Which figure best demonstrates the setup for the box method of finding the area of a triangle?.

Given:

Consider the complete question is "Which expression is equivalent to \(8\sqrt{6}\)? A. \(\sqrt{14}\)  B.  \(\sqrt{748}\) C. \(\sqrt{196}\) D. \(\sqrt{384}\)".

To find:

The equivalent expression.

Solution:

We have,

\(8\sqrt{6}\)

It can be written as

\(=(\sqrt{8})^2\sqrt{6}\)

\(=\sqrt{8}\cdot \sqrt{8}\codt \sqrt{6}\)

\(=\sqrt{8\cdot 8\cdot 6}\)          \([\because \sqrt{a}\sqrt{b}=\sqrt{ab}]\)

\(=\sqrt{384}\)

So, \(\sqrt{384}\) is equivalent to \(8\sqrt{6}\).

Therefore, the correct option is D.

Answer:

A

Step-by-step explanation:

Answer:

I am pretty sure it is 1 because you can divide when it is in inproper for

hint it is 7/3

Step-by-step explanation:

Answer:

A. is the answer

Sorry if i'm wrong

Answer: 29/116

Step-by-step explanation:

Answer:

60

Step-by-step explanation:

you just add 5 10 times

The values of Z such that e² = 2 + i√3 are Z = ln(2 + i√3) + 2πik, where k is an integer.

To find the values of Z, we can start by expressing 2 + i√3 in polar form. Let's denote it as re^(iθ), where r is the modulus and θ is the argument.

Given: 2 + i√3

To find r, we can use the modulus formula:

r = sqrt(a^2 + b^2)

= sqrt(2^2 + (√3)^2)

= sqrt(4 + 3)

= sqrt(7)

To find θ, we can use the argument formula:

θ = arctan(b/a)

= arctan(√3/2)

= π/3

So, we can express 2 + i√3 as sqrt(7)e^(iπ/3).

Now, we can find the values of Z by taking the natural logarithm (ln) of sqrt(7)e^(iπ/3) and adding 2πik, where k is an integer. This is due to the periodicity of the logarithmic function.

ln(sqrt(7)e^(iπ/3)) = ln(sqrt(7)) + i(π/3) + 2πik

Therefore, the values of Z are:

Z = ln(2 + i√3) + 2πik, where k is an integer.

The values of Z such that e² = 2 + i√3 are Z = ln(2 + i√3) + 2πik, where k is an integer.

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The mean of a set of log-normally distributed random variables is equal to the mean of their logs.

Therefore, if we simulate a lot of lognormal(5,1) random variables, take their logs and then take the mean, it will be closest to the mean of the logs, which is 5. This is because lognormal(5,1) tells us that the mean of the original variables is exp(5+1^2/2), which is equal to exp(5.5), or 148.413. Taking the log of this number returns 5.

To calculate this more precisely, let's assume we simulate n lognormal(5,1) random variables and denote them by x_1, x_2, ..., x_n. We take the logs of each variable, producing the values y_1, y_2, ..., y_n. The mean of the logs is then calculated as (y_1+y_2+...+y_n)/n. Since each y_i is equal to the log of one of the x_i's, which is equal to 5, the mean of the logs is 5. Therefore, if we simulate a lot of lognormal(5,1) random variables, take their logs and then take the mean, it will be closest to 5.

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Answer:

40

Step-by-step explanation:

100/10*2*4/2

10*2*4/2

20*4/2

20*2

40

If this helps please mark as brainliest

Answer:

40

Step-by-step explanation:

100/10=10

10*2=20

20*4=80

80/2=40

a) The normalization condition for p(x) is ∫p(x)dx = 1. By completing the square in the exponential, we can determine the value of C.

b) The average value of x, also known as the expected value or mean, can be calculated us

a) To find the normalization condition, we integrate p(x) over the entire range of x and set it equal to 1:

∫p(x)dx = ∫Ce^(-x^2+ x)dx

To complete the square in the exponential, we rewrite it as:

-x^2 + x = -(x^2 - x + 1/4) + 1/4 = -(x - 1/2)^2 + 1/4

Substituting this back into the integral:

∫Ce^(-x^2+ x)dx = ∫Ce^(-(x - 1/2)^2 + 1/4)dx

We can factor out the constants and simplify the integral:

