factor the gcf: 12x3y 6x2y2 − 9xy3. 3x2y(4x2 2xy − 3) 3xy(4x 2xy − 3y2) 3xy(4x2 2xy − 3y2) 3x2y(4x3y − 2x2y2 − 3xy3)

The largest two-digit integer a = 99 satisfies the congruence 3x ≡ 4

To obtain the largest two-digit integer that satisfies the congruence 3x ≡ 4 (mod 7), we need to find an integer value for x that satisfies the congruence equation.

First, we can rewrite the congruence as an equation:

3x = 4 + 7k, where k is an integer.

Next, we can iterate through two-digit integers in descending order to find the largest value of a that satisfies the equation.

Starting with a = 99, we substitute it into the equation:

3(99) = 4 + 7k

297 = 4 + 7k

By trying different values of k, we can see that k = 42 satisfies the equation:

297 = 4 + 7(42)

Therefore, x = 99 is a solution to the congruence equation 3x ≡ 4 (mod 7).

However, we need to find the largest two-digit integer, so we continue the iteration.

Next, we try a = 98:

3(98) = 4 + 7k

294 = 4 + 7k

By trying different values of k, we can see that k = 42 also satisfies the equation:

294 = 4 + 7(42)

Therefore, x = 98 is another solution to the congruence equation 3x ≡ 4 (mod 7).

Since we have found the largest two-digit integer a = 99 that satisfies the congruence, we can conclude that a = 99 is the answer to the problem.

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Answer:

A

Step-by-step explanation:

A secant is a line that goes through a circle at two points.

Answer:

A

Step-by-step explanation:

a straight line that cuts a curve in two or more parts.

Answer:

37

Step-by-step explanation:

Given:

The graph of a relations.

To find:

Whether the relation is a function or not.

Solution:

A relation is a function if there exist unique outputs for each input. If a graph passes the vertical line test then is a function.

Vertical line test: Each vertical line intersect the curve at most one.

In the given figure draw a vertical at [tex]x=1[/tex] as shown in the below figure.

From the below graph, it is clear that the vertical line intersects the curve at two points. So, it does not pass the vertical line test.

Therefore, the given relation is not a function.

The ratio of the surface area to the volume is 14/15 in²/in³.

To find the ratio of surface area to volume for a given object, we divide the surface area by the volume. In this case, we have a ratio of 280 in² to 300 in³.

The surface area represents the total area of all the exposed surfaces of the object, while the volume represents the amount of space occupied by the object.

To calculate the ratio, we divide the surface area by the volume:

Ratio = Surface Area / Volume

Ratio = 280 in² / 300 in³

Simplifying the ratio:

Ratio = (280/300) in²/in³

Ratio = (28/30) in²/in³

Ratio = (14/15) in²/in³

Therefore, the ratio of the surface area to the volume is 14/15 in²/in³.

This ratio represents the relationship between the amount of surface area and the amount of volume for the given object. It indicates that, for every 15 cubic inches of volume, there are 14 square inches of surface area.

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The week se will give a 12-minute speech is week 22 (option A)

The table is a linear table. This is because the variables change by a fixed amount.

Rate of change = change in length of speech / change in week

(180 - 150) / (4 - 3)

= 30 / 1 = 30 seconds  

The next step is to convert minutes to seconds

1 minute = 60 seconds

60 x 12 = 720 seconds

Length of speech in week 2 = 150 - 30 = 120 seconds

Length of speech in week 1 = 120 - 30 = 90

Week the speech would have a length of 720 seconds = 90 +[ 30 x (week number - 1)]

720 = 90 + [30 x (x -1)

720 = 90 + 30x - 30

720 = 60 + 30x

720 - 60 = 30x

660 = 30x

x = 660 / 30

x = 22

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Answer: 12.5

Step-by-step explanation:

So, basically there is 5 minutes and the number of times occuring is 5 so its 5x1 which is 5 and then there is 10 minutes and the number of times occuring  is twice so its 2x10 which is 20. Next, is 15 minutes and the times occuring  is 2 so, 2x15 which is 30. And finally theres  20 minutes and the amount of times occuring  is once so 20x1 which is 20. Then you add them all up which is 5+20+30+20=75. And the formula for mean/average is The sum of all number/ The amount of numbers there are so 75/6 is 12.5.

