group is more likely to have an annual income of at least $50,000​ is a correct statement.

We have given that,

A sociologist polled a random sample of people and asked them their age and annual income.

annual income is the amount of money you receive during the year into your bank account, before any deductions.

The two-way frequency table below shows the results.

Annual income vs. age group Less than $50,000 At least $50,000

Total Ages 44 and under 148 68 216 Ages 45 and above 38 94 132

Total 186 162 348

A person with an annual income of at least

\(=\$50000\\=$50,000dollar\)

Therefore statement b is true.

A person in the 44 and under age group is more likely to have an annual income of at least $50,000​.

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The equation that represents the situation in which case the percent of successful free throws is 75% during the game is; ( 8 + x ) = 0.75 ( 15 + x ).

It follows from the task content that the equation which represents the situation in which case the percent of successful free throws rises to 75% is to be determined.

As evident in the task content; Initially, Shaq made 8 out of 15 free throws he attempted.

Therefore, let the number of consecutive free throws he has to make to raise the success rate to 75% be; x.

Therefore, it follows from percent proportion that;

( 8 + x ) / ( 15 + x ) = 75 / 100

The simplified equation which represents the required situation is therefore;

( 8 + x ) = 0.75 ( 15 + x ).

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In this case, the area of rectangle EHGF is 24 square units. To find the perimeter, we first need to know the dimensions of the rectangle, specifically, its length and width.

Since we don't have any specific information about the length and width, we can consider the possible dimensions that could result in an area of 24 square units. Here are some possibilities:

1. Length (L) = 1 unit, Width (W) = 24 units
2. Length (L) = 2 units, Width (W) = 12 units
3. Length (L) = 3 units, Width (W) = 8 units
4. Length (L) = 4 units, Width (W) = 6 units

Once you identify the correct dimensions, you can calculate the perimeter using the formula for the perimeter of a rectangle:

Perimeter = 2(L+W)

Simply plug in the values for the length and width, and calculate the perimeter. Here's an example using the first set of dimensions:

Perimeter = 2(1+24) = 2(25) = 50 units

Please note that the correct dimensions will depend on the context or specific information given in the problem. With the available information, we can't pinpoint the exact perimeter, but we have a general understanding of how to approach this type of problem.

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The sample space of rolling a six-faced dice is {1, 2, 3, 4, 5, 6}.

The sample space of an experiment is the set of all possible outcomes of the experiment. In the experiment of rolling a six-faced dice, the sample space is the set of all possible values that the dice can show when it is rolled.

Since a dice has six faces, each with a distinct number (1, 2, 3, 4, 5, or 6), the sample space of the experiment of rolling a six-faced dice is {1, 2, 3, 4, 5, 6}. This means that any time the dice is rolled, it can only show one of these six values.

It's important to note that the sample space is a finite set, as there are only a limited number of possible outcomes. In this case, the sample space contains only six elements, each representing one of the faces of the dice that can show up when it is rolled.

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Answer:

cubic polynomial

Step-by-step explanation:

3 is highest exponent and 3 is cubic

there are more than 3 terms, so it's polynomial (poly=many)

Answer:

24.5 inches

Step-by-step explanation:

The ratio of the length of the toucan’s bill to the length of its body can be expressed as

(1: 3.5)

Then if the bill is 7 inches, then the Length of the body will be X

1: 3.5

7: X

If we cross multiply, we have

X= 7×3.5

X= 24.5

Hence, the Length of the body is 24.5 inches

The Laplace transform of \(t^2 * sin(6t)\) is calculated using the properties of Laplace transforms and the formula for the Laplace transform of \(t^n * sin(at).\)

To find the Laplace transform of \(t^2 * sin(6t)\), we can use the formula for the Laplace transform of\(t^n * sin(at)\), which is given by:

\(L{t^n * sin(at)} = (2 * a^n * n!) / (s^(n+1) * (s^2 + a^2)^2)\)

In this case, n = 2 and a = 6. Plugging in these values into the formula, we get:

\(L{t^2 * sin(6t)} = (2 * 6^2 * 2!) / (s^(2+1) * (s^2 + 6^2)^2)\)

\(= (72 * 2) / (s^3 * (s^2 + 36)^2)= (144) / (s^3 * (s^2 + 36)^2)\)

Therefore, the Laplace transform of \(t^2 * sin(6t)\) is (144) / \((s^3 * (s^2 + 36)^2)\).

