(a) The null space of PA is a subset of the null space of A.
(b) The null space of [tex](AQ)^T[/tex] is a subset of the null space of AT.
How to prove that rank(PA) = r?(a) To prove that rank(PA) = r, we need to show that PA has the same null space as A.
Let x be a vector in the null space of A, then we have Ax = 0. Multiplying both sides by P, we get PA x = 0. This shows that x is also in the null space of PA. Therefore, the null space of A is a subset of the null space of PA.Conversely, let y be a vector in the null space of PA.Then we have PA y = 0, which implies that y is in the null space of A. To see this, we can multiply both sides of PA y = 0 by [tex]P^-1[/tex], which gives us Ay = 0.Therefore, the null space of PA is a subset of the null space of A.
How to prove that rank(AQ) = r?(b) To prove that rank(AQ) = r, we need to show that (AQ)T has the same null space as AT.
Let x be a vector in the null space of AT, then we have AT x = 0. Multiplying both sides by [tex]Q^-1[/tex] on the right, we get [tex]A(Q^-1x)[/tex] = 0.This shows that [tex]Q^-1x[/tex] is in the null space of AQ^T.Therefore, the null space of AT is a subset of the null space of (AQ)^T.
Conversely, let y be a vector in the null space of [tex](AQ)^T[/tex].
Then we have [tex](AQ)^{Ty} = 0[/tex], which implies that y is in the null space of AT. To see this, we can multiply both sides of [tex](AQ)^{Ty} = 0[/tex] by [tex]Q^-T[/tex]on the left, which gives us [tex]A^{T(Q^{-T y})} = 0[/tex].Therefore, [tex]Q^{-T y}[/tex] is in the null space of AT. Therefore, the null space of [tex](AQ)^T[/tex] is a subset of the null space of AT.
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Use Strassen's algorithm to compute the matrix product C = AB, where A = [1 3 7 5] and B = [6 8 4 2]. Show intermediate results.
The intermediate result is [tex]C= \left[\begin{array}{cc}18&14\\62&66\end{array}\right][/tex].
In linear algebra, the Strassen algorithm, named after Volker Strassen, is an algorithm for matrix multiplication. It is faster than the standard matrix multiplication algorithm for large matrices, with a better asymptotic complexity, although the naive algorithm is often better for smaller matrices.
To use Strassen's algorithm to compute the matrix product C = AB, we need to divide the matrices into smaller submatrices and apply the algorithm recursively.
First, we need to pad both matrices A and B with zeros to make them both 2x2 matrices:
[tex]A= \left[\begin{array}{cc}1&3\\7&5\end{array}\right][/tex]
[tex]B= \left[\begin{array}{cc}6&8\\4&2\end{array}\right][/tex]
Next, we divide each matrix into four submatrices of size 1x1:
A11 = 1, A12 = 3
A21 = 7, A22 = 5
B11 = 6, B12 = 8
B21 = 4, B22 = 2
We can then apply Strassen's algorithm to compute the product C = AB:
P1 = A11 * (B12 - B22) = 1 * (8 - 2) = 6
P2 = (A11 + A12) * B22 = (1 + 3) * 2 = 8
P3 = (A21 + A22) * B11 = (7 + 5) * 6 = 72
P4 = A22 * (B21 - B11) = 5 * (4 - 6) = -10
P5 = (A11 + A22) * (B11 + B22) = (1 + 5) * (6 + 2) = 48
P6 = (A12 - A22) * (B21 + B22) = (3 - 5) * (4 + 2) = -12
P7 = (A11 - A21) * (B11 + B12) = (1 - 7) * (6 + 8) = -84
Using these intermediate results, we can compute the submatrices of C:
C11 = P5 + P4 - P2 + P6 = 48 - 10 - 8 - (-12) = 18
C12 = P1 + P2 = 6 + 8 = 14
C21 = P3 + P4 = 72 - (-10) = 62
C22 = P5 + P1 - P3 - P7 = 48 + 6 - 72 - (-84) = 66
Finally, we can combine these submatrices to obtain the matrix C:
[tex]C= \left[\begin{array}{cc}18&14\\62&66\end{array}\right][/tex]
Therefore, the product C = AB using Strassen's algorithm is:
[tex]C= \left[\begin{array}{cc}18&14\\62&66\end{array}\right][/tex]
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suppose that you wanted to predict the price of a house based on where the house was located (east, west or central) as well as square footage. how many indicator variables would you need? 0 3 1 2 4
Indicator variables are used to indicate whether an observation belongs to a discrete category or not. They are important for some statistical models where factor variables must be converted to a set of indicator variables. The general rule is to use one fewer indicator variables than categories.