Answer:

24 days

Step-by-step explanation:

Given

[tex]Pages = 528[/tex]

[tex]Rate = 22[/tex]

Required

The number of days

The number of days (d) is calculated as:

[tex]d = \frac{Pages}{Rate}[/tex]

[tex]d = \frac{528}{22}[/tex]

[tex]d = 24[/tex]

Step-by-step explanation:

-.5f -5 < -1

-.5f < 4

f > -8

hope this helps

Answer:

b. y=19x

Step-by-step explanation:

The given binary operations are defined as below:

For the integers x and y, the binary operations are defined as: x + y = max (x, y) and x – y = min (x, y)

a) Commutative Property: The commutative property holds for both binary operations, + and – , because: For any x, y ∈ Z,x + y = y + x and x – y = -(y – x)Therefore, both + and – are commutative.

Associative Property: Associativity can also be shown for both binary operations, + and –, as follows: For any x, y and z ∈ Z,x + (y + z) = max (x, max(y, z)) = max (max(x, y), z) = (x + y) + z(x – y) – z = min (x, min (y, z)) = min (min(x, y), z) = (x – y) – z

Therefore, both + and – are associative.

Distributive Property: The distributive property can be shown for these binary operations, + and –, as follows: For any x, y, and z ∈ Z,x + (y – z) = max (x, min (y, z)) = min (max(x, y), max(x, z)) = (x + y) – (x + z)Therefore, both + and – are distributive.

b) For the given operations, the element that violates the property x + 0 = x is:0If x = 2, then x + 0 = max (2, 0) = 2If x = 0, then x + 0 = max (0, 0) = 0So, the property x + 0 = x holds for all integers except for 0.

For this particular element, the value of x + 0 is always 0.

Therefore, the given property fails to hold.

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This question is incomplete, the complete question is;

Alicia would like to know if there is a difference in the average price between two brands of shoes. She selected and analyzed a random sample of 40 different types of Brand A shoes and 33 different types of Brand B shoes. Alicia observes that the boxplot of the sample of Brand A shoe prices shows two outliers. Alicia wants to construct a confidence interval to estimate the difference in population means.

Is the sampling distribution of the difference in sample means approximately normal?

a) Yes, because Alicia selected a random sample

b) Yes, because for each brand it is reasonable to assume that the population size is greater than ten times its sample size.

c) Yes, because the size of each sample is at least 30

d) No, because the distribution of Brand A shoes has outliers

e) No, because the shape of the population distribution is unknown.

Answer:

the sampling distribution of  the difference in sample means is approximately normal because the size of each sample is al least 30

Option (c) Yes, because the size of each sample is at least 30 ) is the correct answer.

Step-by-step explanation:

Given the data in the question;

sample size of brand A shoes [tex]n_A[/tex] = 40

sample size of brand B shoes [tex]n_B[/tex] = 33

As we can notice,

[tex]n_A[/tex] = 40 > 30

[tex]n_B[/tex] = 33 > 30

Hence, we consider both samples to be large.

Therefore, the sampling distribution of  the difference in sample means is approximately normal because the size of each sample is al least 30.

Option (c) Yes, because the size of each sample is at least 30 ) is the correct answer.

The curvature of the function y = x³ at x = 1 is 6

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

y = x³

To start with, we need to differentiate the function

So, we have

y' = 3x²

Next, we differentiate the function for the second time to calculate the curvature of the function

So, we have

y'' = 6x

The value of x is 1

So, we have

y'' = 6(1)

Evaluate

y'' = 6

Hence, the curvature of the function y = x³ at x = 1 is 6

See attachment for the graph

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The required differential equation of the tractrix is dy/dx = (2y - y²)/s. This is obtained by substituting s = x/cos x into the differential equation 2y dy/dx - y² = 0 and simplifying. The differential equation dy/dx = (2y - y^2)/s is solved as shown below:

Solving the given differential equation In the given figure, let (x, y) be the coordinates of the point P. The weight is at (0, s) and the length of the rope is s. At a point P(x, y) on the tractrix, the tangent to the curve is parallel to the x-axis.The slope of the tangent is given by the differential coefficient dy/dx. We can determine dy/dx in terms of x and y by differentiating y^2 + x^2 = s^2 using implicit differentiation to get 2y dy/dx + 2x = 0.Differentiating again, we have d²y/dx² + y = 0.This differential equation is a second-order linear differential equation, with characteristic equation r² + 1 = 0. This yields r = ±i. Therefore, the general solution is y = c1cos x + c2 sin x, where c1 and c2 are constants. To find c1 and c2, we use the given initial condition that y = s when x = 0. This gives c1 = s and c2 = 0. Therefore, the solution to the differential equation is: y = s cos x. We can now eliminate x from the equation x² + y² = s² by substituting y = s cos x to obtain x² + s² cos² x = s². Solving for s, we have:s = x/cos x. Substituting this expression into the differential equation 2y dy/dx - y² = 0 yields the following equation:dy/dx = (2y - y²)/s This is the required differential equation of the tractrix.

