SHOW STEPS! I WILL MAKE U BRAINLIST
x-y=11
2x+y=19
What is the solution?

The time available for additional sport is the fraction 1/6.

By simply dividing the favourable number of possibilities by the entire number of potential outcomes, the probability of an occurrence may be determined using the probability formula. Because the favourable number of outcomes can never exceed the entire number of outcomes, the chance of an event occurring might range from 0 to 1. Moreover, the proportion of positive outcomes cannot be negative. All probability added together equal one.

Given that, the soccer is changed to 1/3.

The additional time left from 1/2 is:

1/2 - 1/3 = (3 - 2)/6 = 1/6

Hence, the time available for additional sport is the fraction 1/6.

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So the most reasonable estimate for the final price of a bag of chips after the 10% discount is $4.54.

what is an algebraic expression?

In mathematics, an algebraic expression is a combination of variables, constants, and mathematical operations (such as addition, subtraction, multiplication, and division) that represents a quantity.

Algebraic expressions can include numbers, variables, and exponents, but they do not include an equals sign (=) or a relational operator (such as > or <), so they are not equations or inequalities.

To find the final price of a bag of chips, we need to first calculate the markup price and then apply the 10% discount.

The markup price is 152% of the original price, which is equivalent to multiplying the original price by (100% + 152%) = 2.52. Therefore, the markup price of a bag of chips is:

$2 x 2.52 = $5.04

Next, we apply the 10% discount to the markup price by multiplying it by 0.9:

$5.04 x 0.9 = $4.54

Therefore, So the most reasonable estimate for the final price of a bag of chips after the 10% discount is $4.54.

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(Ǝx)(Ǝy)(∀z)¬P(x, y, z) is the simplified statement with no quantifier lying within the scope of a negation

The given statement is ¬(∀x)(∀y)(Ǝz)P(x, y, z). To simplify this statement so that no quantifier lies within the scope of a negation, we need to use the rules of negation.

First, we can use the rule of negation for universal quantifiers, which states that ¬(∀x)P(x) is equivalent to (Ǝx)¬P(x). Applying this rule to the given statement, we get:

(Ǝx)¬(∀y)(Ǝz)P(x, y, z)

Next, we can use the same rule for the second universal quantifier:

(Ǝx)(Ǝy)¬(Ǝz)P(x, y, z)

Finally, we can use the rule of negation for existential quantifiers, which states that ¬(Ǝx)P(x) is equivalent to (∀x)¬P(x). Applying this rule to the last existential quantifier, we get:

(Ǝx)(Ǝy)(∀z)¬P(x, y, z)

This is the simplified statement with no quantifier lying within the scope of a negation. It is equivalent to the original statement, but it is written in a simpler form.

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The last price value for a bond table is $ 918.85.

Finding the last price:

The fomrula used to find the last price is given by

        Net amount = number of bonds x price x (100% - rate)

Where the rate is the percentage of the face value paid as commission, and the price is the price per bond as a percentage of the face value.

Here we have

A bond table for a sale of 40 bonds that resulted in a net amount of 36,754

To find the last price we need to use the formula:

Net amount = number of bonds × price × (100% - rate)

We can rearrange this formula to solve for the price:

Price = Net amount / (number of bonds x (100% - rate))

Substitute the given values, and we get:

Price = $36,754 / (40 x (100% - 0%)) = $918.85  

Therefore,

The last price value for a bond table is $ 918.85.

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Population having expression Pn+1 = 1.04(Pn - G), there were approximately 220,863 moose in Alaska at the beginning of 2017.

What exactly are expressions?

In mathematics, an expression is a combination of numbers, variables, and operators that represents a mathematical quantity or relationship. Expressions can be composed of numbers, variables, and operators such as addition, subtraction, multiplication, division, and exponentiation.

Now,

We are given that:

At the beginning of 2015 (n = 0), P0 = 200,000

At the beginning of 2016 (n = 1), P1 = 200,720

The formula for calculating the number of moose at the start of the following year is Pn+1 = 1.04(Pn - G), where G is a constant.