An indicator variable can also be defined as a random variable that takes the value 1 for some desired outcome and the value 0 for all other outcomes5.
To predict the price of a house based on its location (east, west, or central) and square footage, you would need 2 indicator variables.
One variable would represent the location (with 3 categories: east, west, and central), and the other variable would represent the square footage of the house.
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The number of indicator variables required to predict the price of a house based on location and footage are 3 in count. So, option (b) is right one.
An indicator variable is a random variable that takes the value 1 for some desired outcome or success and the value 0 for all other outcomes like failure. They tell you if a topic is in a category (hence the name). More specifically, it is defined by the variable X, as X = { 1 desired event 0 other event. Logical variables are an example of an indicator variable. Let we want to predict the price of a house based on where the house was located (east, west or central) as well as square footage. Here since total number of levels of indicator =4 ( east, west or central and square footage)
Therefore, number of indicator level needed = 4-1 = 3
Hence, required value is 3.
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use continuity to evaluate the limit. lim x → π 5 sin ( x sin x )
To evaluate the limit lim x → π 5 sin ( x sin x ) using continuity, we need to check if the function 5 sin ( x sin x ) is continuous at x = π.
First, we note that sin ( x sin x ) is continuous everywhere since it is a composition of continuous functions.
Next, we check the continuity of 5 sin ( x sin x ) at x = π. We need to show that lim x → π 5 sin ( x sin x ) = 5 sin ( π sin π ) = 0.
To do this, we can use the fact that sin x is bounded between -1 and 1 for all x, so sin ( x sin x ) is also bounded between -1 and 1. Therefore, we have:
-5 ≤ 5 sin ( x sin x ) ≤ 5 for all x.
Using the squeeze theorem, we can conclude that lim x → π 5 sin ( x sin x ) = 0, since the upper and lower bounds both approach 0 as x approaches π.
Thus, we have evaluated the limit lim x → π 5 sin ( x sin x ) using continuity.
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Calculate the total and average sales by region . In cell B10, use a SUMIF formula to calculate the sum of AnnualSales where the valuc in the 5. Regions named range is equal to the region listed in cell B9 b Copy the formula to cells C10:110 c In cell B11, use an AVERAGEIF formula to calculate the average of AnnualSales where the value im the Regions named range is equal to the region listed in cell B9 d Add a rounding function to the formula in cell B11 to round the result to two decimal places Copy the formula to cells Cl1:111 E If nocessary, format cells B10:111 with the default Accounting Number Format
The sum of Annual Sales for a particular region and using an AVERAGEIF formula to calculate the average of Annual Sales for the same region.
How to calculate the total and average sales by region?To calculate the total and average sales by region, follow these steps:
a. In cell B10, use the SUMIF formula to calculate the sum of Annual Sales where the value in the 5.Regions named range is equal to the region listed in cell B9. The formula would look like this:
=SUMIF(5.Regions,B9,AnnualSales)
Here, "5.Regions" refers to the named range containing the region names, "B9" is the cell containing the region you want to sum, and "Annual Sales" is the range containing the sales figures.
b. Copy the formula in cell B10 to cells C10:110 by selecting cell B10, pressing Ctrl + C to copy, selecting cells C10:110, and pressing Ctrl + V to paste.
c. In cell B11, use the AVERAGEIF formula to calculate the average of Annual Sales where the value in the Regions named range is equal to the region listed in cell B9. The formula would look like this:
=AVERAGEIF(5.Regions,B9,AnnualSales)
Here, "5.Regions" refers to the named range containing the region names, "B9" is the cell containing the region you want to average, and "Annual Sales" is the range containing the sales figures.
d. Add a rounding function to the formula in cell B11 to round the result to two decimal places. The formula would look like this:
=ROUND(AVERAGEIF(5.Regions,B9,AnnualSales),2)
Here, the ROUND function rounds the result of the AVERAGEIF formula to two decimal places.
e. Copy the formula in cell B11 to cells C11:111 by selecting cell B11, pressing Ctrl + C to copy, selecting cells C11:111, and pressing Ctrl + V to paste.
f. If necessary, format cells B10:111 with the default Accounting Number Format by selecting the range B10:111, right-clicking, selecting "Format Cells," and choosing the Accounting Number Format.
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you and your friend are in the studio audience on a television game show. from an audience of 300 people, 2 people are randomly selected as contestants. what is the probability that you and your friend are chosen? express your answer as a fraction in simplest form.
The probability that you and your friend are chosen as the 2 contestants out of a total of 300 people is 1/135,700.
The probability that you and your friend are chosen as the 2 contestants out of a total of 300 people is calculated as follows:
First, the probability of you being chosen is 1/300, since there is only one of you in the audience of 300 people.
Next, the probability of your friend being chosen is 1/299, since there is now one less person in the audience to choose from.
