Let $n$ be a positive integer. Let $r$ be the remainder when $n^2$ is divided by $n 4.$ How many different values can $r$ take on

a) The variables are the variable unit cost of candles and the fixed cost of shipping.

b) The information given in the question include the total costs for 4 candles and 12 candles, including their shipping costs.

c) An equation representing the situation is given as

d) Allie can get 18 candles for $60.

An equation is a mathematical statement showing that two or more algebraic expressions are equal or equivalent.

Algebraic expressions combine variables, values, constants, and numbers without the equal symbol (=), but they use the mathematical operands.

The total cost for ordering 4 candles = $35

The total cost for ordering 12 candles = $49

The difference = 8 candles = $14

Unit cost of candles = $1.75 ($14 ÷ 8)

Shipping (fixed) cost for the first order = $28 ($35 - $1.75 x 4)

Shipping (fixed) cost for the second order = $28 ($49 - $1.75 x 12)

For ordering candles for $60, the number of candles Allie can get = 18 ($60 - $28) ÷ $1.75

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Answer:

D.

Step-by-step explanation:

20/40 = 1/2

40/40 = 1

Difference between 20 and 36 = 16

Difference between 36 and 40 = 4

Therefore, the difference between 36/40 and 40/40 is smaller.

Answer:

:Both Tim and Jasmine calculated the volume properly. Tim gave the exact answer with pi in it, whereas Jasmine multiplied by pi and then rounded

Step-by-step explanation:

Answer:  Both Tim and Jasmine calculated the volume properly. Tim gave the exact answer with pi in it, whereas Jasmine multiplied by pi and then rounded to the nearest hundreth.

Step-by-step explanation:

Answer:

Step-by-step explanation:

(9a⁴b⁷) x (6a³b⁸)

(9 x 6) x (a⁴ x a³) x (b⁷ x b⁸)

54 x a⁷ x b¹⁵ = 54a⁷b¹⁵

Answer:

(9a⁴b⁷).(6a³b⁸)

=54a⁷b¹⁵

The interval in which the solution of the given initial value problem is certain to exist is t > 3. We can use the existence and uniqueness theorem for first-order differential equations.

The existence and uniqueness theorem states that if a differential equation is of the form y' = f(t, y) and the function f(t, y) and its partial derivative with respect to y, ∂f/∂y, are continuous on some rectangular region R = {(t, y) : |t - t0| ≤ a, |y - y0| ≤ b}, where (t0, y0) is the initial point, then there exists a unique solution y(t) that passes through the initial point and is defined on an interval containing t0.

In this case, the given differential equation is:

(t − 3)y′ (ln t)y = 2t

Comparing with y' = f(t, y), we have:

f(t, y) = (2t)/((t - 3)ln t)y

To apply the existence and uniqueness theorem, we need to check that f(t, y) and its partial derivative with respect to y, ∂f/∂y, are continuous on some rectangular region R that contains the initial point (1, 2).

Since t - 3 and ln t are both continuous functions for t > 0, f(t, y) is continuous for t > 0 and y ≠ 0. To check the continuity of ∂f/∂y, we differentiate f(t, y) with respect to y:

∂f/∂y = (2t)/((t - 3)ln t)(ln t)

Since ln t > 0 for t > 0, we have:

|(2t)/((t - 3)ln t)(ln t)| ≤ 2/(t - 3)

Therefore, ∂f/∂y is also continuous for t > 3.

Thus, by the existence and uniqueness theorem, there exists a unique solution to the given initial value problem that passes through the initial point (1, 2) and is defined on an interval containing 1 as long as t > 3.

Therefore, the interval in which the solution of the given initial value problem is certain to exist is t > 3.

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Answer:

One solution

Step-by-step explanation:

I added a photo of my solution

Answer:

The system has one solution: (0, 4).

The box measures 100.79 cm by 100.79 cm by 50.40 cm in size, which minimizes the quantity of material used.

Given,

Let the side of the square base be x

h be the height of the box

Volume V = x²h

512000 = x²h

h = 512000/x² ..... 1

Surface area =  x² + 2xh + 2xh

Surface area S =  x² + 4xh  ...... 2

Substitute 1 into 2;

From 2;

S =  x² + 4xh

S = x² + 4x(512000/x²)

S = x² + 2048000/x

To minimize the amount of material used;

dS/dx = 0

dS/dx = 2x - 2048000/x²

0 = 2x - 2048000/x²

0 = 2x³ - 2048000

2x³ = 2048000

x³ = 1024000

x = ∛1024000

x = 100.79 cm

Since Volume = x²h

512000 = 100.79²h

h = 512000/10158.62

h = 50.40 cm

Hence the dimensions of the box that minimize the amount of material used is 100.79 cm by 100.79 cm by 50.40 cm

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Gender and marital status and Job title.

