What is m∠N? here is a quadrilateral MNPQ in which side MN is parallel to side PQ and side NP is parallel to side MQ. The measure of angle NMQ is (6x-2) degrees and the measure of angle NPQ is (4x+36) degrees

In linear regression, the variance around the regression line represents the variability of the dependent variable (response variable) that is not explained by the regression model.

It measures the dispersion of the actual data points around the predicted values from the regression line.

The variance around the regression line can vary with different values of the predictor variable. This is because the relationship between the predictor variable and the response variable may not be constant across the entire range of the predictor variable. In other words, the spread or dispersion of the data points around the regression line may change as the predictor variable changes.

By examining the residuals (the differences between the observed values and the predicted values from the regression line) and calculating their variances, you can assess the variability of the data points around the regression line. This variability is an important aspect of understanding the goodness of fit of the regression model and the accuracy of the predictions.

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Answer:

2/9

Step-by-step explanation:

y2 - y1 / x2 - x1

8 - 6 / 6 - (-3)

2 / 9

= 2/9

Answer:

I think it's A.)

Step-by-step explanation:

Answer:

Step-by-step explanation:

x + 44x = 180

45x = 180

x = 4

180-x = 176

answer is 4

Answer:

B

Step-by-step explanation:

Answer:

it's B

Step-by-step explanation:

the sum of two interior angle of a triangle are always equal to the degree of the adjoined angle to the 3 angle

sorry i don't know how to explain this better

Answer:

N/A

Step-by-step explanation:

What's the question to the question?

The point- slope form of the line is y-5 = -0.25(x+6).

What is line?

A line is an one-dimensional figure. It has length but no width. A line can be made of a set of points which is extended in opposite directions to infinity. There are straight line, horizontal, vertical lines or may be parallel lines perpendicular lines etc.

A line with a slope of –1/4 passes through the point (–6,5)

Any line in point - slope form can be written as

y - y₁= m(x -x₁) -------(1)

where,

y= y coordinate of second point

y₁ = y coordinate of first point

m= slope of the line

x= x coordinate of second point

x₁ = x coordinate of first point

In the given problem (x₁ , y₁) = (-6,5) and m= -1/4

Putting all these values in equation (1) we get,

y-5= (-1/4) (x- (-6))

⇒ y-5 = -0.25(x+6)

Hence, the point- slope form of the line is y-5 = -0.25(x+6).

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Answer:

No, it is not.

Explanation:

One x value (-3) has more than one corresponding y value (1 and 0). In a proper function, each x value has one and only one corresponding y value.

Answer:

C

Step-by-step explanation:

Answer:

  2. 26.2 m

  3. 117.2 cm

Step-by-step explanation:

You want the perimeters of two figures involving that are a composite of parts of circles and parts of rectangles.

The circumference of a circle is given by ...

  C = πd . . . . . where d is the diameter

The length of the semicircle of diameter 12.6 m will be ...

  1/2C = 1/2(π)(12.6 m) = 6.3π m ≈ 19.8 m

The two lighted sides of the rectangle have a total length of ...

  3.2 m + 3.2 m = 6.4 m

The length of the light string is the sum of these values:

  19.8 m + 6.4 m = 26.2 m

The length of the string of lights is about 26.2 meters.

The perimeter of the figure is the sum of four quarter-circles of radius 11.4 cm, and 4 straight edges of length 11.4 cm.

Four quarter-circles total one full circle in length, so we can use the formula for the circumference of a circle:

  C = 2πr

  C = 2π·(11.4 cm) = 22.8π cm ≈ 71.6 cm

The four straight sides total ...

  4 × 11.4 cm = 45.6 cm

The perimeter of the figure is the sum of the lengths of the curved sides and the straight sides:

  71.6 cm + 45.6 cm = 117.2 cm

The design has a perimeter of about 117.2 cm.

__

Additional comment

The bottom 12.6 m edge in the figure of problem 2 is part of the perimeter of the shape, but is not included in the length of the light string.

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Answer:

(1, 3)

Step-by-step explanation:

The solution to any system is where the two lines meet. It would be no solution if the lines were parallel. An ordered pair is written out in the format (x, y).

The sum of first 20 multiples of 7 will be 1470.

Equation modelling is the process of writing a mathematical verbal expression in the form of a mathematical expression for correct analysis, observations and results of the given problem.