To find the value of P2 (the number of moose at the beginning of 2017), we can use the formula with n = 1:

P2 = 1.04(P1 - G)

We need to find the value of G. We can use the information given to us at the beginning of 2015 and 2016 to form two equations:

P1 = 1.04(P0 - G)

P2 = 1.04(P1 - G)

We can use substitution to solve for G. Rearranging the first equation, we get:

P0 - G = (P1/1.04)

G = P0 - (P1/1.04)

Substituting this expression for G into the second equation, we get:

P2 = 1.04(P1 - (P0 - (P1/1.04)))

Simplifying and solving for P2, we get:

P2 = 220,863 (rounded to the nearest whole number)

Therefore,

                there were approximately 220,863 moose in Alaska at the beginning of 2017.

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Answer:

[tex]4*(7x+8)^2*{(7x+8)^{\frac{1}{4} }} \\[/tex]

Step-by-step explanation:

[tex]\sqrt[4]{625(7x+8)^9} \\\\\sqrt[4]{625}*\sqrt[4]{(7x+8)^9} \\4*(7x+8)^{\frac{9}{4} }\\4*(7x+8)^2*{(7x+8)^{\frac{1}{4} }} \\[/tex]

Answer:

A Rotation 270° counterclockwise around the origin.

Step-by-step explanation:

The solution to limx→2(5x2 - 5x - 1) is 9.

To solve limx→2(5x2 - 5x - 1), we can use the direct substitution method. This method involves plugging in the value that x is approaching into the function and simplifying.

Step 1: Plug in x = 2 into the function:
limx→2(5(2)2 - 5(2) - 1)

Step 2: Simplify the expression:
limx→2(20 - 10 - 1)

Step 3: Combine like terms:
limx→2(9)

Step 4: Since the limit is a constant, the answer is simply the constant:
9

Hence, the limit of 5x^2 - 5x - 1 as x approaches 2 is equal to 9.

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The real equation a^2 = 1 has exactly two solutions: a = -1 and a = 1.

For any given n, the n × n identity matrix Iₙ has precisely one square root, which is Iₙ. Then, let A be any real invertible 3 x 3 matrix. Then, it has two square roots: A and -A. Therefore, 3 x 3 matrices that satisfy the matrix equation A^2 = I3 will be either A or -A.Let A be a 3 x 3 matrix whose entries are given by:A = [0 b c] [d 0 f] [g h 0]

The first diagonal element is A11 = 0, the second diagonal element is A22 = 0, and the third diagonal element is A33 = 0. Therefore, the matrix A is one such matrix that satisfies A² = I3. A second matrix that satisfies A² = I3 is given by -A, which is given by:-A = [0 -b -c] [-d 0 -f] [-g -h 0

]A third matrix that satisfies A² = I3 is given by the identity matrix, which is given by:I3 = [1 0 0] [0 1 0] [0 0 1]Let J be the 3 x 3 matrix whose entries are given by:J = [0 1 0] [0 0 1] [1 0 0]The matrix J satisfies J³ = I3. Therefore, J² also satisfies the matrix equation A² = I3. J² is given by:J² = [1 0 0] [0 1 0] [0 0 1] = I3

Similarly, J³ satisfies the matrix equation A² = I3, and it is given by:J³ = [0 0 1] [1 0 0] [0 1 0]Therefore, the following eight matrices satisfy A² = I3:A = [0 b c] [d 0 f] [g h 0]-A = [0 -b -c] [-d 0 -f] [-g -h 0]I3 = [1 0 0] [0 1 0] [0 0 1]J = [0 1 0] [0 0 1] [1 0 0]J² = [1 0 0] [0 1 0] [0 0 1]-J = [0 0 1] [1 0 0] [0 1 0]-J² = [0 1 0] [0 0 1] [1 0 0]

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The rate of change of the area for each of the given circumferences is:
a. 30π
b. 180π
c. 45π
d. 30π
e. 90π

The rate of change of the radius of a circle is given as 3. We can use this information to determine the rate of change of the area for each of the given circumferences.

When the circumference is 10π, the radius is 10π/2π = 5. The area of the circle is πr^2 = π(5)^2 = 25π. The rate of change of the area is 2πr * rate of change of radius = 2π(5) * 3 = 30π.

When the circumference is 60π, the radius is 60π/2π = 30. The area of the circle is πr^2 = π(30)^2 = 900π. The rate of change of the area is 2πr * rate of change of radius = 2π(30) * 3 = 180π.

When the circumference is 15π, the radius is 15π/2π = 7.5. The area of the circle is πr^2 = π(7.5)^2 = 56.25π. The rate of change of the area is 2πr * rate of change of radius = 2π(7.5) * 3 = 45π.

