Brian is remodeling his standy. He bought a desk for $437.50, a chair for $92.75, a filing cabinet for $99.99, and a bookshelf for $309.75. He bought everything at the same store and paid a 7 percent sales tax. How much did he spend on his study? ​

Answer: cinq minutes

Step-by-step explanation: 1h est egalle à 60minutes. Alors I'll a couru 12km Dan's 60minute. La duration pour 1km sera 60÷12=5 minutes

Answer:

1. 70°

2. 42°

Step-by-step explanation:

they are both isosceles Triangles as both diagrams state that they have two sides the same length. therefore their base angles will be equal.

1. 55 + 55 = 110

180 - 110 = 70°

2. 180 - 96 = 84

84/2 = 42°

The provided sentence "Five less than the quotient of eighteen" is converted into an algebraic expression as 18/x - 5.

Five less than the quotient of eighteen and a number

We must convert the provided sentence into an algebraic expression.

Let "x" represent the number.

Therefore,

Five less than that of the quotient of eighteen and "x"

The term quotient refers to the outcome of division.

Therefore,

18/x - 5

That is, Five less than the quotient of eighteen and x.

As a result, the provided sentence is converted into an algebraic expression.

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Check the picture below.

Answer:

D. (A U C) ∩ B

Step-by-step explanation:

12.

A = {-2, -1, 0, 1}

B = {0, 1, 2, 3}

C = {-2, 0, 2, 4}

A. A U B = {-2, 1, 0 , 1, 2, 3}

B. A U (C ∩ B} = A U {0, 2} = {2, 1, 0, 1, 2}

C. B ∩ C = {0, 2}

D. (A U C) ∩ B = {-2, -1, 0, 1, 2, 4} ∩ {0, 1, 2, 3} = {0, 1, 2}

Maximum = n/n+1 and Minimum = 1/n+1

What is the uniform distribution?

Probability distributions with uniform distributions have outcomes that are all equitably likely. Results are discrete and have the same probability in a discrete uniform distribution. Results are continuous and infinite in a continuous uniform distribution. Data near the mean occur more frequently in a normal distribution.

Here, we have

Given: X1, X2,..., Xn is independent and identically distributed random variables having uniform distributions over (0,1).

a) Z = max{X₁, X₂....Xₙ}

Since Z is maximum so it is greater than X1, X2...Xn so cdf of Z will be

F(Z) = P(Z≤z) = P(X₁, X₂....Xₙ≤z) = P(X₁≤z, X₂≤z.....Xₙ≤z)

= P(X₁≤z)P(X₂≤z)....P(Xₙ≤z) = Fₓ(z)Fₓ(z).....Fₓ(z)

F(Z) = zⁿ

So pdf of Z is

F(z) = F'(z) = nzⁿ⁻¹

Expectation of Z is

F(Z) = \(\int\limits^0_1 {zf_Z(z)} \, dz\) = \(n\int\limits^0_1 {} \,\)zⁿdz

= n[zⁿ⁺¹/n+1]₀¹ = n/n+1

b) Let Y = min{X₁, X₂....Xₙ}

Since Y is the minimum so it is less than X1, X2...Xn so the cdf of Y will be

F(Y) = P(Y≤y) = 1 - P(Y>y) = 1- P(X₁, X₂....Xₙ≤z) = 1- P(X₁≤z, X₂≤z.....Xₙ>y)

= 1- P(X₁>y)P(X₂>y)....P(Xₙ>y) = 1- Fₓ(y)Fₓ(y).....Fₓ(y)

= 1 - [1-y]ⁿ

So pdf of Y is

F(y) = F'(y) = n[1-y]ⁿ⁻¹

The expectation of Y is

E(Y) = \(\int\limits^0_1 {yf_Y(y)} \, dy\) = ∫₀¹ yn(1-y)ⁿ⁻¹dy = 1/n+1

Hence, Maximum = n/n+1 and Minimum = 1/n+1

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The area bounded by the first quadrant loop of the curve x^5 + y^5 = 3xy is approximately 0.536 square units.

To find the area bounded by the curve x^5 + y^5 = 3xy in the first quadrant, we can use the double integral. However, this particular curve is quite complicated to work with directly. Instead, we can use a change of variables to simplify the equation.

