What is score-based generative modeling through stochastic differential equations?

The length of the segment BC which is having midpoint DE is 6cm.

Given that GH is the mid-segment of triangle DEF and DE is the mid-segment of triangle ABC. From this relation, we can obtain the length of the segment BC easily.

The relation between the base and mid-segment of the triangle is,

base = 2 x mid-segment

So, in triangle DEF the length of the base DE = 2 x GH

                                                                      DE = 2 x 1.5 cm = 3 cm.

Similarly, in triangle ABC the length of the base BC = 2 x DE

                                                                                BC = 2 x 3 cm = 6 cm.

From the above analysis, we can conclude that the length of the segment BC is 6cm.

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Answer:

Let x be the length of one side of the square. Then, the perimeter of the square is 4x.

Each rectangle has two sides that are equal in length to the side of the square, and two sides that are equal in length to some other value y. Therefore, the perimeter of each rectangle can be expressed as:

2x + 2y = 20

Simplifying this equation, we get:

x + y = 10

We can use this relationship to solve for y in terms of x:

y = 10 - x

The area of each rectangle is given by:

Area = x*y

Substituting y = 10 - x, we get:

Area = x*(10-x)

Area = 10x - x²

To find the area of each rectangle, we need to solve for x. We know that the area of the square is 44 cm², so:

x² = 44

x ≈ 6.63 cm

Now we can find the area of each rectangle:

Area = 10x - x²

Area = 10(6.63) - (6.63)²

Area ≈ 38.85 cm²

Therefore, each rectangle has an area of approximately 38.85 cm².

The population density is 5750.2 people per square mile.

The radius of the circular region is 6 mile

Area of the circular region [tex]A =\pi r^2[/tex] sq. units.

Now, put the value of radius in above formula

[tex]A = \pi(6)^2[/tex]

A = 3.14 × 36

A = 113.04 [tex]mile^2[/tex]

Calculate the population density

The population of the circular region is

The formula for the population density of the region is : [tex]\frac{Population}{Area}[/tex]

Substitute 650,000 for population and area as 113.04 [tex]mile^2[/tex]

So, Population density = 650000/113.04

Population density = 5750.2 people per square mile

Hence, population density is 5750.2 people per square mile.

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Answer:

180°

Step-by-step explanation:

We Know

Each square represents 90°

We see the △STU rotate counterclockwise about the origin through 2 squares. So, the  △STU rotated 180° counterclockwise to produce △S'T'U.

We can check by using the rule rotate about the origin 180° state that:

(x , y) → (-x , -y)

Pick point U (-2,1)

(-2,1) → (2, -1)

This meet the rule, so rotate 180° is the correct answer.

Answer:

  see attached

Step-by-step explanation:

You want the values that complete the table when y is inversely proportional to the cube of x.

A proportional relation has the form ...

  y = kx

where k is the constant of proportionality.

If y is inversely proportional to x, that is the same as saying y is proportional to the inverse of x:

  y = k(1/x)

Here, y is proportional to the inverse of the cube of x:

  y = k/x³

The value of k can be found as ...

  k = x³·y = (10³)(0.6) = 600

So the relation to use with the table is ...

  y = 600/x³

The y-value can be found using the equation ...

  y = 600/x³

  y = 600/5³ = 600/125 = 24/5

  y = 4.8

Solving the proportion for x gives ...

  x³ = k/y

  x = ∛(k/y) = ∛(600/75) = ∛8

  x = 2

Now, we can fill the table as in the attachment.

Answer: x = 3

Step-by-step explanation:

1:simplify the expression and divide both sides by the same factor

x^2-6x+9=0

2:use the quadratic formula

ax2 + bx + c = 0

a=1    b=-6   c=9

3:Evaluate the exponent

Multiply the numbers

Subtract the numbers

Evaluate the square root

Add zero

Multiply the numbers

Cancel terms that are in both the numerator and denominator

x=3

Answer:

the dilation in g(x) as it relates to the parent function f(x) is a vertical dilation by a factor of 15000.