When the circumference is 10π, the radius is 10π/2π = 5. The area of the circle is πr^2 = π(5)^2 = 25π. The rate of change of the area is 2πr * rate of change of radius = 2π(5) * 3 = 30π.

When the circumference is 30π, the radius is 30π/2π = 15. The area of the circle is πr^2 = π(15)^2 = 225π. The rate of change of the area is 2πr * rate of change of radius = 2π(15) * 3 = 90π.

In conclusion, the rate of change of the area for each of the given circumferences is:
a. 30π
b. 180π
c. 45π
d. 30π
e. 90π

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The solution of the real matrix A for the fundamental set of solution y' =Ay is given by y1(t) = e^(5t)(cos(3t) − sin(3t)) and y2(t) = e^(5t)(cos(3t) + sin(3t)).

Let A be written as A = P + iQ, where P and Q are real matrices,

Since the eigenvalue is complex, the matrix A must have complex entries.

We have,

Av = λv,

where v = −1 + i is the eigenvector.

(Av)

= (P + iQ)(−1 + i)

= (−P − Q) + i(-Q + P)

= λ(−1 + ii)

= (−5 + 3i)(−1 + i)

= 2 − 8i.

Equating real and imaginary parts, we get two equations,

−P − Q = 2,

- Q + P = −8.

Solving for P and Q, we get

-2Q = -6

⇒ Q = 3

⇒ P = -5

P = 3 − i,

Q = 1 + 3i.

Thus, A = P + iQ

            = -5 + 3i

Now, we can find the real and imaginary parts of e^(λt)v,

e^(λt)v

= e^((5 + 3i)t)(−1 + i)

= e^(5t)e^(3it)(−1 + i)

= e^(5t)(cos(3t) − sin(3t) + i(cos(3t) + sin(3t))).

The real part of e^(λt)v gives a real solution,

so we take y1(t) = Re(e^(λt)v),

y1(t) = e^(5t)(cos(3t) − sin(3t)).

The imaginary part of e^(λt)v gives an imaginary solution,

so we take y2(t) = Im(e^(λt)v).

y2(t) = e^(5t)(cos(3t) + sin(3t)).

Thus, a fundamental set of solutions for y′ = Ay consisting entirely of real solutions is given by y1(t) and y2(t) is equal to e^(5t)(cos(3t) − sin(3t)) and

e^(5t)(cos(3t) + sin(3t)) respectively.

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The investment will be worth approximately $4,667.27 after 30 years of compounding at a rate of 6.5% per annum, compounded weekly.

Assuming that the investment compounds weekly for 30 years, the formula to calculate the future value of the investment would be:

FV = [tex]PV * (1 + r/n)^{(n*t)}[/tex]

where:

PV = present value (initial investment) = $1,000

r = annual interest rate = 6.5%

n = number of times the interest is compounded per year = 52 (weekly)

t = number of years the money is invested = 30

Plugging in these values, we get:

FV = [tex]\$1,000 * (1 + 0.065/52)^{(52*30)}[/tex]

FV = [tex]\$1,000 * (1 + 0.00125)^{1560}[/tex]

FV = [tex]\$1,000 * 2.693[/tex]

FV = $2,693.29

Therefore, the investment will be worth $2,693.29 when Mike turns 30.

The formula for compound interest is a mathematical formula that calculates the final value of an investment based on the initial investment amount, the interest rate, and the number of compounding periods per year. It takes into account the effect of compounding on the investment, which means that the interest earned is added back to the principal amount, so that the investment grows at an accelerating rate over time.

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Answer:

m∠A = 35°

Step-by-step explanation:

∠A and ∠B are complementary angles, meaning their sum is 90°.

(2x-13) + (2x + 7) = 90

4x - 6 = 90

4x = 96

x = 24

Substitute 24 for x in ∠A's equation.

2(24) - 13 = 48 - 13 = 35°

m∠A = 35°

Answer:

Angle A = 35

Step-by-step explanation:

When angles are complementary, this means they add up to 90 degrees. If we want to find the value of angle A, we need to first find out what the unknown variable x is before we can evaluate for A. We can do this by writing an equation.

[tex](2x-13)+(2x+7)=90[/tex]

Combine like terms

[tex]4x-6=90[/tex]

Add 6 to both sides

[tex]4x=96[/tex]

Divide by 4

[tex]x=24[/tex]

X is 24. Now we just plug in 24 and evaluate for A.