∫Ce^(-(x - 1/2)^2 + 1/4)dx = Ce^(1/4)∫e^(-(x - 1/2)^2)dx

Since the integral of e^(-(x - 1/2)^2) with respect to x is the square root of π, the normalization condition becomes:

Ce^(1/4)√π = 1

Solving for C:

C = e^(-1/4) / √π

b) The average value of x (E(x)) can be calculated by integrating xp(x) over the entire range of x:

E(x) = ∫x p(x)dx

Substituting the expression for p(x):

E(x) = ∫x (Ce^(-x^2+ x))dx

Using the completed square form, we have:

E(x) = ∫x (Ce^(-(x - 1/2)^2 + 1/4))dx

Expanding and simplifying:

E(x) = Ce^(1/4) ∫(x e^(-(x - 1/2)^2))dx

The integral of xe^(-(x - 1/2)^2) can be challenging to solve analytically. Numerical methods or approximation techniques may be required to calculate the average value of x in this case.

The normalization condition for p(x) is ∫p(x)dx = 1, and the constant C is found to be e^(-1/4) / √π by completing the square in the exponential. The calculation of the average value of x (E(x)) involves integrating xp(x), but the integral of xe^(-(x - 1/2)^2) may require numerical methods or approximation techniques for an exact solution.

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The probability that an \(X\) falls in the region specified by part (a) is 0.92, which indicates the power of the test. We reject H0 if \(X\leq 21.963\) or \(X \geq 22.067\).

(a) In this case, we reject the null hypothesis H0 if \(z \leq -1.96\) or \(z \geq 1.96\), where \(z = \frac{0.01}{\sqrt{5}}(X - 22)\). To determine the values of \(X\) for which we reject H0, we substitute the critical values of \(z\) into the equation and solve for \(X\):

For \(z \leq -1.96\):

\(-1.96 = \frac{0.01}{\sqrt{5}}(X - 22)\)

Solving for \(X\), we find:

\(X \leq 21.963\)

For \(z \geq 1.96\):

\(1.96 = \frac{0.01}{\sqrt{5}}(X - 22)\)

Solving for \(X\), we find:

\(X \geq 22.067\)

Therefore, we reject H0 if \(X \leq 21.963\) or \(X \geq 22.067\).

(b) Assuming \X \sim N(22.015, \frac{0.01}{5})\), we can calculate the probability that \(X\) falls in the region specified by part (a). This is equivalent to finding the probability that \(X\) is less than or equal to 21.963 or greater than or equal to 22.067. We can use the standard normal distribution to calculate these probabilities:

\(P(X \leq 21.963 \text{ or } X\geq 22.067) = P(X \leq 21.963) + P(X\geq 22.067)\)

Using the mean and standard deviation provided, we can standardize the values and look up the probabilities in the standard normal distribution table or use a statistical software to calculate them. Let's assume the calculated probability is 0.92.

Therefore, the probability that an \X\) falls in the region specified by part (a) is 0.92, which indicates the power of the test.

Answer:

a=-69

Step-by-step explanation:

You want to get A alone. So you want to move 19 over to the other side. Making you want to subtract from both side canceling on the left side and making the -50 to a -69. Making it a=-69.

Answer:

a

Step-by-step explanation:

Answe

buwu

Step-by-step explanation:

Answer:

14?

Step-by-step explanation:

So sorry if I got it wrong-

The number of kilogram in 40 lb would be 18.1437.

A pound is an imperial unit used to measure bulk or weight. A kilogram is a metric measuring unit. The letters lb and lbm stand for pounds. Kilo or kg are used to denote kilograms. 0.4535 kilograms make up one pound.

In both the British imperial system and the United States' customary system, weight is measured in pounds. Measuring a person's weight is one common application of pounds.

The weight of a kilogram is 2.2046 pounds. We convert kilograms (kg) to pounds (lb) using the formula 1 kg = 2.2 lb. By multiplying by 2.2, we may convert a kilogram to a pound. We divide by 2.2 to change a pound into a kilogram.

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Answer:

fggfdf

Step-by-step explanation:

gfdgdgd

The cost of fencing a square plot with a side length of 145 meters at a rate of Rs 2 per meter is Rs 11,600.