Step-by-step explanation:

The function g(x) is a dilation of the parent function f(x) = 0.98^x.

Specifically, g(x) = 15000(0.98)^x is a vertical dilation of f(x) = 0.98^x by a factor of 15000.

This means that the graph of g(x) is the same shape as the graph of f(x), but it is vertically stretched by a factor of 15000.

In other words, the values of g(x) are 15000 times greater than the values of f(x) for any given value of x.

For example, when x = 1, f(1) = 0.98 and g(1) = 15000(0.98) = 14700. This means that the number of visitors to the water park in the second month (x = 1) is 15000 times greater than the number of visitors in the first month.

Overall, the dilation in g(x) as it relates to the parent function f(x) is a vertical dilation by a factor of 15000.

Answer:

62

Step-by-step explanation:

180-70-48=62

Angles in a triangle add up to 180.

Answer: (a) tan 16°:

Using a calculator, we get:

tan 16° ≈ 0.291

Rounding to three significant figures, tan 16° ≈ 0.291.

(b) tan 89°:

Using a calculator, we get:

tan 89° ≈ 57.290

Note that this value is extremely large and not meaningful in most contexts.

(c) tan 18.7°:

Using a calculator, we get:

tan 18.7° ≈ 0.340

Rounding to three significant figures, tan 18.7° ≈ 0.340.

(d) tan 45.4°:

Using a calculator, we get:

tan 45.4° ≈ 1.058

Rounding to three significant figures, tan 45.4° ≈ 1.06.

Step-by-step explanation:

The type II error occurs in the example.

According to the given conditions, the null thesis( H0) is that the food is safe, while the alternative thesis( Ha) is that the food isn't safe. The food inspectors are trying to test this thesis by examining samples of the food product and looking for any signs that it may be unsafe.

A type II error occurs when the examination process fails to descry an unsafe condition in the food product, indeed though it actually exists. In other words, the sample data suggests that the food is safe, but it's not. This means that the null thesis( H0) is accepted when it should have been rejected in favor of the alternative thesis( Ha).

In practical terms, a type II error in this environment could be dangerous, as it could affect unsafe food being distributed to consumers, potentially leading to illness or indeed death. thus, it's important for food inspectors to minimize the threat of type II crimes by using applicable slice styles and testing procedures to ensure the safety of the food products they check.

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Answer:

A is correct.

A'B' = (1/2)(34) = 17

The measure of angle A stays the same, 28°.

The measure of angle QVR is 24°

Angles in parallel lines are angles that are created when two parallel lines are intersected by another line called a transversal.

Since angle UWT and QVR are on indirectly opposite to each other , they will be equal.

This means that UWT = QVR

= 3x+12 = x+20

collecting like terms;

3x-x = 20-12

2x = 8

divide both sides by 2

x = 8/2

x = 4

angle QVR = x+20

therefore QVR = 4+20

= 24°

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Answer:

the range describes the distance from the minimum to the maximum value of the entire data set.

The values of x and z are 2 and 90 degrees respectively

The properties of transversals are;

From the information given, we have that;

The angle z is right angles, that is, it is equal to 90 degrees

z = 90 degrees

Since interior angles are supplementary, we have then that;

14x + 72 + 80 = 180

collect the like terms

14x + 152 = 180

14x = 28

Make 'x' the subject of formula

x =2

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There are 130 positive integers less than 1000 that are divisible by 7 but not by 11.