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Answer:

x(4)+y(1.25)=500

Step-by-step explanation:

x=popcorn bags and y=cans of juice

$ 4 is the price for popcorn bags and $1.25 for the cans of juice

so just multiply the price of each item by the number or in this case symbol, that equals the total of what is needed

Answer:

below

Step-by-step explanation:

P (at least one defective) = 1 - P(all calculators work)

=  

1

−

(

30

46

×

29

45

×

28

44

×

27

43

)

=

1

−

1827

10879

=  

9052

10879

≈

0.832

⇒F(-5) just simply means that in the function F in the place of x you have to plug in -5 and simplify if possible.

\(f(-5)= -(-5)^{2} +9(-5)-17\\f(-5)=-(25)-45-17\\f(-5)=-25-45-17\\f(-5)=-70-17\\f(-5)=-87\)

the answer to f(-5) is -87

GOODLUCK!!

Given

in ∆ABC

AB = 5cm

BC = 6 cm

AC = 7 cm

now,

as we have to make ∆PQR inside it

than, possibly ∆PQR's sides length is,

PQ = ½ of CB (according to mid point theorem)

→ CB = 6cm → 6/2 → 3cm

PQ = 3cm

QR = ½ of AC

→ AC/2 = 7/2 = 3.5 cm

RP = ½ of AB

→ AB/2 = 5/2 = 2.5cm

therefore, length of

PQ = 3cm

QR = 3.5 cm

PR = 2.5 cm

Hope this answer helps you dear!

Answer:

$4.99

Step-by-step explanation:

Take twenty and subtract 5.03 from it, then divide by 3. You can use this equation, (20 - 5.03) /3.

Answer:

4.99$

Step-by-step explanation:

20 -  5.03 / 3 =4.99

Answer:

the standared deviation =2.87 so it is small

Step-by-step explanation:

to calculate the standard deviation of those numbers:

1- Work out the Mean (the simple average of the numbers)

(7+6+5+8+9+1+0+4+3+2)/10=4.5

Then for each number: subtract the Mean and square the result.

[(7 - 4.5)^2 + ... + (2 - 4.5)^2]/10=8.25

√8.25=2.87≅3

Answer:

A. Small

Step-by-step explanation:

Consequently, a restaurant pays $27.50 for 50 pounds of red potatoes and $22.50 for 30 pounds of sweet potatoes, with the red potatoes costing $0.485 more per pound.

The cost function's generic formula is C(x)=F+V. (x) F is the sum of all fixed costs, V is the sum of all variable costs, x is the number of units, and C(x) is the sum of all production costs.

given

50 pounds of red potatoes are purchased by a restaurant for $27.50.

The price of 30 pounds of sweet potatoes is $22.50,

hence the price per pound of red potatoes is = 50/27.50 = $1.818

while the price per pound of sweet potatoes is = 30/22.50 = $1.33 .

The red potatoes are therefore more expensive per pound by the amount of $1.818 - $1.33 = $0.485.

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The interior angles of the second triangle will be 50°, 55°, and 75° respectively.

What are similar triangles?

If the corresponding angles and sides of two triangles are congruent, then the two triangles are said to be similar.

In other words, similar triangles have the same shape but may differ in size.

Given, a triangle has interior angles of 50°, 55°, and 75° with side lengths of 3.1 m, 3.2 m, and 4.0 m.

And, a second triangle has side lengths of 10.3 m, 10.5 m, and 12.0 m.

The corresponding sides of both triangles are 3.1 m, 3.2 m, and 4.0 m with  10.3 m, 10.5 m, and 12.0 m respectively.

Then, the angles of both triangles are also corresponding to each other.

So we get, the interior angles of the second triangle will be 50°, 55°, and 75° respectively.

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a) By strong induction, any amount of postage worth 8 cents or more can be made from 3-cent or 5-cent stamps.

b) By strong induction, any amount of postage worth 24 cents or more can be made from 7-cent or 5-cent stamps.

c) By strong induction, any amount of postage worth 12 cents or more can be made from 3-cent or 7-cent stamps.

a. Prove that any amount of postage worth 8 cents or more can be made from 3-cent or 5-cent stamps.

Base case: For postage worth 8 cents, we can use two 4-cent stamps, which can be made using a combination of one 3-cent stamp and one 5-cent stamp.

Induction hypothesis: Assume that any amount of postage worth k cents or less, where k is greater than or equal to 8, can be made from 3-cent or 5-cent stamps.