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Answer:

772 hamburgers sold on monday

Step-by-step explanation:

add the problem

Answer:

Step-by-step explanation:

B

x is approximately equal to 6.

The ratios of the sides of a right triangle are called trigonometric ratios.

Three common trigonometric ratios are the sine (sin), cosine (cos), and tangent (tan).

sin = Perpendicular/ Hypotenuse

Cos = Base / Hypotenuse

tan = Perpendicular/Base

In the figure attached with the answer we can see that in Triangle ABC ,

tan 50.1 = x / 5

5(tan 50.1) = x

Therefore x is approximately equal to 6.

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Answer:

The answer is 16.99

Answer:

It's 17!

Step-by-step explanation:

The preparation of the stockholders' equity statement for the year ended December 31, 2021, is as follows:

Court Casuals

Statement of stockholers' equity

December 31, 2021

Common Stock                        $116,000

Additional Paid-in Capital  $5,326,000

Retained Earnings              $3,677,000

Treasury Stock                         $-2,000

Total stockholders' equity  $9,117,000

The stockholders' equity statement includes the common stock, additional paid-in capital, retained earnings, and the subtraction of the treasury stock.

Court Casuals

Stockholers' equity on January 1, 2021

Common Stock $90,000

Additional Paid-in Capital $4,100,000

Retained Earnings $3,000,000

Net income for the year, 2021, = $900,000

May 18: Cash $1,300,000 Common Stock $26,000 Additional Paid-in Capital $1,274,000 (26,000 x $50 - $26,000)

May 31: Treasury Stock $4,500 Additional Paid-in Capital $175,500 (4,500 x $40 - $4,500) Cash $180,000

Jul 1: Cash Dividend $223,000 (90,000 + 26,000 - 4,500) x $2 Dividends Payable $223,000

July 31: Dividends Payable $223,000 Cash $223,000

August 18: Cash $130,000 Treasury Stock $2,500 Additional Paid-in Capital $127,500 (2,500 x $52 - $2,500)

Statement of Retained Earnings, December 31, 2021:

Beginning balance  $3,000,000

Net income                 $900,000

Dividends                     -223,000

Ending balance        $3,677,000

Beginning balance  $90,000

May 18: Cash              26,000

Ending balance       $116,000

Beginning balance  $4,100,000

May 18: Cash              1,274,000

May 31: Cash                -175,500

August 18: Cash            127,500

Ending balance      $5,326,000

May 31: Cash                 $4,500

August 18: Cash             -2,500

Ending balance            $2,000

Thus, we can summarize from the stockholders' equity that Court Casuals has outstanding shares of 114,000 (116,000 - 2,000) at $1 par, which is the difference between the ending balances of the common stock and the treasury stock accounts, after reflecting the equity transactions for the year.

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Multiple linear regression model is used to fit the observed data, in order to characterize the relationship between the response variable y and i covariates of interest.

The following regression model is used:

                [tex]y=Xβ+ ε[/tex]

where:y denotes the n × 1 response vector.

[tex]β[/tex]represents the [tex]p × 1[/tex]parameter vector consisting of [tex]p = k + 1[/tex] regression coefficients.

[tex]X[/tex]denotes the n × p model matrix.

[tex]ε[/tex] denotes the [tex]n × 1[/tex] vector of error terms.

The entries of the first column of X are all equal to 1.

The entries of other columns of X correspond to the i covariates of interest.

It is given that the model matrix X is an n × p full-column-rank matrix.

The least squares estimator of [tex]β[/tex] is given by:

                     [tex]β^ = (X'X)^-1X'y[/tex]

The error terms are assumed to be independent and identically distributed normal random variables.

The variance-covariance matrix of the least squares estimator is given by:

                   [tex]Var(β^) = σ^2(X'X)^-1[/tex]

It is given that all covariates have the same variance-covariance structure.

Hence,

           [tex]σ^2 = σ0^2[/tex] for every i.

It is also given that a, σ0 for every i, where a represents a positive constant.