The rule used to find the number of positive integers less than 1000 that are divisible by 7 but not by 11 is the inclusion-exclusion principle. This principle states that if we want to find the number of elements that belong to at least one of two sets, we can add the sizes of the sets together and then subtract the size of the intersection of the sets. In this case, we first find the number of positive integers less than 1000 that are divisible by 7 (which is 142), then the number that are divisible by 11 (which is 90), and finally the number that are divisible by both (which is 12). Using the inclusion-exclusion principle, we can find the number of positive integers less than 1000 that are divisible by 7 but not by 11 by subtracting the number that are divisible by both from the number that are divisible by 7: 142 - 12 = 130. Therefore, there are 130 positive integers less than 1000 that are divisible by 7 but not by 11.
To find the number of positive integers less than 1000 that are divisible by 7 but not by 11, you can use the Inclusion-Exclusion Principle.

First, find the number of integers divisible by 7: there are 999/7 = 142.71, so 142 integers are divisible by 7.

Next, find the number of integers divisible by both 7 and 11 (i.e., divisible by their LCM, which is 77): there are 999/77 = 12.97, so 12 integers are divisible by 77.

Now, apply the Inclusion-Exclusion Principle: Subtract the number of integers divisible by both 7 and 11 from the number of integers divisible by 7: 142 - 12 = 130.

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Answer: Incomplete question

(i think)

Step-by-step explanation:

Answer:

-5

Step-by-step explanation:

Answer: it would be -5

Step-by-step explanation: it is negative 5 because that is where there is no x intercept

The quotient of the rational expression is 3(x + 5)/2(2x+5)

From the question, we have the following parameters that can be used in our computation:

3x/2x+5 divided by 2x/x+5

Express properly

So, we have

3x/2x+5 ÷ 2x/x+5

Express as products

So, we have

3x/2x+5 * x+5/2x

Cancel out x

So, we have

3/2x+5 * x+5/2

Evaluate the products

3(x + 5)/2(2x+5)

Hence. the solution is 3(x + 5)/2(2x+5)

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The value of x is 21.75 unit.

We have,

Scale Factor = 1 : 7 1/4 = 1 : 29/4

and, the dimension of Figure A = 3 unit

Now, Using the scale factor

 x/3 = 29/4

4x = 87

x= 87/4

x = 21.75 unit

Thus, the dimension of Figure B is 21.75 unit.

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Answer:

Step-by-step explanation:

4i+19=47

4i=28

i=28/4

i=7

4i+19=47

4i=28

i=28/4

i=7

Answer: 53

Step-by-step explanation:

90 - 37

Answer

53 is the angle

Further explanation

Complementary angles sum to 90°.

We can build an equation to find any two complementary angles.

x + y = 90

Either x or y is given. Notice how they're interchangeable which means it doesn't make any difference what value you plug in for x or y.

Allow me to demonstrate.

In our case we have an angle that measures 37.

We can plug in 37 for x or y, but I always plug it in for x.

37 + y = 90

Subtract 37 on each side

y = 53°

The measures of the exterior angles are:

a = 80°

b = 115°

c = 40°

d = 50°

e = 75°

The sum of the measure of an interior angle and the correspondent exterior angle is always 180°, then we can write:

a + 100° = 180°

a = 180° - 100° = 80°

b + 65° = 180°

b = 180° - 65° = 115°

c + 140° = 180°

c = 180° - 140°  =40°

d + 130° = 180°

d = 180° - 130° = 50°

e + 105° = 180°

e = 180° - 105° = 75°

These are the angles.

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The equations without  absolute value for the given conditions is shown below:

y = 2x - 5, if x > 5

y = -2x, if x < -5

y = |2x - 5|, if -5 < x < 5

We take notice the three stated conditions and simplify with regards to that.

Condition  1: x > 5

Condition  2: x < -5

Condition 3: -5 < x < 5

For condition 1, where  x > 5 the expression inside the absolute value bars will be simplified  to:

y = x - 5 + x + 5 = 2x - 5.