Let's make the substitution u = x^5 and v = y^5. Then, we can express the curve equation in terms of u and v:

u + v = 3uv

This is a much simpler equation to work with. Now, let's find the limits of integration for u and v. Since we are considering the first quadrant, both u and v must be positive. From the original equation, we can see that when x = 0, y = 0, and when y = 0, x = 0. Therefore, the limits of integration for u and v are both from 0 to 1.

Now, we can calculate the area using the double integral:

A = ∬R dA

A = ∫∫R du dv

A = ∫[0,1] ∫[0,1] du dv

A = ∫[0,1] u=0 to 1 v=0 to 1 du dv

A = ∫[0,1] (v/2 + v^2/3) u=0 to 1 dv

A = ∫[0,1] (1/2 + v/3) dv

A = (1/2)v + (1/6)v^2 from 0 to 1

A = (1/2)(1) + (1/6)(1^2) - (1/2)(0) - (1/6)(0^2)

A = 1/2 + 1/6

A = 3/6 + 1/6

A = 4/6

A ≈ 0.667 square units

Therefore, the area bounded by the first quadrant loop of the curve x^5 + y^5 = 3xy is approximately 0.667 square units.

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Horizontal translation: 2 units right.

Vertical translation: 5 units up

Stretch/compression: stretched by a factor of 3.

Reflection: not reflected.

Answer:

164.93 I believe is the answer

Beta had a higher percentage of listeners, higher mean number of hours listening and had a higher range of number of hours listening.

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Answer:

$3,550 dollars

Step-by-step explanation:

6.4% off savings

- a total of $2,400

The percentage is that indicating the hundredth total amount in the saving account before withdrawing such that 6.4% of that is $2400 is $37500.

The percentage is defined as a given amount in every hundred. It is a fraction with 100 as the denominator percentage is represented by the one symbol %.

Let's suppose the total amount of money before withdrawing x.

It is known that x% of y is given as, (x/100)y.

Therefore,6.4% of x is (6.4/100)x = 0.064x

The 6.4% of the x is 2400

So, 0.064x = 2400

x = 2400/0.064 = $37500

Hence "The percentage is that indicating the hundredth total amount in the saving account before withdrawing such that 6.4% of that is $2400 is $37500.".

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Answer:

See attached

Step-by-step explanation:

Refer to attachment

One relation is a linear function

Second one is a graph of non-functional relation

Answer:

below

Step-by-step explanation:

For the first graph, we can use a linear function as it means "straight" lines so when we use the vertical lines test, it passes it (shown below).

We can use a U shaped line so it when we use the vertical line test, it passes through two points to resemble its not a function (shown below).

Best of Luck!

Answer:

(B) 10

Step-by-step explanation:

Rate of change is basically slope

So the slope here is 10/1 so the rate of change is 10

Answer:

(B) 10

Step-by-step explanation:

P1 = (4, 40)

P2 = (0, 0)

m = (y2 -y1) / (x2 - x1)

= 40/4

= 10

(a) To see the graph.

(b) B: A = \(A_{2}C_{2} : A_{1}C_{1}\) ≈ 6mm : zmm = 3.

(c) A: B = \(A_{1}C_{1} :A_{2}C_{2}\) ≈ z mm : 6mm = 1/3.

(d) B : C = \(A_{2}C_{2}:A_{3}C_{3}\) ≈ 6mm : 10mm = 3/5.

Given,

In the question:

Triangles B and C have been built by dilating Triangle A. ABC

(a) To find the the center of dilation.  

Now, According to the question:

Point O is the center of dilation .

{When you look the center of dilation of any similar figure, you can use the method of connecting their vertices, where the lines meet is the center.}

Using a ruler to measure the scale of similar line segments on the original graph is the scale factor between each graph.

In the below, To see the attachment.

(b)To find the Triangle B is a dilation of A with approximately what scale factor  ?

To see the attachment :

B: A = \(A_{2}C_{2} : A_{1}C_{1}\) ≈ 6mm : zmm = 3

(c) To find the Triangle A is a dilation of B with approximately what scale factor?  

To see the attachment:

A: B = \(A_{1}C_{1} :A_{2}C_{2}\) ≈ z mm : 6mm = 1/3

(d) To find the Triangle B is a dilation of C with approximately what scale factor?

To see the attachment:

B : C = \(A_{2}C_{2}:A_{3}C_{3}\) ≈ 6mm : 10mm = 3/5

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Answer:

3200/4500=71.1%

so a 29.9% drop

Hope This Helps!!!