Given are first 20 multiples of 7.

The sequence of first 20 multiples of 7 will be -

7, 14, 21, 28, ..... , 140

Sum = (20/2)[2 x 7 + (19) x 7]

Sum = 10[14 + 133]

Sum = 10 x 147

Sum = 1470

Therefore, the sum of first 20 multiples of 7 will be 1470.

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[Question translation in english -

find the sum of the first 20 multiples of 7 with procedure.]

Answer:

\(50-5+19-15 \\ 49\)

Step-by-step explanation:

please mark me brainliest

answer:

50-5 = 45

45 + 19 = 64

64 - 15 = 49

Answer:

  (b)  1 - x - 2 + 2 x = 10 x - 25 - x

Step-by-step explanation:

The distributive property tells you the factor outside parentheses multiplies each term inside parentheses.

__

  1 -(x +2) +2x = 5(2x -5) -x

The minus sign outside the parentheses on the left side of the equal sign represents a factor of -1. Using the distributive property there, we have ...

  1 +(-1)(x) +(-1)(2) +2x = 5(2x -5) -x

  1 -x -2 +2x = 5(2x -5) -x . . . . . . simplify a bit

The 5 outside parentheses on the right side of the equal sign similarly multiplies each of the terms inside those parentheses:

  1 -x -2 +2x = (5)(2x) +(5)(-5) -x

  1 -x -2 +2x = 10x -25 -x . . . . . . . . matches the second choice

_____

Additional comment

It might be helpful to think of parentheses as a "bag." The factor outside tells you how many bags you have. (All are the same.) When you eliminate the parentheses, you are essentially dumping the contents of the bags into one pile. Since all of the bags have the same contents, the total is the product of what's in a bag and the number of bags.

1. The population does not need to be normally distributed for the sampling distribution of x bar to be approximately normally distributed because the Central Limit Theorem states that regardless of the shape of the underlying population, the sampling distribution of ¯x becomes approximately normal as the sample size, n, increases. Therefore, the correct option is A.

2. The sampling distribution of ¯x is normal or approximately normal with μx = 68 and σx = 0.464. Therefore, the correct option is A.

1. The answer to 'does the population need to be normally distributed for the sampling distribution of x bar to be approximately normally distributed?' is no  because the Central Limit Theorem states that regardless of the shape of the underlying population, the sampling distribution of ¯x becomes approximately normal as the sample size, n, increases. In this case, the sample size is 41, which is large enough for the Central Limit Theorem to apply.

2. The sampling distribution of ¯x is normal or approximately normal with μx = 68 and σx = 0.464. The standard error (SE) of the sampling distribution of the sample mean is σ/√n. Here, the mean of the population is 68 and the standard deviation is 3.

The formula for the sampling distribution of the mean can be used to find the standard error. SE= σ/√n = 3/√41 = 0.464. The sampling distribution of ¯x is normally distributed, and since the sample size is 41, the sample mean can be considered approximately normally distributed.

Thus, the sample mean follows the normal distribution with μ = 68 and σx = 0.464. Therefore, the answer is: The sampling distribution of ¯x is normal or approximately normal with μx = 68 and σx = 0.464.

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Answer:

x = ± \(\sqrt{7}\)

Step-by-step explanation:

Given

x² - 7 = 0 ( add 7 to both sides )

x² = 7 ( take the square root of both sides )

x = ± \(\sqrt{7}\)

Answer:

2008

Step-by-step explanation:

when she is 12 (4 years after 2004) her dad will be 36, which is 3 x 12, therefore, in 2008 he will be able to say this.

Answer:

  2008

Step-by-step explanation:

The difference in their ages is ...

  4m -m = 3m = 3(8) = 24 . . . years (Marika was 8 in 2004)

__

Marika's age when the father was 3 times her age was ...

  3m -m = 2m = 24 . . . years

  m = 12 . . . . . Marika was 12 when her father was 3 times her age

That is 12 -8 = 4 years from 2004, so it will be true in 2008.

To prove that 3x ≤ y, assume the opposite, that is, 3x > y, rearrange the inequality substitute x < 2y and simplify, contradict the given condition that x < 2y, therefore, concluding that 3x ≤ y.