Multiplying these probabilities together gives:
(1/300) x (1/299) = 0.00000111
Therefore, the probability that you and your friend are chosen as the 2 contestants are approximately 0.00000111, or 0.000111%.
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Evaluate the iterated integral by converting to polar coordinates.
o 8
o −8
o 64 − x2 o sin(x^2 + y^2) dy dx
o 0
The value of the iterated integral is (-π/4) (cos(64) - 1).
To evaluate the iterated integral by converting to polar coordinates, we need to use the given terms:
- The integral bounds: 0 to 8 (for x), -8 to 8 (for y)
- The integrand: sin(x² + y²)
- The conversion factor for polar coordinates: rdrdθ
First, we convert the integral bounds and integrand to polar coordinates.
In polar coordinates, x = rcos(θ) and y = rsin(θ), so:
x² + y² = r²cos²(θ) + r²sin²(θ) = r²
because sin²(θ) + cos²(θ) = 1.
Now we rewrite the integrand in polar coordinates:
sin(x² + y²) = sin(r²)
The integral bounds in polar coordinates are 0 to 8 (for r), and 0 to π/2 (for θ) as the given bounds cover the first quadrant.
Now we can write the iterated integral:
[tex]\int_{0} ^ {\pi/2} \int_{0} ^{ 8} sin(r^2) r dr d\theta[/tex]
To evaluate the inner integral with respect to r, we use substitution:
Let u = r², so du = 2r dr
Now the inner integral becomes:
[tex](1/2)\int_{0}^ {64} sin(u) du[/tex] = (-1/2) [cos(64) - cos(0)] = (-1/2) (cos(64) - 1)
Now we evaluate the outer integral with respect to θ:
[tex]\int_{0}^ {\pi/2} ((-1/2) (cos(64) - 1)) d\theta[/tex] = (-1/2) (cos(64) - 1) [θ] evaluated from 0 to π/2
Finally, substitute the limits of integration:
((-1/2) (cos(64) - 1)) (π/2 - 0) = (-π/4) (cos(64) - 1)
So, the value of the iterated integral is (-π/4) (cos(64) - 1).
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(5.46 x 103 m 1.54 x 103 m)(9.54 x 10-3 m – 7.34 x 10-3 m)
To solve this problem, we need to use the distributive property of multiplication. First, we'll multiply the two numbers inside the parentheses:
(9.54 x 10-3 m - 7.34 x 10-3 m) = 2.20 x 10-3 m
Now we'll multiply this result by the first set of numbers:
(5.46 x 103 m) (2.20 x 10-3 m) = 12.012 m
So the final answer is 12.012 m.
To solve the given expression, you'll need to perform the calculations as shown:
(5.46 x 10^3 m * 1.54 x 10^3 m) * (9.54 x 10^-3 m - 7.34 x 10^-3 m)
First, calculate the products and the difference:
(8,408.44 m^2) * (2.20 x 10^-3 m)
Finally, multiply the two results together:
18.497 m
So the answer is 18.497 m.
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Complete question:
Solve: (5.46 x 103 m 1.54 x 103 m)(9.54 x 10-3 m – 7.34 x 10-3 m)
The point ( ¯ 2 , 6 ) is plotted on a coordinate plane. Which statements are true? Select the two statements that are true. A. The reflection point across the y-axis is ( 2 , 6 ) . B. The reflection point across the y-axis is ( 2 , ¯ 6 ) . C. The reflection point across the y-axis is ( ¯ 2 , ¯ 6 ) . D. The reflection point across the x-axis is ( 2 , 6 ) . E. The reflection point across the x-axis is ( 2 , ¯ 6 ) . F. The reflection point across the x-axis is ( ¯ 2 , ¯ 6 ) .
Answer:The two true statements are:
A. The reflection point across the y-axis is (2, 6).
D. The reflection point across the x-axis is (¯2, ¯6).
Step-by-step explanation:
Answer:
The two true statements are:
A. The reflection point across the y-axis is (2, 6).
D. The reflection point across the x-axis is (¯2, ¯6).
Step-by-step explanation:
Please help me!!!!!!!
The value of arc AEC is 240°
What is circle geometry?A circle is a special kind of ellipse in which the eccentricity is zero and the two foci are coincident.