Hence,

       [tex]σ^2 = σ0^2[/tex]

             = [tex]a^2[/tex]

Show that Var and the equality would be attained if

       [tex]X'X = a^2I[/tex],

where [tex]β^[/tex] represents the ith entry of the least square estimator [tex]β[/tex].

From the given data, the variance-covariance matrix of the least squares estimator is given by:

     [tex]Var(β^) = σ^2(X'X)^-1[/tex]

                  [tex]= (a^2/n)(X'X)^-1[/tex]

It is given that all covariates have the same variance-covariance structure.

Hence,

             [tex]σ^2 = σ0^2[/tex] for every i.

It is also given that a, σ0 for every i, where a represents a positive constant.

Hence,

           [tex]σ^2 = σ0^2[/tex]

                    [tex]= a^23[/tex]

Hence,

               [tex]Var(β^) = (a^2/n)(X'X)^-1[/tex]

Now, let the diagonal entries of (X'X) be d1, d2, ..., dp.

Hence,

         (X'X) = [dij]

i=1,2,...,p;

X¹ = [0, 0, ..., 1, ..., 0]'

Let X¹ denote the ith column of X.

Hence, X¹ is given by:

                                 [tex]X¹ = [0, 0, ..., 1, ..., 0]'[/tex]

where 1 is in the ith position.

Hence, the ith diagonal entry of X'X is given by:

              [tex](X'X)ii = Σj(Xj¹)^2[/tex]

where the sum is over all i.

From the given data, the entries of the first column of X are all equal to 1.

Hence, [tex]X1¹ = [1, 1, ..., 1]'.[/tex]

Hence, [tex](X'X)ij = nai[/tex] and

[tex](X'X)ij = nai[/tex] for [tex]i ≠ j.[/tex]

Hence,[tex](X'X) = a^2I + n11'[/tex]

The inverse of (X'X) is given by:

             [tex](X'X)^-1 = (1/n)(I - (1/n)a^-2(1 1'))[/tex]

Hence,

     [tex]Var(β^) = (a^2/n)(X'X)^-1[/tex]

                     =[tex]a^2[(1/n)(I - (1/n)a^-2(1 1'))][/tex]

The variance of the ith entry of the least square estimator is given by:

 [tex]Var(β^i) = ai^2[(1/n)(I - (1/n)a^-2(1 1'))]ii[/tex]

Hence,

           [tex]Var(β^i)[/tex]= [tex]ai^2[(1/n)(1 - (1/n)a^-2)][/tex]

                          = [tex]a^2/n[/tex]

Therefore, the variance of the ith entry of the least square estimator is given by:

        [tex]Var(β^i) = a^2/n[/tex]

The equality would be attained if

          [tex]X'X = a^2I,[/tex]

where β^i represents the ith entry of the least square estimator β^. Therefore, the required result has been obtained.

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Answer:

4 horas para compatir com su familia

Answer: -32

Step-by-step explanation: -14 - 6 - 12 = -32

Answer:

Her results changed because the power increased due to the fact that she reduced variability in her data as a result of using a sample that had lesser variability.

Step-by-step explanation:

She is trying to find if there is difference between people who consumed 1 versus 2 beverages.

Now initially she surveyed 40 people between ages 18 to 38 years and of huge weights and found out there was no difference. However, she decided to run the survey again on only women aged 18 to 22 years with normal body weight.

Since she used a lesser sample size and surveyed only women, it means there will be lesser variability in the result as the sample is smaller and streamlined to only women.

Thus, her result changed because the power increased due to the fact that she reduced variability in her data as a result of using a sample that had lesser variability.

(a) To prove that there exists a neighborhood W of the origin such that f(x) ≤ y for all x ∈ W, given that f is bounded above on a neighborhood V of the origin, we can use the linearity of f.

Since f is a linear function, it satisfies the following properties:

f(0) = 0

f(rx) = rf(x) for any scalar r and vector x

f(x + y) = f(x) + f(y) for any vectors x and y

Given that f is bounded above on V, there exists a positive number M such that f(x) ≤ M for all x ∈ V. Now, let's consider the neighborhood W defined as follows:

W = {x ∈ X | ||x|| < M}

We claim that for any x ∈ W, f(x) ≤ y.

Let x ∈ W. Since x is in the neighborhood W, we have ||x|| < M. By linearity, we can express x as x = rx' for some scalar r and vector x' with ||x'|| = 1.

Now, consider f(x):

f(x) = f(rx') = rf(x')

Since ||x'|| = 1, we have ||rx'|| = |r| ||x'|| = |r|.