Therefore, we have the expression as  y = 2x - 5, if x > 5

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The requried domain restriction for both functions is x ≥ 0.

a. f(x) = x² {x ≥ 0}

To make f(x) one-to-one, we need to restrict the domain so that each output (y-value) of the function fits only one input (x-value). The domain is x ≥ 0, which means that the function starts from the point (0,0) and goes up as x increases. Therefore, the domain restriction for f(x) to be one-to-one is x ≥ 0.

b. g(x) = √x

The square root function g(x) is already one-to-one for x ≥ 0. To make certain that the inverse function also exists, we can restrict the domain to x ≥ 0.

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Answer:

of a polynomial function with integer coefficients.

Rational Zeros Theorem:

If the polynomial ( ) 1

1 1 ... n n P x ax a x ax a n n

− = + ++ − + 0 has integer

coefficients, then every rational zero of P is of the form

p

q

where p is a factor of the constant coefficient 0 a

and q is a factor of the leading coefficient n a

Example 1: List all possible rational zeros given by the Rational Zeros Theorem of

P(x) = 6x

4

+ 7x

3

- 4 (but don’t check to see which actually are zeros) .

Solution:

Step 1: First we find all possible values of p, which are all the factors

of . Thus, p can be ±1, ±2, or ±4. 0 a = 4

Step 2: Next we find all possible values of q, which are all the factors

of 6. Thus, q can be ±1, ±2, ±3, or ±6. n a =

Step 3: Now we find the possible values of p

q by making combinations

of the values we found in Step 1 and Step 2. Thus, p

q will be of

the form factors of 4

factors of 6 . The possible p

q are

12412412412

, , , , , , , , , , ,

11122233366

± ± ± ± ± ± ± ± ± ± ±±

4

6

Example 1 (Continued):

Step 4: Finally, by simplifying the fractions and eliminating duplicates,

we get the following list of possible values for p

q .

1124 1, 2, 4, , , , , 2333

±± ± ± ± ± ± ±

1

6

Now that we know how to find all possible rational zeros of a polynomial, we want to

determine which candidates are actually zeros, and then factor the polynomial. To do this

we will follow the steps listed below.

Finding the Rational Zeros of a Polynomial:

1. Possible Zeros: List all possible rational zeros using the Rational Zeros

Theorem.

2. Divide: Use Synthetic division to evaluate the polynomial at each of the

candidates for rational zeros that you found in Step 1. When the

remainder is 0, note the quotient you have obtained.

3. Repeat: Repeat Steps 1 and 2 for the quotient. Stop when you reach a

quotient that is quadratic or factors easily, and use the quadratic formula

or factor to find the remaining zeros.

Example 2: Find all real zeros of the polynomial P(x) = 2x

4

+ x

3

– 6x

2

– 7x – 2.

Solution:

Step 1: First list all possible rational zeros using the Rational Zeros

Theorem. For the rational number p

q to be a zero, p must be a

factor of a0 = 2 and q must be a factor of an = 2. Thus the

possible rational zeros, p

q , are

1 1, 2, 2

±± ±

Example 2 (Continued):

Step 2: Now we will use synthetic division to evaluate the polynomial at

each of the candidates for rational zeros we found in Step 1.

When we get a remainder of zero, we have found a zero.

Since the remainder is not zero,

12 1 6 7 2

23 31

0

23 3

1 is not a

10 12

zer

o

←

+

−− −

− −

−− −

Since the remainder is zero,

12 1 6 7 2

21

1 is

5 2

21520

a zero

− −−−

−

− −

−

− ←

This also tells us that P factors as

2x

4

+ x

3

– 6x

2

– 7x – 2 = (x + 1)(2x

3

– x

2

– 5x – 2)

Step 3: We now repeat the process on the quotient polynomial

2x

3

– x

2

– 5x – 2. Again using the Rational Zeros Theorem, the

possible rational zeros of this polynomial are

1 1, 2, 2

±± ± .

Since we determined that +1 was not a rational zero in Step 2,

we do not need to test it again, but we should test –1 again.