AB = BC , therefore,
25x +85 = 30x +90
collect like terms
25x-30x = 90-85
-5x = 5
divide both sides by -5
x = 5/-5
x = -1
Therefore angle B = -120x
= -120 × - 1 = 120°
the theorem that says angle at the center is twice the angle at circumference can now be applied.
therefore the value of arc AEC = 2× 120 = 240°
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If the whole of a clock is 12 than what is the angle of 1/12
Step-by-step explanation:
A whole clock is 360 degrees
1/12 of 360 degrees is 30 degrees
the vector space of all lower triangular 7×77×7 matrices
The vector space of all lower triangular 7x7 matrices consists of all 7x7 matrices in which all elements above the main diagonal are zero. In a lower triangular matrix, non-zero elements are found on or below the main diagonal. This specific vector space has a dimension of 28, as there are 28 independent elements in a lower triangular 7x7 matrix.
The vector space of all lower triangular 7x7 matrices is a subspace of the vector space of all 7x7 matrices. This subspace includes matrices where all entries above the main diagonal are zero.
As a vector space, it satisfies the properties of closure under addition and scalar multiplication, and contains a zero vector (the matrix with all entries being zero).
The basis for this subspace is the set of matrices with a single non-zero entry in each row below the diagonal, which has dimension 28. Therefore, any lower triangular 7x7 matrix can be represented as a linear combination of these basis matrices.
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solve the following initial-value problems starting from y 0 = 5 . d y/ d t = e^ 4 ta) y =at what time does y increase to 100 or drop to 1? round your answer to four decimal places.b) t =
The function y(t) that satisfies the differential equation dy/dt = e^(4t) and the initial condition y(0) = 5 is given by y(t) = (1/4)e^(4t) + 4.
The time when y(t) increases to 100 is approximately t = 1.6583. However, y(t) never drops to 1 as the equation 1 = (1/4)e^(4t) + 4 has no real solutions.
Step-by-Step Explanation:
Given the differential equation dy/dt = e^(4t) and the initial condition y(0) = 5, we need to find the function y(t) that satisfies the differential equation.
Integrate both sides of the differential equation with respect to t to get ∫(dy/dt) = ∫(e^(4t) dt).
Using the chain rule, we can simplify the left-hand side of the equation to get ∫(dy/dt) dt = ∫dy = y + C, where C is the constant of integration.
Integrating the right-hand side with respect to t results in (1/4)e^(4t) + K, where K is another constant of integration. Substituting this into the equation from step 3, we get y + C = (1/4)e^(4t) + K.
Using the initial condition y(0) = 5, we can solve for the constant of integration C: y(0) + C = (1/4)e^(4*0) + K. Simplifying this equation, we get C + K = 5.
Substituting the values of C and K into the equation from step 4, we get y(t) = (1/4)e^(4t) + 4.
To find when y(t) increases to 100, we set the equation from step 6 equal to 100 and solve for t:
100 = (1/4)e^(4t) + 4
(1/4)e^(4t) = 96
e^(4t) = 384
4t = ln(384)
t = ln(384)/4 ≈ 1.6583 (rounded to four decimal places)
To find when y(t) drops to 1, we set the equation from step 6 equal to 1 and solve for t:
1 = (1/4)e^(4t) + 4
(1/4)e^(4t) = -3
e^(4t) = -12
Since e^(4t) is always positive, there are no real values of t that satisfy this equation. Therefore, y(t) never drops to 1.
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When using the chi-square goodness of fit test, the smaller the value of the chi-square test statistic, the more likely we are to reject the null hypothesis.
Select one: True or False
The statement "When using the chi-square goodness of fit test, the smaller the value of the chi-square test statistic, the more likely we are to reject the null hypothesis." is false.
When using the chi-square goodness of fit test, the larger the value of the chi-square test statistic, the more likely we are to reject the null hypothesis.
A small chi-square value indicates that the observed data fits the expected distribution well, which supports the null hypothesis.
A large chi-square value on the other hand indicates a poor fit between the observed data and the expected distribution, which suggests that the null hypothesis is not true and we should reject it.
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On graph paper, graph Triangle ABC if A(3,0), B(2,7), and C(6,4). Investigate the slopes and side lengths of triangle ABC and make a list of any special relationships that they have
part a. )
Side AB:
Length: dAB = √[(2 - 3)² + (7 - 0)²] = √(10) = 3.16
Slope: mAB = (7 - 0) / (2 - 3) = -7
Side AC:
Length: dAC = √[(6 - 3)² + (4 - 0)²] = √(25) = 5
Slope: mAC = (4 - 0) / (6 - 3) = 4/3
Side BC:
Length: dBC = √[(6 - 2)² + (4 - 7)²] = √(20) = 4.47
Slope: mBC = (4 - 7) / (6 - 2) = -3/4
part b.)
some special relationships between the sides of Triangle ABC:
Side AB is shorter than side BC, but has a steeper slope.Side AC is the longest side of the triangle.We will use the distance formula and slope formula.