Induction step: Consider any amount of postage worth (k+1) cents. Since k is greater than or equal to 8, we can use the induction hypothesis to make k cents using 3-cent or 5-cent stamps. Then, we can add one more stamp to make (k+1) cents. If the last stamp we added was a 3-cent stamp, we can replace it with a 5-cent stamp to get the same value. If the last stamp we added was a 5-cent stamp, we can replace it with two 3-cent stamps to get the same value. Therefore, any amount of postage worth (k+1) cents can be made from 3-cent or 5-cent stamps.

b. Prove that any amount of postage worth 24 cents or more can be made from 7-cent or 5-cent stamps.

Base case: For postage worth 24 cents, we can use three 8-cent stamps, which can be made using a combination of one 7-cent stamp and one 5-cent stamp.

Induction hypothesis: Assume that any amount of postage worth k cents or less, where k is greater than or equal to 24, can be made from 7-cent or 5-cent stamps.

Induction step: Consider any amount of postage worth (k+1) cents. Since k is greater than or equal to 24, we can use the induction hypothesis to make k cents using 7-cent or 5-cent stamps. Then, we can add one more stamp to make (k+1) cents. If the last stamp we added was a 5-cent stamp, we can replace it with two 7-cent stamps to get the same value. If the last stamp we added was a 7-cent stamp, we can replace it with three 5-cent stamps to get the same value. Therefore, any amount of postage worth (k+1) cents can be made from 7-cent or 5-cent stamps.

c. Prove that any amount of postage worth 12 cents or more can be made from 3-cent or 7-cent stamps.

Base case: For postage worth 12 cents, we can use one 3-cent stamp and three 3-cent stamps, which can be made using a combination of two 7-cent stamps.

Induction hypothesis: Assume that any amount of postage worth k cents or less, where k is greater than or equal to 12, can be made from 3-cent or 7-cent stamps.

Induction step: Consider any amount of postage worth (k+1) cents. Since k is greater than or equal to 12, we can use the induction hypothesis to make k cents using 3-cent or 7-cent stamps. Then, we can add one more stamp to make (k+1) cents. If the last stamp we added was a 3-cent stamp, we can replace it with two 7-cent stamps to get the same value. If the last stamp we added was a 7-cent stamp, we can replace it with one 3-cent stamp and two 7-cent stamps to get the same value. Therefore, any amount of postage worth (k+1) cents can be made from 3

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The correct answer is option 2. The radius of the balloon after 9 minutes is approximately 0.7889 centimeters.

To find the radius of the balloon after 9 minutes, we substitute t = 9 into the equation r(t) = 3√t + 6:

r(9) = 3√9 + 6

r(9) = 3 * 3 + 6

r(9) = 9 + 6

r(9) = 15

Therefore, the radius of the balloon after 9 minutes is 15 centimeters.

However, none of the given options match the value of 15 centimeters. It is possible that there was an error in the options provided or in the calculations. If we assume that option 2 corresponds to the radius after 9 minutes, the value of 0.7889 centimeters does not align with the equation r(9) = 15. Therefore, there may be a discrepancy in the options or in the calculations provided.

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To express the volume in cubic inches, we first need to convert from liters to cubic centimeters (\(cm^{3}\)) using the conversion 1l = 1000\(cm^{3}\).

Volume in \(cm^{3}\) = 5.2l x 1000\(cm^{3}\)/l = 5200\(cm^{3}\)

Next, we can use the conversion 1in = 2.54cm to convert from \(cm^{3}\) to cubic inches.

Volume in cubic inches = 5200\(cm^{3}\) x \((1 in/2.54cm)^3\)= 317.01 \(in^3\) (rounded to two decimal places)

Therefore, the volume is 317.01 cubic inches.

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Answer: Option (d).

Step-by-step explanation:

it is given that B is between A and C.

Using segment addition property, we get

\(AC=AB+BC\)

The given values are AB = 46, BC = 12x, and AC = 142.

Substitute these values in the above equation.

\(142=46+12x\)

\(142-46=12x\)

\(96=12x\)

\(\dfrac{96}{12}=x\)

\(8=x\)

\(x=8\)

Therefore, the correct option is (d).

Answer:

Step-by-step explanation:

Growing a culture of bacteria: This example can be modeled by an exponential function, as the growth of bacteria is typically rapid and continues to accelerate over time.The distance a jet can travel at a certain speed over a given period of time: This example can be modeled by a linear function, as the distance traveled by the jet is directly proportional to the time and speed.Kicking a ball into the air: This example can be modeled by a quadratic function, as the trajectory of the ball follows a parabolic path.Running a race at a constant speed: This example can be modeled by a linear function, as the distance covered is directly proportional to the time and speed.A pumpkin decaying: This example can be modeled by an exponential function, as the rate of decay is typically accelerated over time.Jumping from a high dive: This example can be modeled by a quadratic function, as the trajectory of the diver follows a parabolic path.

The Standard Error of mean is 217.71 .