[tex]A=2(24)-13[/tex]

Multiply

[tex]A=48-13[/tex]

Subtract

[tex]A=35[/tex]

To check our work, we can also find B and evaluate

[tex]B=55[/tex]

[tex]55+35=90[/tex]

The limit of the function as x approaches 0 is 0.

The limit of a function is the value that the function approaches as the input (or x-value) approaches a certain number. In this case, we are asked to find the limit of the function as x approaches 0.

The function given is: x^3 - 3x

To find the limit, we simply plug in the value of x that we are approaching into the function:

lim x→0 x^3 - 3x = (0)^3 - 3(0) = 0 - 0 = 0

Hence, the function approaches 0 as x gets closer to 0.

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The general solution of the given fourth order linear homogeneous differential equation is: y(x) = c₁x³e^4x + c₂v₁(x) + c₃v₂(x) + c₄v₃(x)where v₁(x), v₂(x), v₃(x) are the solutions of the differential equation v⁽⁴⁾(x) + 36xv'''(x) + (216x² + 72)v''(x) + (432x³ + 324x² + 36)v'(x) + (144x⁴ + 216x³ + 108x²)v(x) = 0, and c₁, c₂, c₃, c₄ are constants.

The given fourth order linear homogeneous differential equation for y(x) with real numbers as coefficients is

y^4 + a3y^3 + a2y^2 + a1y + a0 = 0

where a3, a2, a1 and a0 are real numbers.

Let x³e^4x be one solution of the above differential equation.

Therefore, the other solution should be of the form y(x) = v(x) x³e^4x

Now we have to find the fourth order linear homogeneous differential equation for v(x).

For this, let's differentiate y(x) twice.

y''(x) = [v''(x) + 6xv'(x) + 12v(x)] x³e^4x

Now differentiate y(x) thrice.

y'''(x) = [v'''(x) + 18xv''(x) + (54x² + 12)v'(x) + (36x³ + 36x²)v(x)] x³e^4x

And differentiate y(x) four times.

y⁽⁴⁾(x) = [v⁽⁴⁾(x) + 36xv'''(x) + (216x² + 72)v''(x) + (432x³ + 324x² + 36)v'(x) + (144x⁴ + 216x³ + 108x²)v(x)] x³e^4x

We substitute these values in the original differential equation

y⁽⁴⁾(x) + a3y'''(x) + a2y''(x) + a1y'(x) + a0y(x) = 0

to get

[v⁽⁴⁾(x) + 36xv'''(x) + (216x² + 72)v''(x) + (432x³ + 324x² + 36)v'(x) + (144x⁴ + 216x³ + 108x²)v(x)] x³e^4x + a3[v'''(x) + 18xv''(x) + (54x² + 12)v'(x) + (36x³ + 36x²)v(x)] x³e^4x + a2[v''(x) + 6xv'(x) + 12v(x)] x³e^4x + a1[v'(x) + 3x²e^4xv(x)] x³e^4x + a0[v(x)] x³e^4x = 0

We can divide both sides by x³e^4x to get

[v⁽⁴⁾(x) + 36xv'''(x) + (216x² + 72)v''(x) + (432x³ + 324x² + 36)v'(x) + (144x⁴ + 216x³ + 108x²)v(x)] + a3[v'''(x) + 18xv''(x) + (54x² + 12)v'(x) + (36x³ + 36x²)v(x)] + a2[v''(x) + 6xv'(x) + 12v(x)] + a1[v'(x) + 3x²v(x)] + a0[v(x)] = 0

This is the fourth order linear homogeneous differential equation for v(x).

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The Laplace transform of [tex]f(t)[/tex]  is [tex]L{f(t)} = {4/s (1 - e^{(-2s)})} + 0= 4/s (1 - e^{(-2s)})[/tex]

Laplace Transform is an important mathematical tool that converts a function from a time-domain function to a frequency-domain function. In order to find L{f(t)}, we can use the Laplace transform definition 4.1.1.