Start by assuming the opposite, that is, 3x > y.

From the given inequality,\(7xy \leq 3x^2 + 2y^2,\), we can rearrange to get:
\(7xy - 3x^2 \leq 2y^2\)

We can substitute \(x < 2y\) into this inequality:
\(7(2y)x - 3(2y)^2 \leq 2y^2\)

Simplifying, we get:
\(y(14x - 12y) \leq 0\)

Since y is a real number, this means that either y ≤ 0 or 14x - 12y ≤ 0.

If y ≤ 0, then 3x ≤ y is trivially true.

If 14x - 12y ≤ 0, then we can rearrange to get:
3x ≤ (12/14)y
3x ≤ (6/7)y
3x < y (since we assumed 3x > y)

But this contradicts the given condition that x < 2y, so our assumption that 3x > y must be false.

Therefore, we can conclude that 3x ≤ y.

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The z statistic that is associated with the class's average is 1.25.

In the given question the population of all standardized ex scores has a mean of 500 and a standard deviation of 100 . Twenty-five students in a professor's class take the ex , and their average score is 525.

We have to find z statistic that is associated with the class's average.

n = 25

μ = 500

σ = 100

\(\bar X\) = 525

Then ,

z statistic = \(\frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt n} }\)

putting the value in the formula:

z statistic = (525-500)/(100/\sqrt 25)

z statistic = (25)/(100/5)

z statistic = (25)/(20)

z statistic = 1.25

Therefore,

The z statistic that is associated with the class's average is 1.25.

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The total number of ways to split the group of 10 people into two groups of size 3 and one group of size 4 is 120 * 35 * 1 = 4,200. To split a group of 10 people into two groups of size 3 and one group of size 4, we can use the concept of combinations.

The number of ways to split the group can be calculated by determining the number of combinations of selecting 3 people from 10 for the first group, then selecting 3 people from the remaining 7 for the second group, leaving the remaining 4 people for the third group.

To split the group of 10 people into two groups of size 3 and one group of size 4, we can calculate the number of ways using combinations. The first group of size 3 can be formed by selecting 3 people from the total of 10 people. This can be represented as C(10, 3) = 10! / (3!(10-3)!).

Evaluating this expression:

C(10, 3) = 10! / (3! * 7!) = (10 * 9 * 8) / (3 * 2 * 1) = 120.

After selecting the first group, we are left with 7 people. From these 7 people, we need to select another group of size 3, which can be represented as C(7, 3) = 7! / (3!(7-3)!).

Evaluating this expression:

C(7, 3) = 7! / (3! * 4!) = (7 * 6 * 5) / (3 * 2 * 1) = 35.

Lastly, we have 4 people remaining, and they will form the third group of size 4. Since there is only one group left, there is only one way to assign the remaining 4 people to this group.

Therefore, the total number of ways to split the group of 10 people into two groups of size 3 and one group of size 4 is 120 * 35 * 1 = 4,200.

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The probability that at least 2 babies out of a cluster of 5 babies will speak tomorrow is 47/128.

The probability that less than 2 babies will speak is

5C0 * (0.25)^0 * (0.75)^5 +  5C1 * (0.25)^1 * (0.75)^4

= (0.75)^5 +  5 * 0.25 * (0.75)^4

(0.75)^5+5*0.25*(0.75)^4 = 81/128

Therefore,

probability that 2 or more will speak is  

=  1- 81/128 = 47/128

The binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a series of n independent experiments, each asking a yes-or-no question and having its own Boolean-valued outcome: success (with probability p) or failure (with probability display style q=1-pq=1-p).

This distribution is used in probability theory and statistics. A Bernoulli trial, or experiment, is another name for a single success-or-failure experiment, and a Bernoulli process is another name for a series of results.