The critical value , Z꜀= 1.96 < Z , it is then concluded that the null hypothesis is rejected.

Hence, we can say that the the mean earnings of men in the General Social Survey is not representative of the earnings of the entire U.S. male population.

We have given that

A researcher working with the General Social Survey (GSS) on a project focusing on income and health.

Sample size, N = 1500

Sample mean , X-bar =$31,607

male population mean , μ = $33,405

Standard deviations, σ = $8,432

Std. error of mean , S.E = σ /√N

= 8432/√1500

= 217.71

Consider null and alternative hypothesis as

H₀ : μ = 33405

Hₐ : μ not equal to 33,405

Rejection region: Based on information provided the significance level,alpha = 0.05 and the critical value for two tailed test , Z꜀ = 1.96.

The rejection region for two tailed test R is

{ z : |z|> 1.96}.

Test Statistic:

Z- statistic test formula,

Z = (X-bar - u)/std. error

=> Z = ( 31,607 - 33,405)/217.71

=> Z = - 8.259

Decision: Since, it is observed that |z| = 8.259 > Z꜀ = 1.96 , it is concluded that Null hypothesis is rejected .

Conclusion: It is concluded that Null hypothesis H₀ is rejected. So there is sufficient evidence to support the claim that the population mean ,u is different from 33,405 at α = 0.05 significance level.

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.

Answer:

coefficient is 1/6

Step-by-step explanation:

=-2a/3+5a/6-1/6

= a/6 -1/6

= 1/6(a -1)

coefficient is 1/6

Answer: \(-\frac{3}{10}\)

Step-by-step explanation:

Answer:

-3/10

Step-by-step explanation:

Step 1: Use this rule a/b x c/d = ac/bd

-3x2/4x5

Step 2: Simplify 3x2 to 6

-6/4x5

Step 3: Simplify 4x5 to 20

-6/20

Step 4: Simplify 6/20 to 3/10.

-3/10

The corresponding probabilities are: A. P(R) = 0.72 B. P(R') = 0.28 C. P(A ∩ R) = P(R) * P(A | R) = 0.72 * 0.51 = 0.367 D. P(A | R) = 0.51 E. P(A ∩ R') = P(R') * P(A | R') = 0.28 * 0.10 = 0.028 F. P(A' ∩ R) = P(R) * P(A' | R) = 0.72 * 0.49 = 0.352 G. P(A' ∩ R') = P(R') * P(A' | R') = 0.28 * 0.90 = 0.252

(a) The tree diagram with the corresponding probabilities is:

A (0.51) A' (0.49) A (0.10) A' (0.90)

(b) The proportion of students who earn an A and don't attend class regularly is:

P(A ∩ R') = 0.028

(c) The chance that a randomly chosen student will earn an A in the class is:

P(A) = P(A ∩ R) + P(A ∩ R')

= 0.367 + 0.028

= 0.395

(d) Given that a student earned an A, the chance they attend class regularly is:

P(R | A) = P(A ∩ R) / P(A)

= 0.367 / 0.395

≈ 0.9291 (rounded to four decimal places)

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Complete question:

A professor has noticed that students hat attend class regularly, mss no more than two classes per term, generally get better grades. For he class, the overall percent o students who attend regularly s 72% or those who come to class on a regular basis, 51% receive A's. Of those who don't attend regularly, only 10% get A's. (b)Among all students what proportion earn an A and don't attend class regularly?

The equilibrium point corresponding to the constant input u = 0 is (0,0). The other eigenvalue of the linearized system is -15.

The nonlinear system is given by:

x' = -ax + u

y' = ay

The equilibrium point corresponding to the constant input u = 0 is found by setting x' = y' = 0. This gives the equations:

-ax = 0

ay = 0

The first equation implies that x = 0. The second equation implies that y = 0. Therefore, the equilibrium point is (0,0).The linearized system around the equilibrium point is given by:

x' = -ax

y' = ay

The A matrix of the linearized system is given by:

A = [-a 0]

   [0 a]

The eigenvalues of A are given by the solutions to the equation:

|A - λI| = 0

This equation factors as:

(-a - λ)(a - λ) = 0

The solutions are λ = 0 and λ = -a. Since a = 15, the other eigenvalue is -15.

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A and B are typically subitized, which means that they can be immediately recognized without counting. C and D are more likely to be counted.

Counting is the process of determining the number of items in a set by successively adding one. Subitizing, on the other hand, is the ability to immediately recognize the number of items in a small set without counting. The capacity for subitizing is limited to small sets typically ranging from 1 to 5 items, depending on the individual's age and mathematical ability. The answer to this question would be option C, 3. For most individuals, recognizing the number of items in a set of 3 would require counting rather than subitizing, as 3 is at the upper limit of the typical range for subitizing.

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