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Answer:

D and E because they are bigger than 28. C is 28 but there isn't a greater OR EQUAL sign next to it  

Answer:

21,704

Step-by-step explanation:

Find the volume of both the top of the silo (hemisphere) and the end bit of the silo (cylinder)

V = \(\frac{2}{3}\) \(\pi\)\(r^{3}\) is the formula for finding the volume of a hemisphere

V = \(\pi\)\(r^{2}\)h is the formula for finding the volume of a cylinder

Let's find the hemisphere first.

We're using 3.14 for pi, so the equation would look like this

V = \(\frac{2}{3}\) x 3.14 x \(12^{3}\)

V = \(\frac{2}{3}\) x 3.14 x 1728

V = 3617.28 \(ft^{3}\)

Now let's find the volume of the cylinder.

V = 3.14 x \(12^{2}\) x 40

V = 3.14 x 144 x 40

V = 18,086.4 \(ft^{3}\)

Last, you add both volumes together.

3617.28 + 18,086.4 = 21,703.68

Round it to make it 21,704.

The total number of cubic feet of grain or volume of the grain that the silo can hold is 28947.41 .

The area that any three-dimensional solid occupies is known as its volume. These solids can take the form of a cube, cuboid, cone, cylinder, or sphere.

Given that the top of a silo is a hemisphere

Radius(r) = 12 ft.

Then, the volume of a hemisphere is,

V = 2/3πr³

V = 2/3 x 3.14 x 12³

V = 10851.84 feet³

And, the bottom of the silo is a cylinder with

height (h) = 40ft.

the volume of a cylinder is,

V = πr²h

V = 3.14 x 12² x 40

V = 18095.57

So, the total volume is

V = 10851.84 + 18095.57

V = 28947.41

Then, total volume = total number of cubic feet of grain

Hence, the required answer is, 28947.41 .

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Answer:

608

Step-by-step explanation:

because I think I'm right

Monotone convergence theorem states that if A∈A and (An​)n∈N​⊆A, if An​⊆An+1​ for n∈N and A=⋃n∈N​An​, then limn→[infinity]​μ(An​)=μ(A). Proof: Define B1​=A1​, Bn​=An​∖An−1​ for all n∈N, then we have ⋃n∈N​An​=⋃n∈N​Bn​ and Bn​∩Bm​=∅ for all n,m∈N and n≠m.

Then we can write μ(⋃n∈N​An​)=∑n=1∞μ(An​)since μ is countably additive, and we have∑n=1∞μ(An​)=limn→[infinity]∑k=1nμ(Ak​)=limn→[infinity]μ(⋃k=1nAk​)=μ(⋃n∈N​An​).Therefore, limn→[infinity]​μ(An​)=μ(A).  We can use monotone convergence/dominated convergence to show that if μ(A1​)<[infinity], An​⊇An+1​ for n∈N and A=⋂n∈N​An​, then limn→[infinity]​μ(An​)=μ(A).Proof: Define C1​=A1​, Cn​=Cn−1​∩An​ for all n∈N, then we have⋂n∈N​An​=⋂n∈N​Cn​ and Cn​⊆Cn+1​ for all n∈N.

