You will write a program that lets a teacher convert exam grades for his/her class from scores to letter-grades, then calculate how many grades are in each letter-grade category, and finally visualize the letter-grade distribution in the form of a histogram.

x + y = 15------(1)

x/y = 2------(2)

y = 15 - x

put this value in eq 2

x/(15-x) = 2

x = 2 (15 - x)

x = 30 - 2x

2x + x = 30

3x = 30

x = 10

y= 15 - x

y = 15 - 10

y = 5

Solution:

Given:

Let’s simplify the second equation to find out the value of x.

Now, let’s substitute the value of x into the first equation.

Now, let’s substitute the value of y into the simplified second equation.

Hence, the value of x and y are 5 and 10 respectively.

Answer:

\(\dfrac{7}{9}\)

Step-by-step explanation:

\(\dfrac{x+1}{y+1}=\dfrac{4}{5}\\\Rightarrow 5x-4y=-1\)

\(\dfrac{x-5}{y-5}=\dfrac{1}{2}\\\Rightarrow 2x-y=5\)

Putting it in matrix form

\(\begin{bmatrix}a_{1}&b_{1}\\a_{2}&b_{2}\end{bmatrix}{\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}{c_{1}}\\{c_{2}}\end{bmatrix}\\\Rightarrow\begin{bmatrix}5 & -4\\2 & -1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}-1\\ 5\end{bmatrix}\)

From Cramer's rule we have

\(x=\dfrac{\begin{vmatrix}c_1 &b_1 \\ c_2 & b_2\end{vmatrix}}{\begin{vmatrix}a_1 &b_1 \\ a_2& b_2\end{vmatrix}}\\\Rightarrow x=\dfrac{\begin{vmatrix}-1 &-4 \\ 5 & -1\end{vmatrix}}{\begin{vmatrix}5 &-4 \\ 2& -1\end{vmatrix}}\\\Rightarrow x=\dfrac{1+20}{-5+8}\\\Rightarrow x=7\)

\(y=\dfrac{\begin{vmatrix}a_1 &c_1 \\ a_2 & c_1\end{vmatrix}}{\begin{vmatrix}a_1 &b_1 \\ a_2& b_2\end{vmatrix}}\\\Rightarrow y=\dfrac{\begin{vmatrix}5 &-1 \\ 2 & 5\end{vmatrix}}{\begin{vmatrix}5 &-4 \\ 2& -1 \end{vmatrix}}\\\Rightarrow y=\dfrac{25+2}{-5+8}\\\Rightarrow y=9\)

Verifying the results

\(\dfrac{7+1}{9+1}=\dfrac{8}{10}=\dfrac{4}{5}\)

\(\dfrac{7-5}{9-5}=\dfrac{2}{4}=\dfrac{1}{2}\)

Hence, the fraction is \(\dfrac{7}{9}\).

Answer: \(400^{2}\)

Step-by-step explanation:

First, write the formula:

A = (b × h) / 2

Next, multiply the base and the height of the triangle.

20 x 40 = 800

Finally, divide 800 by 2.

800/2=400

Therefore, the answer is \(400^{2}\).

In a linear time-invariant system, the zero-state response (ZSR) is the output of the system when the input is zero, assuming all initial conditions (such as initial voltage or current) are also zero.

(a) For H1(s) = 1, the zero-state response Yzs(s) is simply the product of the transfer function H1(s) and the input Ei(s):

Yzs(s) = H1(s) * Ei(s) = (45+2)

To find the time-domain zero-state response yzs(t), we need to take the inverse Laplace transform of Yzs(s):

yzs(t) = L^-1{Yzs(s)} = L^-1{(45+2)} = 45δ(t) + 2δ(t)

where δ(t) is the Dirac delta function.