Definition 4.1.1:Let [tex]f(t)[/tex] be a continuous function defined for [tex]t \geq 0[/tex]. Then the Laplace transform of [tex]f(t)[/tex] is defined to be [tex]L{f(t)} = ∫_0^{\infinity} f(t)e^{(-st)} dt[/tex]provided the integral exists, where s is a complex variable.For the given function [tex]f(t)[/tex], we have to find [tex]L{f(t)}.1. f(t )= -1, 0 \leq t < 11, t \geq 1[/tex]Let's consider the first part where 0 ≤ t < 1.

We know that [tex]f(t) = -1 , 0 \leq t < 1[/tex]. So,[tex]L{f(t)} = ∫_0^1 -e^{(-st)} dt= [-1/s e^{(-st)}]_0^1= [-1/s e^{(-s)} + 1/s e^0]= [1/s (1 - e^{(-s)})][/tex]The Laplace transform for 1,[tex]t \geq 1[/tex] is [tex]L{f(t)} = ∫_1^∞ e^(-st) dt= [-1/s e^(-st)]_1^∞= 0 + 1/s e^(-s) = 1/s e^(-s)[/tex] Therefore, the Laplace transform of f(t) is[tex]L{f(t)} = {1/s (1 - e^{(-s)})} + {1/s e^{(-s)}}= 1/s (1 - e^{(-s)} + e^{(-s)})= 1/s2[/tex]. [tex]f(t ) = 4[/tex], [tex]0 \leq t < 20, t \geq 2[/tex]

For the first part where [tex]0 \leq t < 2[/tex], we know that [tex]f(t) = 4[/tex]. So,[tex]L{f(t)} = ∫_0^2 4e^{(-st)} dt= [4/s e^({-st)}]_0^2= [4/s e^{(-2s)} - 4/s e^0]= [4/s (1 - e^{(-2s)})][/tex]For the second part where [tex]t \geq 2[/tex], we know that[tex]f(t) = 0[/tex]. So,[tex]L{f(t)} = ∫_2^∞ 0 e^(-st) dt= 0[/tex]Therefore, the Laplace transform of [tex]f(t)[/tex] is [tex]L{f(t)} = {4/s (1 - e^{(-2s)})} + 0= 4/s (1 - e^{(-2s)})[/tex]

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If you multiplied out 82!, it would end in 19 zeros by factoring out from the product of all integers from 1 to 82.

To determine the number of zeros at the end of 82!, we need to find the number of times 10 can be factored out from the product of all integers from 1 to 82.

This is because a zero is created whenever a factor of 10 is present in a number, which occurs when a multiple of 10 or a multiple of 5 and 2 is present in the product.

We can find the number of multiples of 5 and 2 by dividing 82 by 5 and then by 25, since each multiple of 5 contributes at least one factor of 5 and each multiple of 25 contributes an additional factor of 5.

82 ÷ 5 = 16 with a remainder of 2, so there are 16 multiples of 5 between 1 and 82.

82 ÷ 25 = 3 with a remainder of 7, so there are 3 multiples of 25 between 1 and 82.

Therefore, the total number of factors of 5 in 82! is 16 + 3 = 19. Since every factor of 5 can be paired with a factor of 2, there will always be enough factors of 2 to create a factor of 10. Thus, there are 19 zeros at the end of 82!.

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Mary can prove that the angles of a triangle always add up to 180° by taking three rays and placing them side by side to form a triangle, and then measuring the angles. The sum of the angles will always be 180°.

To prove that the angles of a triangle always add up to 180°, Mary can follow these steps:

1. Draw three rays that are not in the same straight line.

2. Place them side by side to form a triangle.

3. Measure the angles of the triangle using a protractor.

4. Add the three angles together.

5. The sum of the angles should be 180°.

6. If the sum of the angles is not 180°, check your measurements.

7. If the measurements are correct, the angles must add up to 180°.

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Therefore , the solution of the given problem of graphs comes out to be  scatter plot makes the individual data points easier to understand.

Theoretical physicists use graphs to visually and analytically depict claims rather than values. A graph point typically depicts the connection between several distinct objects. A specific type of pas de train construction made up of groups but also lines is called a graph. Glue should be used to connect the channels, also referred as the boundaries. The boundaries of this network also contained the numbers (2.5), (3.5), and (4.5).

Here,

The best way to show and describe "New Machine" is to use a graph with the y-axis labelled "hot chocolate amount in fluid ounces" going from 0 to 18, the x-axis labelled "time in seconds" going from 0 to 28, and points at => (4, 4), (8, 5.5), (12, 7), (16, 9), (20, 12), and (24, 15).