Therefore, ||rx'|| < M implies |r| < M.

Using the fact that f is bounded above on V, we have f(x') ≤ M.

Combining these results, we get:

|f(x)| = |rf(x')| = |r| |f(x')| ≤ M

Since this inequality holds for any x ∈ W and |r| < M, we have shown that f(x) ≤ y for all x ∈ W, where W is a neighborhood of the origin.

(b) To prove that f is continuous, we can show that f is bounded above on any compact set in X. Let K be a compact set in X.

Since K is compact, it is also closed and bounded. By the linearity of f, we have:

f(K) = {f(x) | x ∈ K}

Since K is bounded, there exists a positive number M such that ||x|| ≤ M for all x ∈ K. By the linearity of f, we have:

f(K) = {f(x) | x ∈ K} ⊆ {f(x) | ||x|| ≤ M}

Thus, f(K) is bounded above by M.

By the previous result in part (a), if f is bounded above on a neighborhood of the origin, then it is bounded above on any neighborhood of the origin. Therefore, f is bounded above on the neighborhood V of the origin.

Since K is compact, it can be covered by finitely many neighborhoods of the origin, say V1, V2, ..., Vk. Thus, f is bounded above on each Vi, i = 1, 2, ..., k.

Now, consider the open cover {V1, V2, ..., Vk} of K. By compactness, there exists a finite subcover {V1, V2, ..., Vm}. Therefore, f is bounded above on K.

Since f is bounded above on any compact set K, it follows that f is continuous.

(c) The previous part (b) already proves that if f is bounded above on any compact set, it is continuous. Therefore, if f is bounded above on a set 2 with int(2) ≠ Ø (i.e., the interior of 2 is not empty), then f is continuous.

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9A:

The problem says 35% of the 3560 applications are from boys who lived in other states.

This can be expressed as:

3560*35% = 3560*0.35 = 1246 applications.

9B:

The problem says applications to the university (3560 applications) represented 40% of all applications.

This can be expressed as:

3560 = 40% * A, where A = number of applications received in all.

To find how many applications received in all, just solve for A.

A = 3560 / 0.4 = 8900 applications.

The key to these types of problems is that "of" signals multiplication. For example, 40% of all applications is 40% * all applications.

Answer:

B

Step-by-step explanation:

Answer:

v^ 3 = 1000

so v = cube root of 1000

    v = 10

The Laplace transform of the function f(t) = t sin(4t) +1 is

L[f(t)] = (4 / (s^2 + 16) + 1) / s

To find the Laplace transform of the function f(t) = t sin(4t) + 1, we can use the linearity property of the Laplace transform.

The Laplace transform of t sin(4t) can be found using the derivative property and the Laplace transform of sin(4t). The derivative property states that if F(s) is the Laplace transform of f(t), then sF(s) is the Laplace transform of f'(t).

Taking the derivative of t sin(4t) with respect to t, we get:

f'(t) = 1⋅sin(4t) + t⋅(4⋅cos(4t))

Now, we can find the Laplace transform of f'(t) using the derivative property:

L[f'(t)] = sL[f(t)] - f(0)

Since f(0) = 0, the Laplace transform becomes:

L[t sin(4t)] = sL[t sin(4t) + 1]

Next, we need to find the Laplace transform of sin(4t). The Laplace transform of sin(at) is [tex]a / (s^2 + a^2)[/tex]. Therefore, the Laplace transform of sin(4t) is [tex]4 / (s^2 + 16).[/tex]

Now, substituting these values into the equation, we have:

sL[t sin(4t)] - 0 = sL[t sin(4t) + 1]

Simplifying, we get:

sL[t sin(4t)] = sL[t sin(4t) + 1]

Dividing both sides by s, we have:

L[t sin(4t)] = L[t sin(4t) + 1] / s

Now, we can substitute the Laplace transform of sin(4t) and simplify further:

[tex]L[t sin(4t)] = (4 / (s^2 + 16) + 1) / s[/tex]

Therefore, the Laplace transform is: [tex]L[f(t)] = (4 / (s^2 + 16) + 1) / s[/tex]

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Answer:

I got 38 and 11

Step-by-step explanation:

Answer:

A- In one simulated SRS of 100 students, the average GPA was  = 3.45.

Step-by-step explanation:

Just took the assignment myself, but here's a bit of insight to help anyone later on!

The x-bar means sample statistic. Since it was one sample taken out of the dot plot, 3.45 is a sample statistic.