The distance formula is:
d = √[(x2 - x1)² + (y2 - y1)²]
And the slope formula is:
m = (y2 - y1) / (x2 - x1)
Side AB:
Length: dAB = √[(2 - 3)² + (7 - 0)²] = √(10) ≈ 3.16
Slope: mAB = (7 - 0) / (2 - 3) = -7
Side AC:
Length: dAC = √[(6 - 3)² + (4 - 0)²] = √(25) = 5
Slope: mAC = (4 - 0) / (6 - 3) = 4/3
Side BC:
Length: dBC = √[(6 - 2)² + (4 - 7)²] = √(20) = 4.47
Slope: mBC = (4 - 7) / (6 - 2) = -3/4
The slopes of sides AB and BC are negative, which indicates that these sides slope downwards from left to right on the coordinate plane.
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Suppose that the average country song length in America is 4.75 minutes with a standard deviation of 1.10 minutes. It is known that song length is not normally distributed. Suppose a sample of 25 songs is taken from the population. What is the approximate probability that the average song length will last more than 5.25 minutes? Round to the nearest thousandth. a. 0.488 b. 0.012 c. 0.325 d. 0.175
The approximate probability that the average song length of a sample of 25 songs will last more than 5.25 minutes is 0.012. So, the correct option is b.
To find the approximate probability that the average song length of a sample of 25 songs will last more than 5.25 minutes, we will use the Central Limit Theorem, which states that the distribution of sample means approaches a normal distribution as the sample size increases. In this case, we have:
1. Population mean (μ) = 4.75 minutes
2. Standard deviation (σ) = 1.10 minutes
3. Sample size (n) = 25 songs
Now, we will calculate the standard error (SE) using the formula:
SE = σ / √n
SE = 1.10 / √25
SE = 1.10 / 5
SE = 0.22
Next, we will find the z-score using the formula:
z = (x - μ) / SE
z = (5.25 - 4.75) / 0.22
z = 0.50 / 0.22
z ≈ 2.27
Now, we will use a z-table to find the probability that the z-score is greater than 2.27. Looking up the value in the table, we find the area to the left of the z-score is approximately 0.988. Since we want the probability to the right of the z-score, we subtract the area from 1:
P(Z > 2.27) = 1 - 0.988 = 0.012
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I don’t understand this type of question!
Answer: circle with center at (-11,-8) and a radius of 16
Step-by-step explanation:
Consider the package weight data in Exercise 8.13. Suppose there is a lower specification at 0.985 kg. Calculate an appropriate process capability ratio for this material. What percentage of the packages produced by this process is estimated to be below the specification limit?
To calculate the process capability ratio and the percentage of packages below the specification limit, you will need the following data from Exercise 8.13:
Process mean (μ)
Process standard deviation (σ)
Lower specification limit (LSL)
Step-by-step calculations:
Calculate the process capability ratio (Cpk) using the following formula:
Cpk = min [(USL - μ) / (3σ), (μ - LSL) / (3σ)]
In this case, the lower specification limit is given as 0.985 kg.
Assuming the upper specification limit is not given, we can calculate the Cpk value using only the lower specification limit.
If the calculated Cpk value is less than 1, the process needs improvement. If it is greater than or equal to 1, the process is considered acceptable.
Estimate the percentage of packages below the lower specification limit (LSL) using the following steps:
a. Calculate the z-score for the lower specification limit:
z = (LSL - μ) / σ
b. Look up the z-score in a standard normal distribution table or use an online calculator to find the corresponding probability (area under the curve). This probability represents the percentage of packages estimated to be below the lower specification limit.
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Find the area of the kite 15 and 8
The answer of the given question based on the area of kite is , the area of the kite is 60unit².
What is Diagonal?A diagonal is a straight line connecting two non-adjacent vertices of a polygon or a polyhedron. In other words, it is a line segment that connects two corners of a shape that are not next to each other.
Diagonals play an important role in geometry, as they can be used to determine various properties of shapes. For instance, the length of the diagonal of a rectangle can be used to find its area and perimeter, and the length of the diagonal of a cube can be used to find its volume and surface area.
To find the area of a kite, we need to know the lengths of its two diagonals. Let d1 and d2 be length of two diagonals of kite.
In this case, we know that the two diagonals have lengths 15 and 8. Let's label them as d1 = 15 and d2 = 8.
The area of kite can be calculated using formula:
Area = (1/2) x d1 x d2
Substituting values of d1 and d2, we will get:
Area = (1/2) x 15 x 8
Area = 60
Therefore, the area of the kite is 60unit².
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Draw a line of symmetry through the pentagon by connecting two black points
The line of symmetry through the pentagon by connecting two black points is illustrated below.
A pentagon is a geometric shape that consists of five straight sides and five vertices or corners.