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Answer:

Let's first calculate the amount of interest that would be earned if the interest were compounded annually. The formula for the future value of a single sum is:

F = P * (1 + r/n)^(nt)

Where:

F is the future value

P is the principal (the initial deposit)

r is the annual interest rate

n is the number of compounding periods per year

t is the number of years

For our calculation, we have:

P = $10,000

r = 1.8% = 0.018

n = 1 (annual compounding)

t = 30

So, the future value of the account with annual compounding is:

F = $10,000 * (1 + 0.018/1)^(1 * 30) = $10,000 * (1.018)^30 = $21,784.08

Now, let's calculate the amount of interest that would be earned if the interest were compounded monthly. The formula for the future value of a single sum is the same, but we need to use the monthly compounding rate (r/12) instead of the annual rate and the number of months (12t) instead of the number of years:

F = P * (1 + r/n)^(nt)

Where:

F is the future value

P is the principal (the initial deposit)

r is the annual interest rate

n is the number of compounding periods per year

t is the number of years

For our calculation, we have:

P = $10,000

r = 1.8% = 0.018

n = 12 (monthly compounding)

t = 30

So, the future value of the account with monthly compounding is:

F = $10,000 * (1 + 0.018/12)^(12 * 30) = $10,000 * (1.0015)^360 = $22,254.51

The difference in the two future values is $22,254.51 - $21,784.08 = $470.43.

So, the account would be worth $470.43 more if interest were compounded monthly rather than annually over a period of 30 years. Round to the nearest dollar, the answer is $470.

Step-by-step explanation:

By means of right triangles and trigonometric functions, we find that building 2 is 80 meters high.

Herein we find a geometric system formed by a triangle between two buildings, a triangle generated with an angle of depression and an angle of elevation.

A representation of the system (not in scale) is shown below. We must find the height of building 2 by means of trigonometric functions as the geometric system is also the result of three right triangles.

Then, we proceed to solve the question:

x = 20 / tan 30°

x = 20√3 m

h = 20√3 · (tan 30° + tan 60°)

h = 80 m

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Answer:

1st option

Step-by-step explanation:

a) The geometric and algebraic multiplicity of each eigenvalue of matrix

\( A = \begin{pmatrix} 3& 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\\\end{pmatrix}\)

are one and three respectively.

b) The geometric and algebric multiplicity \( A = \begin{pmatrix} 3& 0 & 0 \\ 1 & 3 & 0 \\ 1& 1 & 3\\\end{pmatrix} \) are equal to 1 and 3.

c) The geometric and algebric multiplicity

\( A = \begin{pmatrix} 3& 0 & 0 \\1 & 3 & 0 \\ 1 & 0 & 3\\\end{pmatrix}\) are equal to 1 and 3.

The algebraic multiplicity of eigenvalue λ is equals to the number of times λ appears as a root of the characteristic polynomial. Dimensions of the λ eigenspace equivalently represents the geometric multiplicity of eigenvalue λ. It is calculated by the dimension of ker(A − λIn). Generally, relation between both is almu(λ) ≥ gemu(λ) for every eigenvalue λ. We have a matrix \( A = \begin{pmatrix} 3& 0 & 0 \\0 & 3 & 0 \\0 & 0 & 3\\ \end{pmatrix}\) the characteristic equation for matrix A is \(\lambda I - A = 0\)

but for determining the algabraic multiplicity the equation is written as

\(det(\lambda I - A) = 0\)

\(det(\lambda \begin{pmatrix} 1& 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\\ \end{pmatrix} - \begin{pmatrix} 3& 0 & 0 \\0 & 3 & 0 \\0 & 0 & 3\\\end{pmatrix})= 0 \)

\(det( \begin{pmatrix} 3 - \lambda& 0 & 0 \\0 & 3 - \lambda & 0 \\0 & 0 & 3 - \lambda\\\end{pmatrix}) = 0\)

=> ( 3 -λ)³ = 0

=>λ= 3,3,3

So, algebraic multiplicity value is 3.

Now, for geometmric multiplicity, we have to determine the eigen vector corresponding to eigen value, lambda = 3. Let \(\begin{pmatrix} x_1 \\x_2 \\x_3\\\end{pmatrix}\)

be the eigen vector for λ = 3. Now solve the following equation,

\(\begin{pmatrix} 3 - \lambda& 0 & 0 \\0 & 3 - \lambda & 0 \\0 & 0 & 3 - \lambda\\\end{pmatrix} \begin{pmatrix} x_1 \\x_2 \\x_3\\\end{pmatrix} = \begin{pmatrix}0 \\0 \\0\\\end{pmatrix}\)

\(\begin{pmatrix} 3 - 3 & 0 & 0 \\ 0 & 3 - 3& 0 \\0 & 0 & 3 - 3\\ \end{pmatrix} \begin{pmatrix} x_1 \\x_2 \\x_3\\\end{pmatrix} = \begin{pmatrix} 0 \\0 \\0\\\end{pmatrix}\)