The best representation is a scatter plot, which enables us to view the various data points and spot any patterns or trends in the data.

This data would also be suitable for a line graph with line segments linking each point to the next, but a scatter plot makes the individual data points easier to understand.

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Answer:

That number is irrational because it cannot be expressed as a simple fraction. The number of numbers after the decimal point is infinite.

The angle theta such that the absolute value of theta is less than or equal to pi and theta is conterminal with 20π/3 is π/3.

Fraction:

In mathematics, a fraction is defined as a part of a whole. A whole can be an object or a group of objects. In real life, when we cut a whole piece of cake, the serving size is the fraction of the cake. Fraction is a word from Latin.

According to the Question:

Subtract 2π from 20π/3

20π/3 - 2π

The resulting angle of 14π/ 3 is positive, less than 2π, and conterminal with 20π/3.

= 14π/3

Subtract 2π from 14π/3.

= 14π/3−2π

The resulting angle of 8π3 is positive, less than 2π, and conterminal with 20π/3. 8π/3

Subtract 2π from 8π/3.8π/3−2π

The resulting angle of 2π/3 is positive, less than 2π, and conterminal with 20π/3.2π/3.

Or, π/3

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Evaluation score is 2.93 based null hypothesis. Test statistic is 6.485. Evaluation score = 2.72 + 0.153 * beauty score. Correlation coefficient is 0.8041.

Let the population slope of the relationship between teacher assessment and attractiveness be 1, and formulate some hypotheses.

A = 0, the null hypothesis (there is no significant linear relationship between teaching evaluation and beauty)

Another possibility is that 1 > 0. (there is a positive linear relationship between teaching evaluation and beauty)

The test statistic is given by the formula: t-statistic = (1 - 0) / SE(1), where SE(1) is the standard error of the slope coefficient.

If the p-value is less than, we reject the null hypothesis at a significance threshold of 0.05.

The output of the regression shows that the p-value is less than 0.0001 and the t-statistic is 6.485. As a result, the null hypothesis can be rejected, and it can be determined that there is a meaningful positive linear association between teacher assessment and attractiveness.

Regression equation:

evaluation score = 2.72 + 0.153 * beauty score

Correlation coefficient:

The square root of[tex]R^2[/tex]is the correlation coefficient. Therefore, the correlation coefficient is [tex]\sqrt{(0.6466)} = 0.8041[/tex].

Evaluation score: If a professor has a standardized beauty score of 1.43, we can enter this value into the regression equation to obtain the evaluation score: evaluation score = 2.72 + 0.153 * 1.43 = 2.93

Thus, we would anticipate this professor to have a rating of roughly 2.93.

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The given equation is x2y3 + x(1+y 2)y' = 0. This is not an exact equation, as the derivative of F(x,y) is not equal to zero. To make this equation exact, we need to multiply it by an integrating factor.

The integrating factor (I.F.) of a differential equation is a function of the independent variable, which can be used to transform an equation into an exact equation. An exact equation is one in which the derivative of F(x,y) is equal to zero. To find the integrating factor, we need to multiply the given equation by I.F. and set the derivative of F(x,y) to zero.

To find the I.F., we need to divide both sides of the equation by x(1+y2):
x2y3 + x(1+y 2)y' / x(1+y2) = 0 / x(1+y2)

Rearranging this equation, we get:
y' + (x2y3 / x(1+y2)) = 0

Now, we can integrate both sides of the equation,
∫y' + (x2y3 / x(1+y2)) dx = 0

This gives us an expression for the I.F.,
I.F. = exp(∫(x2y3 / x(1+y2))dx)

Now, to make the equation exact, we need to multiply the given equation with the integrating factor. This gives us:    I.F. x x2y3 + I.F. x x(1+y2)y' = 0

Simplifying this equation, we get:
F(x,y) = G(x) + H(y) = K

Where K is a constant. Thus, we have now converted the given equation into an exact equation.

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In case of theses statistically independent variables, for (p-q), the expected value is 0, Variance is 40 and expected value of (p-q)^2 is 40.