To draw a line of symmetry through the pentagon by connecting two black points, we first need to identify these points. In a regular pentagon, all sides and angles are equal, and it has five lines of symmetry that pass through the center of the shape and connect opposite vertices.
Once we have identified the two black points that we want to connect, we need to draw a straight line that passes through both of them and divides the pentagon into two halves that are mirror images of each other.
This line is the line of symmetry that we are looking for. To check if it is correct, we can fold the pentagon along the line of symmetry and see if both halves match up perfectly.
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A study has ni independent binary observations{yi1, . . . , yin }when X=xi, i=1, . . . , N, with ∑ini. Consider the model logit(πi)=α+βxi, where πi=P(Yi j=1).
a. Show that the kernel of the likelihood function is the same treating the data as n Bernoulli observations or N binomial observations.
b. .For the saturated model, explain why the likelihood function is different for these two data forms. ( Hint: The number of parameters differs.. Hence, the deviance reported by software depends on the form of data entry)
c. Explain why the difference between deviances for two unsaturated models does not depend on the form of data entry.
d. Suppose that each ni=1. Show that the deviance depends on width= but not yi. Hence, it is not useful for checking model fit
a. The kernel of the likelihood function is the same for n Bernoulli or N binomial observations.
b. The likelihood function is different due to the number of parameters. c. The difference between deviances for unsaturated models does not depend on the form of data entry.
d. Deviance depends on the width of bins, but not individual observations yi when ni=1.
a. The kernel of the likelihood function is the same treating the data as n Bernoulli observations or N binomial observations since each Bernoulli observation can be considered as a binomial observation with N = 1.
b. For the saturated model, the likelihood function is different for these two data forms because the number of parameters differs. In the case of Bernoulli observations, there are n parameters for the n observations, while for binomial observations, there is only one parameter for the entire dataset. Therefore, the deviance reported by software depends on the form of data entry.
c. The difference between deviances for two unsaturated models does not depend on the form of data entry because the number of parameters is the same for both forms of data. Hence, the deviance can be used to compare models regardless of the form of data entry.
d. If ni=1 for all i, then the deviance depends only on the number of observations (n) and the width of the bins used for grouping the data. The individual observations yi do not affect the deviance, and hence it is not useful for checking model fit.
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Find the probability in n = 49, µ = 75, and σ = 6;
P( x_bar < 76. 5) = ?
The probability of the given question under the condition that n = 49, µ = 75, and σ = 6 is 0.9599.
Probability refers to the measure of how many events that occur at random are likely to occur. It is considered as a branch of mathematics which deals with occurrence of unexpected events taking place. Given the event has a probability of 1 then the event will surely happen in some time.
therefor,
z = (x' - μ)/(σ/[tex]\sqrt{n}[/tex])
z = ( 76.5 - 75) /(6/[tex]\sqrt{49}[/tex])
z = 1.75
with the help of standard distribution table we can find
P( x_bar < 76. 5) = 0.9599.
The probability of the given question under the condition that n = 49, µ = 75, and σ = 6 is 0.9599.
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Let W1 and W2 be subspaces of a finite dimensional inner product space. Prove that (W1 + W2)perp = W1perp ∩ W2perp and (W1 ∩ W2)perp = W1perp + W2perp.
Note: "w perp" is a non empty subset of an inner product space V, and is the set of all vectors in V that are orthogonal to every vector in W. aka orthogonal complement
As we have proved that a finite dimensional inner product space is (W1 ∩ W2)⊥ = W1⊥ + W2⊥.
Our goal is to prove that the orthogonal complement of the intersection of W1 and W2 is equal to the sum of the orthogonal complements of W1 and W2. In other words, we want to show that:
(W1 ∩ W2)⊥ = W1⊥ + W2⊥
To prove this, we need to show that every vector in the left-hand side is also in the right-hand side, and vice versa. Let's start with the "⊆" direction:
Suppose v belongs to (W1 ∩ W2)⊥, i.e., v is orthogonal to every vector in W1 ∩ W2. We want to show that v also belongs to W1⊥ + W2⊥, i.e., v can be written as the sum of a vector in W1⊥ and a vector in W2⊥.
Since v is orthogonal to every vector in W1 ∩ W2, it is in particular orthogonal to every vector in W1 and every vector in W2. Therefore, v belongs to both W1⊥ and W2⊥, by definition of orthogonal complement. Hence, we can write v as the sum of a vector in W1⊥ and a vector in W2⊥, which shows that v is in the right-hand side.
Now let's prove the other direction, "⊇":
Suppose v belongs to W1⊥ + W2⊥, i.e., v can be written as the sum of a vector in W1⊥ and a vector in W2⊥. We want to show that v also belongs to (W1 ∩ W2)⊥, i.e., v is orthogonal to every vector in W1 ∩ W2.