Then we can write μ(⋂n∈N​An​)=μ(⋂n∈N​Cn​)and thenμ(⋂n∈N​An​)=limn→[infinity]μ(Cn​) since Cn​⊆Cn+1​, and μ is monotonic, we can apply monotone convergence theorem and conclude thatlimn→[infinity]μ(Cn​)=μ(⋂n∈N​Cn​)=μ(A).Therefore, limn→[infinity]​μ(An​)=μ(A).Dominated convergence theorem states that if limn→[infinity]​1An​​=1A​ almost surely, and there is B∈A such that An​⊆B and μ(B)<[infinity], then μ(A)<[infinity] and limn→[infinity]​μ(An​)=μ(A).Proof: For all n, we have|1An​​|≤1B∈L1, and 1An​​→1A​​ almost surely. Therefore, by the dominated convergence theorem, we have∫|1An​​−1A​​|dμ→0 as n→∞.Since 1An​​→1A​​ almost surely, we haveμ(An​)→μ(A) as n→∞. Therefore, limn→[infinity]​μ(An​)=μ(A).

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Monotone convergence theorem states that if A∈A and (An​)n∈N​⊆A, if An​⊆An+1​ for n∈N and A=⋃n∈N​An​, then limn→[infinity]​μ(An​)=μ(A).

For Proof:

To Define B1​=A1​, Bn​=An​∖An−1​ for all n∈N, we have ⋃n∈N​An​=⋃n∈N​Bn​ and Bn​∩Bm​=∅ for all n,m∈N and n≠m.

Then we will write μ(⋃n∈N​An​)=∑n=1∞μ(An​)

since μ is countably additive, and we have;

∑n=1∞μ(An​)=limn→[infinity]∑k

=1nμ(Ak​)=limn→[infinity]μ(⋃k=1nAk​)

=μ(⋃n∈N​An​).

Therefore, limn→[infinity]​μ(An​)=μ(A).  

We will use monotone convergence/dominated convergence to show that if μ(A1​)<[infinity], An​⊇An+1​ for n∈N and A=⋂n∈N​An​,

then limn→[infinity]​μ(An​)=μ(A).

Proof:

To Define C1​=A1​, Cn​=Cn−1​∩An​ for all n∈N, then ⋂n∈N​An​=⋂n∈N​Cn​ and Cn​⊆Cn+1​ for all n∈N.

Then we can write μ(⋂n∈N​An​)=μ(⋂n∈N​Cn​)and thenμ(⋂n∈N​An​)=limn→[infinity]μ(Cn​) since Cn​⊆Cn+1​, and μ is monotonic,

Now we can apply the monotone convergence theorem and conclude thatlimn→[infinity]μ(Cn​)=μ(⋂n∈N​Cn​)=μ(A).

Therefore, limn→[infinity]​μ(An​)=μ(A).

For convergence theorem states that if limn→[infinity]​1An​​=1A​ almost surely, and there is B∈A such that An​⊆B and μ(B)<[infinity], then μ(A)<[infinity] and limn→[infinity]​μ(An​)=μ(A).

To Proof For all n, we have|1An​​|≤1B∈L1, and 1An​​→1A​​ almost surely.

Therefore, by the dominated convergence theorem, we have∫|1An​​−1A​​|dμ→0 as n→∞.Since 1An​​→1A​​ almost surely μ(An​)→μ(A) as n→∞.

Therefore, limn→[infinity]​μ(An​)=μ(A).

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Answer: Supplementary, 3 and 2, substitution, 2.

Step-by-step explanation: just did it .

The ∠1 is congruent to ∠3 by definition of congruency.

An angle is formed from the intersection of two lines. Types of angles are acute, obtuse and scalene.

m∠1 + m∠2 = 180°  (given)

m∠2 + m∠3 = 180°  (given)

Hence:

m∠1 + m∠2  = m∠2 + m∠3  (equality property)

Subtract m∠2 from both sides to get:

m∠1  = m∠3

The ∠1 is congruent to ∠3 by definition of congruency.

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Answer:

The Human Side of Human Beings

The Fundamentals of Co-Counseling Manual

How to Begin "Re-evaluation Counseling"

The Human Situation

Working Together to End Racism

Guidelines for the Re-evaluation Counseling Communities

Step-by-step explanation:

Answer:

7 po ang tamang sagot welcome

I don't answer the grammar but if you asked how many kids in total (not counting duplicate people that signed up for both) then 1073 student registered for the courses

Answer: y=8x-58

Step-by-step explanation:

1. find the slope

Since the new line is parallel to the first one that means the slope of 8x will remain the same.