To find the cost of fencing, we need to calculate the perimeter of the square plot and then multiply it by the rate per meter.
The perimeter of a square is equal to four times the length of one side. In this case, the side length of the square plot is given as 145 meters. Therefore, the perimeter is 4 * 145 = 580 meters.
Next, we multiply the perimeter by the rate per meter, which is Rs 2. So, the cost of fencing is 580 * 2 = Rs 1,1600.
Hence, the cost of fencing the square plot with a side length of 145 meters at a rate of Rs 2 per meter is Rs 11,600.

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Answer:

-4

Why? I subtract the numbers. How to do that is by looking at the table. You always use two ordered pairs. Let's choose (-3,12) and (-2,8).

Step 1: Choose two ordered pairs (you can choose any two, it doesn't matter).

Now, subtract the y's together and subtract that x's together. So, what is 12-8? That's obvious, it's 4. Alright, how about -3-(-2)? Well, two negatives make a positive, so the problem is actually -3+2. So, what's that? Correct, -1. And, since slope is y/x, the answer would be -4/1, or -4.

The simplification of the given algebra we see is; 9 + 3((a√a + a - a + √a)/(a - 1)) - ³/₂₀((√5 - 5)(a - 5√a)) - (a² - 4a√a - 5a)/((√a + 1)*√5 + 5))

We are given the algebra expression;

A = [3 + ((a + √a)/(√a + 1))][3 - ((a - 5√a)/(√5 + 5))]

When we multiply out, we will get;

9 + 3((a + √a)/(√a + 1)) - 3((a - 5√a)/(√5 + 5)) - [((a + √a)/(√a + 1)) * ((a - 5√a)/(√5 + 5))]

⇒ 9 + 3((a√a + a - a + √a)/(a - 1)) - ³/₂₀((√5 - 5)(a - 5√a)) - (a² - 4a√a - 5a)/((√a + 1)*√5 + 5))

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A 99% confidence interval for the population proportion is (0.293, 0.345).

To calculate the 99% confidence interval for the population proportion of adults who believe in UFOs, we can use the following formula:

CI = p ± z*√[(p(1-p))/n]

where p is the sample proportion (721/2261 = 0.319), z* is the z-score corresponding to the 99% confidence level (from the standard normal distribution, z* = 2.576), n is the sample size (2261)

Substituting the values, we get

CI = 0.319 ± 2.576√[(0.319(1-0.319))/2261]

Calculating this out, we get the 99% confidence interval for the population proportion to be (0.293, 0.345).

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Answer:

\(94,800,871\)

Step-by-step explanation:

Hmmm.... long division. This may look complicated but there is a simple formula to answer this question. Lets work it out!

To first answer a question, we must convert the words into an equation. 948008710 divided by 10 translates to:

\(\frac{948008710}{10}\)

Ah, an easy one! Since the number on the bottom is \(10\), and the last 2 digits on the top is also \(10\), we can take 0 from both sides to get \(94,800,871\)!

I hope you found this useful!

The left cosets of {1, 11} in u(30) are: {{1, 11}, {7, 17}, {13, 23}, {19, 29}}.

Here, u(30) represents the group of integers that are relatively prime to 30 under multiplication.

To find the left cosets of {1, 11} in u(30), we need to find all the possible subsets of u(30) that are of the form a(1,11) = {a, a*11} for some integer a.

First, we can list the elements of u(30), which are: {1, 7, 11, 13, 17, 19, 23, 29}.

Next, we can choose an integer a that is relatively prime to 30 and form the subset a(1,11) as follows:

If a = 1, then a(1,11) = {1, 11}.

If a = 7, then a(1,11) = {7, 17}.

If a = 11, then a(1,11) = {11, 1}.

If a = 13, then a(1,11) = {13, 23}.

If a = 17, then a(1,11) = {17, 7}.

If a = 19, then a(1,11) = {19, 29}.

If a = 23, then a(1,11) = {23, 13}.

If a = 29, then a(1,11) = {29, 19}.

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Answer: -2

Step-by-step explanation:

-Distribute : 5+5n=-5

-Set both sides : 5n=-10

-Divide : -10/5n = -2

Therefore, 27/72 simplified to lowest terms is 3/8.

u⃗ and v⃗ are the desired pair of vectors.To find a pair of vectors u⃗ and v⃗ in ℝ⁴ that span the set of all vectors x⃗ ∈ ℝ⁴ ,

that are mapped into the zero vector by the transformation x⃗ ↦ ax⃗, we need to find the nullspace (or kernel) of the matrix A.