=> x₁ = x₂ = x₃ = 0, so the only one vetor exist that is zero vector so, the geometric multiplicity is 1. With the similar process we can determine the algebraic and geometric multiplicity for other matrix. So,

b) The algebraic mutiplicity for matrix

\( A = \begin{pmatrix} 3& 0 & 0 \\ 1 & 3 & 0 \\ 1 & 1 & 3\\ \end{pmatrix}\)

is 3 ( for λ= 3,3,3 ) and geomtric mutiplicity is 1 ( since, one eigen vector ( 1,-1, 0)).

c) The algebraic multiplicity for matrix

\( A = \begin{pmatrix} 3& 0 & 0 \\1 & 3 & 0 \\1 & 0 & 3\\ \end{pmatrix}\)

is 3 ( for λ= 3,3,3 ) and geomtric mutiplicity is 1 ( since, one eigen vector ( 1,0, 0)).

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Answer:

The first puppy would weigh 8.6 ounces, the second puppy would weigh 9.4 ounces, and the third puppy would weigh 8.5 ounces.

       For C1(Q) = 3 - 2Q.

        For C2(Q) =  4.

     2. The AVC function

           For C1(Q) = 5/Q + 3 + 20/Q - Q.

           For C2(Q) = 3/Q + 4 + 2/Q.

     3. The AFC function

              For C1(Q)= 5/Q - 20/(5 + 3Q + 20/Q - Q)

              For C2(Q) = 0.

     4.  To find the AC function

        For C1(Q) = (5 + 3Q + 202 - Q^2)/Q + 5/Q - 20/(5 + 3Q + 20/Q - Q).

         For C2(Q) = (3 + 4Q + 2)/Q + 3/Q + 4 + 2/Q.

    5.To find the ATC function

    For C1(Q)= 5/Q² + 3/Q + 20/Q² - Q/Q + 5/Q - 20/(5Q + 3Q² + 20 - Q²)

     For C2(Q)= 3/Q² + 4/Q + 2/Q² + 3/Q + 4/Q + 2/Q.

1. To find the MC function, we take the derivative of the cost functions with respect to Q.

For C1(Q) = 5 + 3Q + 202 - Q^2, MC1(Q) = 3 - 2Q.

For C2(Q) = 3 + 4Q + 2, MC2(Q) = 4.

2. To find the AVC function, we divide the cost functions by Q.

For C1(Q), AVC1(Q) = (5 + 3Q + 202 - Q^2)/Q = 5/Q + 3 + 20/Q - Q.

For C2(Q), AVC2(Q) = (3 + 4Q + 2)/Q = 3/Q + 4 + 2/Q.

3. To find the AFC function, we subtract the AVC function from the ATC function.

For C1(Q), AFC1(Q) = (5 + 3Q + 202 - Q^2)/Q - (5 + 3Q + 202 - Q^2)/(5 + 3Q + 20/Q - Q)

= 5/Q - 20/(5 + 3Q + 20/Q - Q).

For

C2(Q), AFC2(Q) = (3 + 4Q + 2)/Q - (3 + 4Q + 2)/(3/Q + 4 + 2/Q) = 0.

4. To find the AC function, we add the AVC function to the AFC function.

For

C1(Q), AC1(Q) = (5 + 3Q + 202 - Q^2)/Q + 5/Q - 20/(5 + 3Q + 20/Q - Q).

For

C2(Q), AC2(Q) = (3 + 4Q + 2)/Q + 3/Q + 4 + 2/Q.

5. To find the ATC function, we divide the AC function by Q.

For

C1(Q), ATC1(Q) = [(5 + 3Q + 202 - Q²)/Q + 5/Q - 20/(5 + 3Q + 20/Q - Q)]/Q

= 5/Q² + 3/Q + 20/Q² - Q/Q + 5/Q - 20/(5Q + 3Q² + 20 - Q²).

For

C2(Q), ATC2(Q) = [(3 + 4Q + 2)/Q + 3/Q + 4 + 2/Q]/Q

= 3/Q² + 4/Q + 2/Q² + 3/Q + 4/Q + 2/Q.

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