Answer:

40 + 6x = 3

Step-by-step explanation:

Multiple all the way through by 8

8( 5 + \(\frac{3}{4}\) x) = (\(\frac{3}{8}\))8

8(5) + \(\frac{8}{1}\)\((\frac{3}{4})\)x = \(\frac{3}{8}\)\((\frac{8}{1})\)

40 + 6x = 3

Answer:

55

Step-by-step explanation:

6m+90=96m

A ranked frequency distribution of data can be created by sorting the data in ascending or descending order and then counting the frequency of each value.

The given data set is 245, 261, 289, 222, 291, 289, 240, 233, 249, and 200. To create a ranked frequency distribution of this data set, we first need to sort it in ascending or descending order. Let's sort it in ascending order:200, 222, 233, 240, 245, 249, 261, 289, 289, 291 Next, we need to count the frequency of each value. We can do this by going through the data set and counting how many times each value occurs. Here is the frequency distribution table:Value Frequency 200 1222 1233 1240 1245 1249 1261 1289 2291 1 From this table, we can see that the most frequent value is 289, which occurs twice. We can also see that the least frequent values are 200, 222, 233, and 240, which each occur only once.

In conclusion, a ranked frequency distribution of data can be created by sorting the data in ascending or descending order and then counting the frequency of each value. This allows us to see which values are most and least frequent in the data set.

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Based on the AA Similarity Theorem, triangles PQR and TSR are proven to be similar to each other because they have two pairs of corresponding angles that are congruent to each other.

The AA (Angle-Angle) Similarity Theorem states that if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. This means that the corresponding sides of the triangles are in proportion to each other.

With the information given, the proof that shows that the two triangles are similar as as follows:

Statements                                         Reasons                                        

1. ∠QPR ≅ ∠STR                                1. Given

2. QR ⊥ PT                                         2. Given

3. ∠QRP ≅ ∠SRT                               3. def. of perpendicular

4. ΔPQR ~ ΔTSR                                4. AA Similarity Theorem

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The matrices A and B are given as:

A = [3/4 0]
[0 3/4]

B = [-4 0]
[0 -4]

2a) Compute AB:

AB = [3/4 0] * [-4 0]
[0 3/4] [0 -4]

= [3/4 * -4 + 0 * 0 3/4 * 0 + 0 * -4]
[0 * -4 + 3/4 * 0 0 * 0 + 3/4 * -4]

= [-3 0]
[0 -3]

2b) Compute BA:

BA = [-4 0] * [3/4 0]
[0 -4] [0 3/4]

= [-4 * 3/4 + 0 * 0 -4 * 0 + 0 * -4]
[0 * 3/4 + -4 * 0 0 * 0 + -4 * 3/4]

= [-3 0]
[0 -3]

2c) How did your answer in part (a) and (b) compare?

The answers in part (a) and (b) are the same. Both AB and BA resulted in the matrix [-3 0] [0 -3].

2d) Will this be true, in general, for any two matrices when you multiply them (assuming their dimensions line up so that they may be multiplied)? If so, explain your reasoning. If not, show an example of two matrices C and D such that CD≠DC.

No, this will not be true in general for any two matrices when you multiply them. The order in which matrices are multiplied matters, and in most cases, AB≠BA. Here is an example of two matrices C and D such that CD≠DC:

C = [1 2]
[3 4]

D = [5 6]
[7 8]

CD = [1 * 5 + 2 * 7 1 * 6 + 2 * 8]
[3 * 5 + 4 * 7 3 * 6 + 4 * 8]

= [19 22]
[43 50]

DC = [5 * 1 + 6 * 3 5 * 2 + 6 * 4]
[7 * 1 + 8 * 3 7 * 2 + 8 * 4]

= [23 34]
[31 50]

As you can see, CD≠DC.

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Answer:

A

Step-by-step explanation:

Edge

Answer:

4 1/3 is the same as 11 decreased by a number.

Step-by-step explanation:

Answer: The picture is the answer to your question

Answer:

4

Step-by-step explanation:

\((20-2^{2} )\) ÷ \((6-2)\)

(20-4) ÷ 4

16 ÷ 4=4

Answer:

2

Step-by-step explanation:

Here we're "evaluating," not solving, the given expression.