Since the remainder is zero,

12 1 5 2

232

1 is again a zero

2 3 2 0

− −−

−

−

−

− − ←

Thus, P factors as

2x

4

+ x

3

– 6x

2

– 7x – 2 = (x + 1)(2x

3

– x

2

– 5x – 2)

= (x + 1) (x + 1)(2x

2

– 3x – 2)

= (x + 1)2 (2x

2

– 3x – 2)

Example 2 (Continued):

Step 4: At this point the quotient polynomial, 2x

2

– 3x – 2, is quadratic.

This factors easily into (x – 2)(2x + 1), which tells us we have

zeros at x = 2 and 1

2

x = − , and that P factors as

2x

4

+ x

3

– 6x

2

– 7x – 2 = (x + 1)(2x

3

– x

2

– 5x – 2)

= (x + 1) (x + 1)(2x

2

– 3x – 2)

= (x + 1)2 (2x

2

– 3x – 2)

= (x + 1)2 (x – 2)(2x + 1)

Step 5: Thus the zeros of P(x) = 2x

4

+ x

3

– 6x

2

– 7x – 2 are x = –1, x = 2,

and 1

2

x = − .

Descartes’ Rule of Signs and Upper and Lower Bounds for Roots:

In many cases, we will have a lengthy list of possible rational zeros of a polynomial. A

theorem that is helpful in eliminating candidates is Descartes’ Rule of Signs.

In the theorem, variation in sign is a change from positive to negative, or negative to

positive in successive terms of the polynomial. Missing terms (those with 0 coefficients)

are counted as no change in sign and can be ignored. For example,

has two variations in sign.

Descartes’ Rule of Signs: Let P be a polynomial with real coefficients

1. The number of positive real zeros of P(x) is either equal to the number

of variations in sign in P(x) or is less than that by an even whole

number.

2. The number of negative real zeros of P(x) is either equal to the number

of variations in sign in P(–x) or is less than that by an even whole

number.

Example 3: Use Descartes’ Rule of Signs to determine how many positive and how

many negative real zeros P(x) = 6x

3

+ 17x

2

– 31x – 12 can have. Then

determine the possible total number of real zeros.

Solution:

Step 1: First we will count the number of variations in sign of

( ) . 3 2 Px x x x =+ −− 6 17 31 12

Step-by-step explanation:

The difference in density, in pounds per cubic foot, between the box A and box B to the nearest whole number is 3. So, correct answer is option b.

The density of the body is the mass per unit volume and to find the difference in density we have to find density of each box.

Density of box A = Mass of box A / Volume of box A

= 24.6 pounds / 4 cubic feet

= 6.15 pounds per cubic foot

Density of box B = Mass of box B / Volume of box B

= 37.4 pounds / 4 cubic feet

= 9.35 pounds per cubic foot

The difference in density,

Density box A - density box B.

= (9.35- 6.15)pounds per cubic foot

= 3.2 pounds per cubic foot

To the nearest whole number, the difference in density is 3. Therefore, the answer is (b) 3.

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a) 32 percentage of the Cherry Hill residents polled supported Zhang.

The relative frequency table shows the results of a survey in two communities, Cherry Hill and Mountain View, regarding their preferred candidate for a local election. The table displays the percentage of residents in each community who support the two candidates, Gartman and Zhang.

To find the percentage of Cherry Hill residents who supported Zhang, we need to look at the second row of the table, which shows the percentages of Zhang supporters in each community. The percentage for Cherry Hill is 0.38 or 38%. Therefore, the answer is A) 32%.

Correct Question :

A survey of two communities asked residents which candidate they supported for a local election. The survey data are shown in the relative frequency table Gartman 0.30 0.18 Total 0.62 Chemy Hill Mountain View Tota Zhang 0.32 0.20 0.,38 1.0 What percentage of the Cherry Hill residents polled supported Zhang?

a) 32%

b) 30%

c) About 6280

d) About 52%

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