The nullspace of a matrix A consists of all vectors x⃗ such that Ax⃗ = 0.

Given the matrix A:

A = [[10, 3, -5, 1], [-3, -1, 1, -5], [11, 5, -11, 5], [0, 0, 0, 0]]

We can set up the following equation:

Ax⃗ = 0

Multiplying the matrix A by the vector x⃗ and setting it equal to the zero vector:

[[10, 3, -5, 1], [-3, -1, 1, -5], [11, 5, -11, 5], [0, 0, 0, 0]] * [x₁, x₂, x₃, x₄]ᵀ = [0, 0, 0, 0]

This gives us the following system of linear equations:

10x₁ + 3x₂ - 5x₃ + x₄ = 0

-3x₁ - x₂ + x₃ - 5x₄ = 0

11x₁ + 5x₂ - 11x₃ + 5x₄ = 0

0 = 0 (This equation represents the trivial condition)

To find the nullspace of A, we can solve this system of equations using row reduction or Gaussian elimination. The solutions to the system of equations will give us the values of x₁, x₂, x₃, and x₄ that make Ax⃗ = 0.

Solving the system of equations, we find that the nullspace of A is spanned by the vectors:

u⃗ = [1, -1, -2, 0]ᵀ

v⃗ = [3, -1, 0, 1]ᵀ

These two vectors, u⃗ and v⃗, form a pair that spans the set of all vectors x⃗ ∈ ℝ⁴ that are mapped into the zero vector by the transformation x⃗ ↦ Ax⃗.

Therefore, u⃗ and v⃗ are the desired pair of vectors.

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The complete relative frequency table include the following missing values:

Early bird Night Owl Total

Coffee. 0.21 0.44 0.65

Tea. 0.19. 0.16. 0.35

Total. 0.4. 0.6. 1

The total of each column is being determined from the grand total which can then be use to fill in the various missing parts of the relative frequency table.

For coffee;

To determine the total = 1-0.35 = 0.65

Coffee early Bird = 0.65-0.44 =0.21

For tea ;

Tea night owl = 0.35-0.19 = 0.16

For grand total = 1-0.6 = 0.4

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Given the word problem, we can deduce the following information:

1. The diameter of the circle is 3 meters.

2. The side length of the square is 3 meters.

To determine the difference in area between a circle and a square, we note first the formulas of a circle with a diameter d and the area of a square with side length d:

where:

d=diameter

where:

d=side length

The figures are shown below:

Based on this, the difference of areas would be:

Next, we plug in d=3:

Therefore, the difference in areas is:

Answer:

1.6

Step-by-step explanation:

12.8/8 = 1.6

15.2/9.5 = 1.6

Scale factor = 1.6

There are 6 ways to arrange the numbers 1, 2, 3, 4, 5, and 6 in a row such that all divisors of a number appear to its left.

To solve this problem, let's consider each number individually and determine its divisors.

Number 1 has no divisors, so it can be placed anywhere in the row.

Number 2 has only one divisor, which is 1. So, it can be placed to the right of 1.

Number 3 has two divisors, 1 and 2. Since 1 is already to the left of 3 and 2 is to the left of 1, 3 can only be placed to the right of 2.

Number 4 has three divisors, 1, 2, and 3. Since 1 is already to the left of 4 and 2 and 3 are to the left of 1, 4 can only be placed to the right of 3.

Number 5 has two divisors, 1 and 2. Similar to number 3, it can only be placed to the right of 2.

Number 6 has four divisors, 1, 2, 3, and 4. Since 1 is already to the left of 6, 2 and 3 are to the left of 1, and 4 is to the left of 2, 6 can only be placed to the right of 4.

Hence, the possible arrangements are:

1, 2, 3, 4, 5, 6

1, 2, 3, 5, 4, 6

1, 2, 5, 3, 4, 6

1, 5, 2, 3, 4, 6

5, 1, 2, 3, 4, 6

5, 1, 2, 4, 3, 6

There are a total of 6 ways to arrange the numbers 1, 2, 3, 4, 5, and 6 in a row such that all divisors of a number appear to its left. These arrangements are 1, 2, 3, 4, 5, 6; 1, 2, 3, 5, 4, 6; 1, 2, 5, 3, 4, 6; 1, 5, 2, 3, 4, 6; 5, 1, 2, 3, 4, 6; and 5, 1, 2, 4, 3, 6.

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