Substitute -2 for x and 6 for y:

20 - (-2)^2      20 - 4

--------------- = -------------- = 16/8 = 2

(6) - (-2)              8

The value of this expression, when x = -2 and y = 6, is 2.

Answer:

(1,3)

Step-by-step explanation:

:)

Answer:

C. 0.75

Step-by-step explanation:

P(A or B) = P(A) + P(B) - P(A and B)

P(A or B) = 0.60 + 0.30 - 0.15

P(A or B) = 0.75

The area of the shaded region in this problem is given as follows:

995.44 cm².

The area of a circle of radius r is given by the multiplication of π and the radius squared, as follows:

A = πr²

The radius of a circle represents the distance between the center of the circle and a point on the circumference of the circle, hence it's measure is given as follows:

r = 21 cm.

Then the area of the entire circle is given as follows:

A = π x 21²

A = 1385.44 cm².

The right triangle has two sides of length 39 cm and 20 cm, hence it's area is given as follows:

A = 0.5 x 39 x 10

A = 390 cm².

Then the area of the shaded region is given as follows:

1385.44 - 390 = 995.44 cm².

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Answer:

13) A,B are the diameter of the circle - the diameter is the distance ACROSS the circle and since the line between A and B is across the circle it is the diameter.

14) It's giving you a circular wheel and make a drawing following the letters in the wheel, I think it means make a triangle following those letters.

Hope this helps!

A. T is an equivalence relation, satisfying reflexivity, symmetry, and transitivity.

B. U is not necessarily an equivalence relation, as it may fail to satisfy transitivity in certain cases.

A. To show that T is an equivalence relation, we need to demonstrate that it satisfies three properties: reflexivity, symmetry, and transitivity.

Reflexivity: For any element x in set A, we need to show that x\(T_x\) holds.

Since R and S are both equivalence relations on A, we know that xRx and x\(S_x\) hold for any x in A. Therefore, the conjunction of x\(R_y\) and x\(S_y\) is true when y is also equal to x. Hence, x\(T_x\) holds, and T is reflexive.

Symmetry: For any elements x and y in set A, if x\(T_y\) holds, then we need to show that y\(T_x\) also holds.

If x\(T_y\) holds, it means that x\(R_y\) and x\(S_y\) are both true. Since R and S are equivalence relations, we know that y\(R_x\) and y\(S_x\) are also true. Therefore, the conjunction of yRx and y\(S_x\) holds, which implies y\(T_x\). Hence, T is symmetric.

Transitivity: For any elements x, y, and z in set A, if x\(T_y\) and y\(T_z\) hold, then we need to show that x\(T_z\) also holds.

If x\(T_y\) and y\(T_z\) hold, it means that x\(R_y\), x\(S_y\), y\(R_z\), and y\(S_z\) are all true. Since R and S are equivalence relations, we can conclude that x\(R_z\) and x\(S_z\) are true. Therefore, the conjunction of x\(R_z\) and x\(S_z\) holds, which implies x\(T_z\). Hence, T is transitive.

Since T satisfies all three properties of an equivalence relation (reflexivity, symmetry, and transitivity), we can conclude that T is indeed an equivalence relation on set A.

B. To show that U is not necessarily an equivalence relation, we need to provide an example where U fails to satisfy one or more of the properties: reflexivity, symmetry, or transitivity.

Let's consider an example where R and S are equivalence relations on set A, but U fails to be an equivalence relation:

Suppose A = {1, 2, 3}, and we define R and S as follows:

R = {(1, 1), (2, 2), (3, 3)}

S = {(1, 2), (2, 1)}

In this case, R is the identity relation on A, and S is a symmetric relation.

Now, let's examine U using the definition x\(U_y\) ⇐⇒ x\(R_y\) ∨ x\(S_y\):

1\(U_1\): Since (1, 1) is in R, we have 1\(U_1\).

1\(U_2\): Since (1, 2) is in S, we have 1\(U_2\).

2\(U_1\): Since (2, 1) is in S, we have 2\(U_1\).

2\(U_2\): Since (2, 2) is in R, we have 2\(U_2\).

3\(U_3\): Since (3, 3) is in R, we have 3\(U_3\).

However, U fails to be transitive:

1\(U_2\) and 2\(U_1\) holds, but 1\(U_1\) does not hold.

Since U fails to satisfy transitivity, it is not an equivalence relation in this example. Therefore, we have demonstrated that U is not necessarily an equivalence relation.

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