Given that x and y are two statistically independent n(0,1) random variables, and p and q are random variables defined as: p = 7x + 2xy and q = x, we can calculate the variance of (p-q) using the following steps:

First, we need to find the expected value of (p-q):
E(p-q) = E(p) - E(q) = E(7x + 2xy) - E(x) = 7E(x) + 2E(x)E(y) - E(x) = 7(0) + 2(0)(0) - 0 = 0

Next, we need to find the expected value of (p-q)^2:
E[(p-q)^2] = E[(7x + 2xy - x)^2] = E[(6x + 2xy)^2] = E[36x^2 + 24xy^2 + 4xy^2] = 36E(x^2) + 24E(x)E(y^2) + 4E(x^2)E(y^2) = 36(1) + 24(0)(1) + 4(1)(1) = 40

Finally, we can calculate the variance of (p-q) using the formula: Var(p-q) = E[(p-q)^2] - [E(p-q)]^2 = 40 - 0^2 = 40. Therefore, the variance of (p-q) is 40.

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Therefore, the size of the angle xyz is approximately 26.2 degrees.

A triangle is a closed two-dimensional geometric shape that is formed by three straight lines that intersect at three points, called vertices. These three lines, known as sides or edges, form three angles where two sides meet, known as the angles of the triangle. The sum of the angles of a triangle is always 180 degrees.

Triangles can be classified based on the lengths of their sides and the sizes of their angles. There are several types of triangles, including equilateral triangles, isosceles triangles, scalene triangles, acute triangles, obtuse triangles, and right triangles.

Triangles have a wide range of applications in mathematics, geometry, physics, and engineering, and they are also used in various fields such as architecture, art, and design.

by the question.

the angle opposite the 15cm side is the right angle, then we can use the Pythagorean theorem to find the length of the third side (xy):

[tex]xy^2 = 15^2 - 6^2[/tex]

[tex]xy^2 = 225 - 36[/tex]

[tex]xy^2 = 189[/tex]

[tex]xy = \sqrt(189)[/tex]

[tex]xy = 13.7[/tex] (rounded to 1 decimal place)

Now we can use trigonometry to find the size of the angle opposite the 6cm side (xyz):

[tex]sin(xyz) = 6/13.7[/tex]

[tex]xyz = sin^-1(6/13.7)[/tex]

[tex]xyz = 26.2 degrees[/tex] (rounded to 1 decimal place)

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The supply of styluses is falling at a rate of 0.0144p² styluses per week.

To determine the exact rate of change in supply, we need to look at the equation provided: 500p²x² = 1996. We can rearrange this equation to solve for x, which represents the supply of styluses:

x² = (1996/500p²)

x = √(1996/500p²)

Now, we can differentiate this equation with respect to time to find the rate of change in supply:

dx/dt = (-1/2)(1996/500p²)(-1000p²/1996)*dp/dt

Simplifying this equation, we get:

dx/dt = (6/25)p²dp/dt

Using the information provided in each situation, we can plug in the appropriate values to find the rate of change in supply:

In the first situation, where the price is rising at 12 cents per week, dp/dt = 0.12. Plugging this into the equation above, we get:

dx/dt = (6/25)(p²)(0.12) = 0.0288p²

This means that the supply of styluses is rising at a rate of 0.0288p² styluses per week.

In the second situation, where the price is falling at 6 cents per week, dp/dt = -0.06. Plugging this into the equation above, we get:

dx/dt = (6/25)(p²)(-0.06) = -0.0144p²

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The number in 2 equivalent forms as a​ fraction, decimal are  19/19 and  1.0 respectively.

There are just two ways to express numbers: fractions and decimals.

The number given as:

19 over 19

It can be written as:

19/19 (equivalent as fraction)

19/19 = 1.0 (equivalent as decimal)

Thus,  the number in 2 equivalent forms as a​ fraction, decimal are  19/19 and  1.0 respectively.

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When dealing with the problem of non-constant variance, the log transformation means using b. LogX as the independent variable instead of X.

A log transformation is a type of data transformation that is often used to address the problem of non-constant variance, also known as heteroscedasticity. This type of transformation involves taking the natural logarithm of the independent variable, X, and using LogX as the independent variable instead of X.

This transformation can help to stabilize the variance of the data and make the relationship between the independent variable and the dependent variable more linear. It is important to note that a log transformation may not always be appropriate, and it is important to carefully consider the characteristics of the data before applying this type of transformation.

In conclusion, when dealing with the problem of non-constant variance, the log transformation means using LogX as the independent variable instead of X. This can help to stabilize the variance and make the relationship between the independent and dependent variables more linear.

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