2. plug it into the point slope formula y-y1=m(x-x1)

which would look like

y-(-2)=8(x-7)

3. solve for y

y-(-2)=8(x-7)

y+2=8x-56

y=8x-58

For every positive integer n, there are n consecutive composite integers.

We will prove the statement using the fact that every integer greater than 1 is either prime or can be factored into a product of primes.

Consider the n consecutive integers starting with (n+1)!+2, which are:

(n+1)!+2, (n+1)!+3, (n+1)!+4, ..., (n+1)!+n+1

We will show that each of these integers is composite.

First, note that (n+1)!+2 is composite, since it is greater than 2 and can be factored as 2*(n+1)!/2 + 1.

Next, for each i between 2 and n+1, inclusive, we have:

(n+1)!+i = i*((n+1)!/i) + i

Since i divides (n+1)!, we have (n+1)!/i as an integer, and since i is between 2 and n+1, we have (n+1)!/i greater than 1. Thus, (n+1)!+i can be factored into a product of at least two integers, i and (n+1)!/i + 1. Since i is between 2 and n+1, we have (n+1)!/i + 1 between 3 and (n+1), inclusive. Therefore, (n+1)!+i is composite.

Thus, each of the n integers (n+1)!+2, (n+1)!+3, ..., (n+1)!+n+1 is composite, as desired. Therefore, for every positive integer n, there are n consecutive composite integers.

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Answer:

0.90

Step-by-step explanation:

Take the 600 and get rid of the zeroes and divide 5.40 by 6.

The unit price of butter per 100 grams is $0.9.

Given that, one can buy 600 grams of butter for $5.40.

Unit rate can be defined as the ratio between two measurements with the second term as 1. It is considered to be different from a rate, in which a certain number of units of the first quantity is compared to one unit of the second quantity.

Here, \(unit \ price = \frac{Total \ cost \ of \ butter}{Total \ quantity \ of \ butter}\)

Now, unit price \(= \frac{5.40}{600}\)

= $0.009

Thus, the unit price of butter per 100 grams is

\(=100\times0.009\)

\(=\$0.9\)

Therefore, the unit price of butter per 100 grams is $0.9.

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Answer:

450

Step-by-step explanation:

Step 1:

18 × 25 = 450

Answer:

450

Hope This Helps :)

Answer:

450 is your answer lol...

18/35 fraction of students are girls who brought their lunch today.

A fraction is a mathematical concept that denotes a component of a whole or, more generally, any number of equal parts. A fraction, such as one-half, eight-fifths, and three-quarters, denotes the number of components of a specific size when used in conversational English.

Let's say,

Total number of students over school = x

Number of girls = (6/7)x

Number of girls who brought their lunch = 3/5 of (6/7)x

⇒ 3/5(6/7)x

⇒ (18/35)x

Hence "18/35 fraction of students are girls who brought their lunch today".

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She could make her approximation in the multiples of 1/6.

Probability is the chance of happening a certain event of all the possible events that can occur.

Statistics is a branch of mathematics where organising and interpretation of data analysing is needed.

A fair 6 sided dice has 6 sides having numbers from 1 to 6 and for a fair dice each side has a probability of 1/6. So she could make her approximation in the multiples of 1/6 for example 6/36 and the numerator she can vary by ±10 % for approximation.

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Answer:

The first is 36,763    second 30 hours

Step-by-step explana

Answer:

i)

Odd numbers:

P(odd and odd) = 3/6*3/6 = 1/4

ii)

Total outcomes:

Sum of 8:

Sum of 12:

P(8 or 12) = 5/36 + 1/36 = 6/36 = 1/6

iii)

Prime numbers:

Odd numbers:

P(prime or odd) = 3/6*3/6 + 3/6*3/6 = 1/4 + 1/4 = 1/2

Answer:

cos(30)

Step-by-step explanation:

Answer:

sin(15)

Step-by-step explanation: