Which set of ordered pairs (x, y) could represent a linear function?

If 25 percent like vanilla, 35 percent like swirl, 40 percent like chocolate, then 5 more people chose chocolate than swirl.

One hundred students were surveyed about their favorite kind of frozen yogurt.

Student like vanilla frozen yogurt = 25% = 0.25

Student like swirl frozen yogurt = 35% = 0.35

Student like chocolate frozen yogurt = 40% = 0.40

Now we determining the number of students who likes different kinds of frozen yogurt.

The number of student who like vanilla frozen yogurt = 100 × 0.25

The number of student who like vanilla frozen yogurt = 25

The number of student who like swirl frozen yogurt = 100 × 0.35

The number of student who like swirl frozen yogurt = 35

The number of student who like chocolate frozen yogurt = 100 × 0.40

The number of student who like chocolate frozen yogurt = 40

So the number of more people who chose chocolate than swirl = The number of student who like chocolate frozen yogurt - The number of student who like swirl frozen yogurt

The number of more people who chose chocolate than swirl = 40 - 35

The number of more people who chose chocolate than swirl = 5

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Answer:its 5 {:

Step-by-step explanation:

i just did it }:

The approximate perimeter of the entire shape, given that it is a one - half circle attached to a square, is 32 inches

First, find the perimeter of the one - half of a circle with the formula:

= ( π x diameter ) / 2

The diameter is the same as the side of the square which is:

= ( π x 7 ) / 2

= 11 inches

The perimeter of the square component will be for 3 sides alone as the last side is covered by the semi - circle :

= 3 x 7

= 21 inches

The perimeter is:

= 11 + 21

= 32 inches

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If the size of an animal population in year "t" is P(t) = (210t + 400)/(t + 2) , then the size of population in year 2004 is 200 thousands of individuals.

The Size of the animal population in year t is given by the equation P(t) = (210t + 400) / (t + 2) ;

For t = 0, which represents the year 2004, we have:

Substituting the value of t = 0 in P(t) , we get ;

⇒ P(0) = (210×0 + 400)/(0 + 2) = 400/2 = 200 .

Therefore ,  the size of the animal population in year 2004 is 200 thousands of individuals.

The given question is incomplete , the complete question is

The size of an animal population in year t is given by P(t)=(210t+400)/(t+2). Here t=0 represents the year 2004 and P(t) is the size of the population in thousands of individuals. Find the size of the population in year 2004 .

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The graph of the function y = -2x² is given by the image presented at the end of the answer.

The function for this problem is defined as follows;

y = -2x².

The parent function is y = x², hence the transformation is defined as follows:

As there was no horizontal/vertical movement, the vertex, intercept and axis of symmetry remain constant, as follows:

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The numeric values of the piecewise function are given as follows:

a) f(4) = -5.

b) f(-5) = -26.

c) f(2) = -7.

d) f(-2) = -11.

e) f(0) = -9.

f) f(-10) = -51.

To find the numeric value of a function or of an expression, we replace each instance of the variable in the function or in the expression by the value at which we want to find the numeric value.

A piecewise function is a function that has different definitions based on the input of the function.

The definitions for the function in this problem are given as follows:

Considering the first definition, the numeric values are given as follows:

For the second definition, the numeric values are given as follows:

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Answer:

*1.5

Step-by-step explanation:

15/10＝1.5

80*1.5＝120

60*1.5＝90

30*1.5＝45

36*1.5＝54

A six-sided dice is rolled twice then the probability that the larger of the two rolls was equal to 3, is 0.1389.

A six-sided dice is rolled twice.

Then the possibility of getting numbers in one rolled = 6

The possibility of getting numbers in second rolled = 6

So the possibility of getting numbers in both rolled = 6 × 6

The possibility of getting numbers in both rolled = 36

Let A be the event that have the larger of the two rolls was equal to 3.

So the possible outcomes;

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

So total number of possible outcomes to getting larger of the two rolls was equal to 3 = 5

Then the probability that the larger of the two rolls was equal to 3 = Number of possible outcome/Total of possible outcome

The probability that the larger of the two rolls was equal to 3 = 5/36

The probability that the larger of the two rolls was equal to 3 = 0.1389

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Answer:2 or 11

Step-by-step explanation:

not one bc ur winning 2 bags and not 10 bc ur winning 2 bags so add one dollar to both 1 and 10

The solution set can be described as the line {(2-4t, -1-t, 3t-2, -5-6t)} in R4.  This line can be defined by a vector (4, 1, 1, -6) and any point on the line (2, -1, 0, -5).

The solution set can be described by the line {(2-4t, -1-t, 3t-2, -5-6t)} in R4. To describe this line, we need a point on the line and a vector. The point on the line can be found by setting t = 0, so the point is (2, -1, 0, -5). The vector can be found by subtracting the point from each of the coordinates of the solution set, so the vector is (4, 1, 1, -6). This vector can be used to describe the line in R4, as any point on the line can be expressed as a multiple of the vector plus the initial point on the line. Therefore, the solution set of the system of equations can be described as the line {(2-4t, -1-t, 3t-2, -5-6t)} in R4, defined by the point (2, -1, 0, -5) and the vector (4, 1, 1, -6).

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The number of ways to order the first, second and third place finishers is simply the number of permutations of 8 runners taken 3 at a time. This can be calculated using the formula for permutations

The formula for permutations, P(n,r), gives the number of ways to choose r items from a set of n items, where order matters. This formula is calculated as:

P(n,r) = n! / (n-r)!

where n! represents the factorial of n (i.e. n multiplied by n-1 multiplied by n-2, etc. down to 1).

In this problem, n = 8 and r = 3, so the number of permutations is:

P(8,3) = 8! / (8-3)! = 8! / 5! = 40320 / 120 = 336

So there are 336 different arrangements of first, second, and third place finishers in the 100-meter sprint.

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Answer:

( 1,-1)

Step-by-step explanation:

To find the x coordinate of the midpoint, add the x coordinates of the endpoint and divide by 2.

( 5+-3) /2 = 2/2 = 1

To find the y coordinate of the midpoint, add the y coordinates of the endpoint and divide by 2.

(4+-6)/2 = -2/2 = -1

The coordinates of the midpoint are ( 1,-1).

The probability that the sum of the numbers shown will be 6 when rolling two is equals to the 1/80. So, the correct answer is option (a).

The probability of a simple event happening is defined as the number of times the event can happen, divided by the number of possible event. The probability of occurrence of the event A is P(A) = n/N. We have a twenty sided die has each side showing numbers from 1-20. Let us consider an event A , A : the event of getting sum of 6 when two dice are rolled. Also, Assume that it is fair dice, we have all the outcomes to be equally likely. We have, to determine the probability of occurrence of the event A, P(A)= No. of favourable outcomes/total num of outcomes = n/N

Thus the favourable outcomes for event A, n = {(1,5),(5,1),(2,4),(4,2),(3,3)} = 5

On rolling a die, total possible outcomes

= 20 so, on rolling two dice, we have total number of outcomes as, N = 20× 20= 400.

Thus the probability of occuring of event A, P(A) = n/M = 5/400 = 1/80.

Hence, required probability is 1/80.

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Complete question:

On a 20-sided die, each side shows a number from 1-20. What is the probability that the sum of the numbers shown will be 6 when rolling two?

a. 1/80

b. 1/6

c. 395/400

d. 1/400

The operation * in the expression  a*b = ab/4 + a + b is associative.

∗ is association if (a∗b)∗c = a∗(b∗c)

(a∗b)*c =  (ab/4*c)/4 = abc /16

​a∗(b*c) =  (a x (bc/4))/4 = abc /16

​Since (a∗b)∗c=a∗(b∗c)∀a,b,cϵQ

∗ is an associative binary operation.

Since addition is also associative.

So here a*b = ab/4 + a + b is also associative.

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This problem involves the concept of parametric differentiation. We are given parametric equations:

1. [tex]\(x(t) = e^{-3t}\)[/tex]

2. [tex]\(y(t) = e^{3t}\)[/tex]

We are asked to find [tex]\(\frac{d^2 y}{d x^2}\), the second derivative of \(y\) with respect to \(x\). Here are the steps to solve this problem:

Step 1: Calculate [tex]\(\frac{dy}{dt}\) and \(\frac{dx}{dt}\)[/tex]

[tex]\(\frac{dy}{dt} = \frac{d}{dt} e^{3t} = 3e^{3t}\)[/tex]

[tex]\(\frac{dx}{dt} = \frac{d}{dt} e^{-3t} = -3e^{-3t}\)[/tex]

Step 2: Calculate [tex]\(\frac{dy}{dx}\)[/tex]

By the chain rule, we can express [tex]\frac{dy}{dx}\)[/tex] as [tex]\frac{dy}{dt} / \frac{dx}{dt}\)[/tex].

Hence,

[tex]\(\frac{dy}{dx} = \frac{3e^{3t}}{-3e^{-3t}} = -e^{6t}\)[/tex]

Step 3: Calculate  [tex]\(\frac{d^2 y}{dx^2}\)[/tex]

Now, we find the second derivative. Here we have to apply the chain rule again, but now it's a bit trickier because [tex]\(\frac{dy}{dx}\)[/tex] itself is a function of t, not x So we need to take [tex]\(\frac{d}{dt}\)[/tex] of [tex]\(\frac{dy}{dx}\)[/tex] and then divide by [tex]\(\frac{dx}{dt}\)[/tex]

[tex]\(\frac{d^2 y}{dx^2} = \frac{d}{dt} (\frac{dy}{dx}) / \frac{dx}{dt}\)[/tex]

Taking the derivative of [tex]\(\frac{dy}{dx} = -e^{6t}\)[/tex] with respect to t, we get:

[tex]\(\frac{d}{dt} (\frac{dy}{dx}) = -6e^{6t}\)[/tex]

So,

[tex]\(\frac{d^2 y}{dx^2} = \frac{-6e^{6t}}{-3e^{-3t}} = 2e^{9t}\)[/tex]

So, the answer is (D)  [tex]2e^{9t}\)[/tex]

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The second derivative d²y/dx² in terms of t is [tex]2e^{(9t)[/tex].

Calculus's essential idea of derivatives is how quickly a function changes in relation to its independent variable. They offer details about how a function is altering for a certain input or point.

We must use the chain rule to determine the second derivative of y with respect to x (d²y/dx²) in terms of t.

According to the chain rule, the derivative of y with respect to x is given by dy/dx = (dy/dt) / (dx/dt) if we have a parametric curve defined by x = f(t) and y = g(t).

In this case, we have [tex]x(t) = e^{(-3t)[/tex] and [tex]y(t) = e^{(3t)[/tex].

First, we'll find the first derivatives dx/dt and dy/dt:

dx/dt = d/dt [tex](e^{(-3t)}) = -3e^{(-3t)[/tex]

dy/dt = d/dt [tex](e^{(3t)}) = 3e^{(3t)[/tex]

Next, we can find dy/dx:

dy/dx = (dy/dt) / (dx/dt)

[tex]= (3e^{(3t)}) / (-3e^{(-3t)})\\= -e^{(6t)[/tex]

Finally, we differentiate dy/dx with respect to x to find d²y/dx²:

d²y/dx² = [tex]d/dx (-e^{(6t)})[/tex]

[tex]= d/dt (-e^{(6t))} \times (dt/dx)\\= -6e^{(6t)} \times (1 / (-3e^{(-3t)}))\\= 2e^{(9t)[/tex]

Therefore, the second derivative d²y/dx² in terms of t is [tex]2e^{(9t)[/tex].

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The null hypothesis for this scenario would be that the percentage of residents in favor of the construction of a new community is less than or equal to 46%.

And the alternative hypothesis would be that the percentage of residents in favor of the construction is greater than 46%. The mayor would need to conduct a hypothesis test to determine if there is sufficient evidence at the 0.01 level to support the claim.

This test would involve collecting data from a sample of the town's residents and determining if the sample proportion greater than or equal to the 46% claimed by the mayor. If the sample proportion is significantly greater than 46%, then the mayor's claim can be supported at the 0.01 level.

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The equation -hx + 4 = 2x - 1 solved for the variable x is:

x = 5/(h + 2)

We want to solve the equation below for x:

-hx + 4 = 2x - 1

To solve this for x, we need to isolate the variable x in one of the sides of the equation.

We can rewrite:

-hx + 4 = 2x - 1

-hx -2x = -4 - 1

We can take x as a common factor now to get:

-(h + 2)*x = -5

x = 5/(h + 2)

That is the equation solved for x.

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Since no two numbers in the range have these characteristics, P(sum 1, product > 1/6) = 0.Let x and y be  probability two randomly generated numbers between 0 and 1.

Since no two numbers in the range 0–1 have these features,

P(sum 1, product > 1/6)

= P(x + y 1, xy > 1/6)

= P(x 1, y 1, xy > 1/6)

= P(x 1, xy > 1/6)

= 0.

There is zero chance that two randomly chosen numbers from the range of 0 to 1 will have a product more than 1/6 and a sum less than 1. (zero). This is thus because no two numbers selected from the range have the same characteristics. We can begin by describing the probability of the desired outcome in terms of the variables x and y, which stand in for the two numbers selected, in order to calculate this. Then, we broaden this expression to take into account the requirements that x and y must both be less than 1 and that their product must be higher than 1/6.

As a result, we can see that the probability is 0, as no two values selected from the range of 0 to 1 can meet all of these requirements at once. Therefore, there is no chance that two numbers chosen at random from the range of 0 to 1 will have a total that is less than 1 and a product that is higher than 1/6.

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Answer: [tex]y=10[/tex]

Step-by-step explanation:

Note that [tex]\angle ACB \cong \angle ACD[/tex] because diagonals of a rhombus bisect the angles from which they are drawn.

Now, because [tex]m\angle DEC=90^{\circ}[/tex] since diagonals of a rhombus are perpendicular, it follows that [tex]m\angle CDB+m\angle ABC=90^{\circ}[/tex].

[tex]6y+2y+10=90\\\\8y+10=90\\\\8y=80\\\\y=10[/tex]

Answer:

a) The first term of an AP can be found using the formula for the nth term of an AP, which is a + (n-1)d, where a is the first term, d is the common difference, and n is the position of the term.

Using this formula, we can substitute the known values: -15 = a + (5-1)d

-15 = a + 4d

To find the first term, we will have to solve for a by isolating it on one side of the equation:

a = -15 - 4d

b) To find the common difference, we can use the relation between the second and fifth term.

The fifth term of an AP is -15, and the second term is 0, so we can use the formula for the nth term of an AP to find the common difference:

-15 = a + (5-1)d

-15 = a + 4d

To find the common difference, we will have to solve for d by isolating it on one side of the equation:

d = (-15 - a) / 4

Total weight of the Durains sold by Rowena is  5.8 kg

One of the four fundamental operations in mathematics is addition, along with subtraction, multiplication, and division. The entire amount or sum of the two whole numbers is obtained by adding them.

The units of mass, the kilogram and the gram, must first be translated into grams before being added. Thereafter, the basic addition procedure must be followed. As with regular numbers, we can add two or more mass units expressed in kilograms and grams. Always write grams as three-digit figures, such as 3g = 003g.

In arithmetic, addition is the process of adding two or more integers together. Addends are the numbers being added, and the result or the final answer we get after the process is called the sum.

The weight of the Durains are 1.3 kg, 1.5kg, 1.4 kg and 1.6kg

Total weight of the Durains sold by Rowena is

(1.3 + 1.5 + 1.4 + 1.6) = 5.8 kg

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The fixed cost to travel is $4 which is option (1).

What is meant by the cost of an item?

A cost is the worth of money that has been expended to produce something or provide a service and is therefore no longer available for use in production, research, retail, and accounting. In the case of an acquisition cost, the money spent on the acquisition is considered the cost.

Given chart,

Distance travelled (miles) (x)          Cost (dollars) (y)

0                                                             4

1                                                             9

2                                                            14

We are asked to find the fixed amount charged.

From the data given, we can say that

The amount charged when no distance is covered is $4.

The amount charged per mile = 9-4 = 14 -9 = $5

A fixed cost is the amount that is charged no matter the distance travelled.

In this case, it is $4, as it is the money needed to get on the bus.

Therefore the fixed cost to travel is $4 which is option (1).

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Answer:

We're required to find s(10).

therefore , the polynomial applicable will be

[tex]s(x) = - 5{x}^{2} + x - 8 \: \\ \: ( \: since \: x > - 1)[/tex]

[tex]\therefore \: s(10) = - 5( {10}^{2}) + 10 - 8 \\ \dashrightarrow \: s(10) = - 5(100) + 2 \\ \dashrightarrow \: s(10) = - 500 + 2 \\ \dashrightarrow \: \boxed{ \: s(10) = - 498}[/tex]

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The solution to the separable differential equation with the initial condition u(0)=3 is u = 3e−3t²/2

solve the separable differential equation for u. dudt=e3u 6t use the following initial condition: u(0)=3

The solution to the separable differential equation is found by separating the variables dudt and t.

dudt = e3u6t

We can then rearrange this equation to get t on one side and u on the other.

t = (e3u)⁄6

We can then integrate both sides of the equation to get the solution.

∫tdt = ∫(e3u)⁄6du

t²/2 = e3u/3 + c

We can then use the initial condition to solve for c.

3²/2 = e3(3)/3 + c

c = 0

Therefore the solution to the differential equation is:

t²/2 = e3u/3

u = 3e−3t²/2

This is the solution to the separable differential equation with the initial condition u(0)=3.

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Complete question:

Given a separable differential equation dudt=e3u 6t with the initial condition u(0)=3, find the solution u(t).

so the sign is made up a rectangular 17x8 piece, and then we cut out a whole with a radius of 2.2 cm, so if we put it on the sidewalk, the sign will end up looking like the one in the picture below.

so if we just get the whole area of the rectangle and then subtract the area of the circle, in effect making a hole in it, what's leftover is their difference.

[tex]\stackrel{\textit{\LARGE Areas}}{\stackrel{rectangle}{(17)(8)}~~ - ~~\stackrel{circle}{\pi (2.2)^2}} \implies 136-4.84\pi ~~ \approx ~~ \text{\LARGE 121}~cm^2[/tex]

The time  it takes Saturn pictured above, to orbit the sun is about 29.2 years

The movements of planets and other celestial bodies in our solar system are described by a group of three empirical laws known as Kepler's laws.

A planet's average distance from the sun is inversely related to the square of its period of revolution around the sun.

By the use of the Kepler's laws, we can see that the equivalent equation for the function is t = √d^3

We can now obtain the time from this equation by the use of the equation as follows;

t = √(9.5)^3

t = 29.2 years

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a) Upper bound for ∫RfdA is 108, lower bound is 0.

b) Overestimate: ∫RfdA≈ 67.5,

underestimate: ∫RfdA≈ 13.5,

average: ∫Rf dA = 40.5.

a) An upper and lower bounds for ∫RfdA without subdividing R, we can use the fact that f(x,y) is a continuous function on the rectangle R.

By the Extreme Value Theorem, there must exist a maximum and minimum value of f(x,y) on R.

Hence, We can find these extreme values by evaluating f(x,y) at each of the vertices of R:

f(0,0) = 0

f(6,0) = 0

f(6,6) = 3

f(0,6) = 0

So, the maximum value of f(x,y) on R is 3, and the minimum value is 0.

Therefore, we have:

lower bound = min(f(x,y))

area(R) = 0 6 × 6 = 0

upper bound = max(f(x,y)) area(R) = 3 6 × 6 = 108

So the lower bound for ∫RfdA is 0, and the upper bound is 108.

b) An ∫RfdA by partitioning R into four sub rectangles, we can divide the rectangle R into four equal squares with vertices (0,0), (3,0), (3,3), and (0,3), (3,3), (6,3), (6,6), and (3,3), (6,3), (6,6), (3,6).

We can then evaluate f(x,y) at its maximum and minimum values on each sub rectangle:

For the sub rectangle with vertices (0,0), (3,0), (3,3), and (0,3): max(f(x,y)) = f(3,3) = sqrt(0.25*9) = 1.5 min(f(x,y)) = f(0,0) = 0

For the sub rectangle with vertices (3,0), (6,0), (6,3), and (3,3):

max(f(x,y)) = f(6,3) = √(0.25 × 18) = 1.5

min(f(x,y)) = f(3,0) = 0

For the sub rectangle with vertices (3,3), (6,3), (6,6), and (3,6):

max(f(x,y)) = f(6,6) = √(0.25×36) = 3

min(f(x,y)) = f(3,3) = √(0.25 × 9) = 1.5

For the sub rectangle with vertices (0,3), (3,3), (3,6), and (0,6):

max(f(x,y)) = f(3,6) = √(0.25×18) = 1.5

min(f(x,y)) = f(0,3) = 0

We can then calculate the area of each sub rectangle:

area(sub rectangle 1) = 3×3 = 9

area(sub rectangle 2) = 3×6 = 18

area(sub rectangle 3) = 3×3 = 9

area(sub rectangle 4) = 3×6 = 18

Finally, we can estimate the integral by averaging the maximum and minimum values of f(x,y) over all four sub rectangles, and multiplying each by its corresponding area:

overestimate: ∫RfdA≈ [(1.5×9) + (1.5×18) + (3×9) + (1.5×18)] = 67.5 underestimate: ∫RfdA≈ [(0×9) + (0×18) + (1.5×9) + (0×18)] = 13.5

The average of these two estimates is:

average: ∫Rf dA = (67.5 + 13.5)/2 = 40.5

So, we estimate that the value of the integral is approximately 40.5.

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El Sr. Jorge va a pagar 810$ por los productos. Con el aumento del IVA del 16%, la cuenta total será 942.56$.

El Sr. Jorge compró un pantalón por 430 dólares y una camisa por 380 dólares en la tienda, lo que significa que el costo total sin incluir el impuesto es de 810 dólares. Sin embargo, se aplicará un 16% de impuesto al valor agregado (IVA) al precio total, lo que significa que la cuenta final será de 942.56 dólares. Este es el costo total que el Sr. Jorge tendrá que pagar.

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The result of operations between two functions is listed below:

In this question we need to determine the result of three operations between two functions, a quadratic equation and a linear equation, both polynomic functions. According to function theory, the domain of polynomials is equal to the set of all real numbers. There are three operations:

Addition

(f + g) (x) = f(x) + g(x)

Subtraction

(f - g) (x) = f(x) - g(x)

Multiplication

(f · g) (x) = f(x) · g(x)

If we know that f(x) = x² and g(x) = 7 · x - 3, then the result of the operations are:

Addition

(f + g) (x) = x² + 7 · x - 3, Domain: All real numbers.

Subtraction

(f - g) (x) = x² - 7 · x + 3, Domain: All real numbers.

Multiplication

(f · g) (x) = x² · (7 · x - 3), Domain: All real numbers.

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Answer:

Step-by-step explanation:

Tessa over the weekend did 15 problems

Is a number that expresses the portion of some number over a total. The number that expresses the portion is known as the numerator and the number that expresses the total is known as the denominator.

The total number of problems assigned = x

On Saturday she did = 3/5x

The remaining problems are = 1 - 3/5x

The remaining problems are = 2/5x

On Sunday she did = 1/3 * (2/5x)

On Sunday she did = 2/15x

The remaining problems= (1 - 1/3)*(2/5x)

The remaining problems= (2/3) * (2/5x)

The remaining problems= 4/15x

As that the final number of problems remaining are 4, we get:

4/15x = 4

x = 4 (15/4)

x = 15

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