Let u be any vector in W1 ∩ W2. Since u is in both W1 and W2, it is orthogonal to v1 and v2, respectively, for some v1 in W1 and v2 in W2. Hence, we have:
⟨v,u⟩ = ⟨v,v1⟩ + ⟨v,v2⟩ = 0
where ⟨·,·⟩ denotes the inner product. The first equality follows from the fact that v is in the orthogonal complement of W1 (hence, orthogonal to every vector in W1), and the second equality follows from the fact that v is in the orthogonal complement of W2. Therefore, we have shown that v is orthogonal to every vector in W1 ∩ W2, which implies that v is in the left-hand side.
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Let an = 5+1+1. Prove that an = O(1). 6.7.15.
To prove that an = O(1), we need to show that there exists a constant c and a positive integer n0 such that for all n >= n0, |an| <= c.
Let's simplify the expression for an:
an = 5+1+1 = 7
Now we can choose c = 15 and n0 = 1.
For all n >= n0, |an| = |7| = 7 <= 15 = c.
Therefore, we have shown that an = O(1).
Based on your question, it seems you want to prove that the sequence an = 5 + 1 + 1 has a constant order of growth, which is represented as O(1).
Let an = 5 + 1 + 1. This simplifies to an = 7. Since an is a constant value (7), its growth rate is not dependent on n. Therefore, the order of growth is constant, and we can say that an = O(1).
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what is the largest interval containing zero on which f(x) = \sin x is one-to-one?
The largest interval containing zero on which f(x) = \sin x is one-to-one is (-\pi/2, \pi/2).
This is because on this interval, sin(x) is strictly increasing from -1 to 1, meaning that there is only one output value for each input value in this interval. Beyond this interval, sin(x) starts to repeat its values, which means that it is no longer one-to-one.
The largest interval containing zero on which f(x) = sin(x) is one-to-one is the open interval (-π/2, π/2). This interval ensures the sine function has a unique output for each input, making it an injective (one-to-one) function.
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Let Y be a random variable such that P(Y = −2) = 1/4, P(Y = 0) = 1/2, and P(Y = 1) = 1/4.
a) Find the mgf of Y .
b) Use the mgf to find E(Y ) and V (Y )
a) The moment-generating function of Y is M(t) = (1/4)e^(−2t) + 1/2 + (1/4)e^(t).
b) The expected value of Y is E(Y) = −1/4, and the variance of Y is V(Y) = 11/16.
a) The moment-generating function (MGF) of a random variable Y is defined as M(t) = E(e^(tY)), where E denotes the expected value. Using the given probabilities, we can compute the MGF of Y as follows
M(t) = E(e^(tY)) = (1/4)e^(−2t) + (1/2)e^(0t) + (1/4)e^(1t)
= (1/4)e^(−2t) + 1/2 + (1/4)e^(t)
b) To find the expected value of Y, we can take the first derivative of the MGF with respect to t and evaluate it at t = 0
E(Y) = M'(0) = d/dt [(1/4)e^(−2t) + 1/2 + (1/4)e^(t)]|t=0
= (−1/2)e^(−2t)|t=0 + 0 + (1/4)e^(t)|t=0
= −1/2 + 1/4
= −1/4
Therefore, the expected value of Y is −1/4.
To find the variance of Y, we need to compute E(Y^2), which can be found by taking the second derivative of the MGF with respect to t and evaluating it at t = 0
E(Y^2) = M''(0) = d^2/dt^2 [(1/4)e^(−2t) + 1/2 + (1/4)e^(t)]|t=0
= (1/2)e^(−2t)|t=0 + 0 + (1/4)e^(t)|t=0
= 1/2 + 1/4
= 3/4
Therefore, the variance of Y is
V(Y) = E(Y²) − [E(Y)]² = 3/4 − [−1/4]² = 3/4 − 1/16 = 11/16
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Verify the following identity. sin^2x + cos2x = cos^2x. To transform the left side into the right side, __ should be changed to ___ and the left side simplified. o 2sin x cos x o cos^2x- sin 2x o sin^2x+ cos²x
To transform the left side into the right side, we need to use the identity that cos2x = cos^2x - sin^2x.
Thus, we can rewrite the left side as sin^2x + cos^2x - sin^2x, which simplifies to cos^2x. Therefore, the left side equals the right side, and the identity is verified.
To verify the identity sin^2x + cos2x = cos^2x, we need to transform the left side into the right side.
To do this, we will use the trigonometric identity for cos2x, which is: cos2x = cos^2x - sin^2x.
Now, replace cos2x in the original equation with this identity:
sin^2x + (cos^2x - sin^2x) = cos^2x.
The sin^2x terms will cancel out:
cos^2x = cos^2x.
Thus, the given identity is verified. To transform the left side into the right side, cos2x should be changed to cos^2x - sin^2x, and the left side simplified.
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(Which word best describes the degree of overlap between the two data sets?
Responses (I WANT EXPLANATION ON HOW TO DO IT AND ANSWER PLEASE TY)
high
moderate
none
low
Two line plots with the same scale and interval, from zero to ten, with tick marks every one unit. The plots are stacked vertically for comparison. Each plot has sixteen data values. Values appear as x marks above the line. Upper plot data values are one x mark above two, one x mark above three, one x mark above four, two x marks above five, four x marks above six, three x marks above seven, two x marks above eight, one x mark above nine, and one x mark above ten. Lower plot data values are one x mark above 1, two x marks above two, three x marks above three, four x marks above four, two x marks above five, two x marks above six, one x mark above seven, and one x mark above eight.
The degree of overlap between the two data sets is moderate. Therefore, the correct answer is B. moderate.
What is the degree of overlap?To determine the degree of overlap between the two data sets, we need to visually compare the line plots. Based on the description provided, we can see that the two line plots have some common data points but also have different data values.
The upper plot has data values that range from 2 to 10, while the lower plot has data values that range from 1 to 8. However, both plots share some common data values such as 2, 3, 4, and 5.
Therefore, we can say that the degree of overlap between the two data sets is moderate. While there are some common data values, the majority of the data values are different between the two plots.
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let λ be an eigenvalue of an invertible matrix a. show that λâ1 is an eigenvalue of aâ1. [hint: suppose a nonzero x satisfies ax=λx.]
To show that λâ1 is an eigenvalue of aâ1, we need to show that there exists a nonzero vector y such that aâ1y = λâ1y.
Let's use the hint and suppose that a nonzero vector x satisfies ax = λx. Then we can multiply both sides of this equation by aâ1 on the left to get:
aâ1(ax) = aâ1(λx)
x = aâ1(λx)
λâ1x = aâ1x
So if we let y = x, we have aâ1y = aâ1x = λâ1x = λâ1y. And since we started with a nonzero x, we know that y is also nonzero.
Therefore, we have shown that λâ1 is an eigenvalue of aâ1, as required.
To show that λ^(-1) is an eigenvalue of A^(-1), we will use the hint provided. Suppose a nonzero vector x satisfies Ax = λx, where λ is an eigenvalue of the invertible matrix A.
Since A is invertible, A^(-1) exists. Now, we can multiply both sides of Ax = λx by A^(-1) on the left:
A^(-1)(Ax) = A^(-1)(λx)
Using the associative property of matrix multiplication, we have:
(A^(-1)A)x = λ(A^(-1)x)
As A^(-1)A is the identity matrix I, the equation becomes:
Ix = λ(A^(-1)x)
Since Ix = x, we have:
x = λ(A^(-1)x)
Now, we want to find the eigenvalue of A^(-1). To do this, we'll divide both sides of the equation by λ:
x/λ = A^(-1)x
Since x is a nonzero vector, we can rewrite the equation as:
A^(-1)x = λ^(-1)x
Here, we see that λ^(-1) is indeed an eigenvalue of A^(-1), as it satisfies the eigenvalue equation A^(-1)x = λ^(-1)x for the nonzero vector x.
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What is the growth factor when something is decreasing by:
30%?
200%?
8%?
0.12%?
Answer: When something is decreasing by 30%, the growth factor is 0.7 (1 - 0.3).
When something is increasing by 200%, the growth factor is 3 (1 + 2).
When something is decreasing by 8%, the growth factor is 0.92 (1 - 0.08).
When something is decreasing by 0.12%, the growth factor is 0.99988 (1 - 0.0012).
if a confidence interval is given from 43.85 up to 61.95 and the mean is known to be 52.90, what is the margin of error?
The margin of error can be calculated using the formula:
Margin of error = (upper limit of the confidence interval - lower limit of the confidence interval) / 2
In this case, a margin of error of 9.55 suggests that the sample mean is quite precise, since it's relatively close to the true population mean (which we know to be 52.90).
In this case, the lower limit of the confidence interval is 43.85 and the upper limit is 61.95.
Margin of error = (61.95 - 43.85) / 2
Margin of error = 9.55
Therefore, the margin of error is 9.55. This means that if the sample size were to be repeated, we would expect the sample mean to be within 9.55 units of the true population mean 95% of the time.
It's worth noting that the confidence interval provides a range of values within which we can be reasonably certain that the true population mean lies. The margin of error, on the other hand, gives us an indication of the precision of our estimate. However, if the margin of error were larger, this would indicate that our estimate is less precise and that we need a larger sample size to obtain a more accurate estimate of the population mean.
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