(b) For H2(s) = 45+1, the zero-state response Yzs(s) is again the product of the transfer function H2(s) and the input E2(s):

Yzs(s) = H2(s) * E2(s) = (45+1)E2(s)

To find the time-domain zero-state response yzs(t), we need to take the inverse Laplace transform of Yzs(s):

yzs(t) = L^-1{Yzs(s)} = L^-1{(45+1)E2(s)} = (45+1)e^(t/2)u(t)

where u(t) is the unit step function.

(c) For H3(s) = ns, the zero-state response Yzs(s) is given by:

Yzs(s) = H3(s) * E3(s) = ns * 542

To find the time-domain zero-state response yzs(t), we need to take the inverse Laplace transform of Yzs(s):

yzs(t) = L^-1{Yzs(s)} = L^-1{ns * 542} = 542L^-1{ns}

Using the inverse Laplace transform from problem 1, we have:

yzs(t) = 542 δ'(t) = -542 δ(t)

where δ'(t) is the derivative of the Dirac delta function.

(d) For H4(s) = 2e^(-4s) / (s+3)(4s+1), the zero-state response Yzs(s) is given by:

Yzs(s) = H4(s) * E4(s) = (2e^(-4s) / (s+3)(4s+1)) * (1/(s+3))

Simplifying the expression, we have:

Yzs(s) = (2e^(-4s) / (4s+1))

To find the time-domain zero-state response yzs(t), we need to take the inverse Laplace transform of Yzs(s):

yzs(t) = L^-1{Yzs(s)} = L^-1{(2e^(-4s) / (4s+1))}

Using partial fraction decomposition and the inverse Laplace transform from problem 1, we have:

yzs(t) = L^-1{(2e^(-4s) / (4s+1))} = 0.5e^(-t/4) - 0.5e^(-3t)

Therefore, the zero-state response for each of the four LTIC systems is:

(a) Yzs(s) = (45+2), yzs(t) = 45δ(t) + 2δ(t)

(b) Yzs(s) = (45+1)E2(s), yzs(t) = (45+1)e^(t/2)u(t)

(c) Yzs(s) = ns * 542, yzs(t) = -542 δ(t)

(d) Yzs(s) = (2e^(-4s) /

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Answer:

110º

Step-by-step explanatio:

6x+20 + 4x+10 = 180

180 - 30= 150

150=10x

10x/150

x = 15

6(15) + 20 = 110

Answer:

list in the explanation

Step-by-step explanation:

BOOT x2

SKINNY x2

RELAXED x2

BOOT, SKINNY

BOOT, RELAXED

SKINNY, BOOT

SKINNY, RELAXED

RELAXED, BOOT

RELAXED, SKINNY

The solution set is given by Set A = { BB , SS , RR , BS , BR , SR } , where B , S and R are boot cut (B), skinny (S), and relaxed fit (R) respectively

The union of two sets A and B is the set of all those elements which are either in A or in B, i.e. A ∪ B, whereas the intersection of two sets A and B is the set of all elements which are common. The intersection of these two sets is denoted by A ∩ B

The union of two sets is a new set that contains all of the elements that are in at least one of the two sets

The intersection of two sets is a new set that contains all of the elements that are in both sets

Given data ,

Let the set be represented as A

Now , the equation will be

Marcus is shopping for pants and the available pants are boot cut (B), skinny (S), and relaxed fit (R)

He wants to buy 2 pair of pants , and the possible ways of 2 pants are

Set A = { BB , SS , RR , BS , BR , SR }

So , the total number of elements in set A = 6 pairs

Therefore , the value of A is 6 pairs

Hence , the set A is { BB , SS , RR , BS , BR , SR }

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Answer:

A.

Step-by-step explanation:

A Foot is 12 inches. The Swimming pool drops by 6 inches, so 0.5 of a feet.

The depth is 4 Feet. 4-0.5 \(\leq\) 4 Makes the most sense out of all the other questions.

Students will board the bus at 1350 even though it is not totally full.

Area is the total amount of space that an object's shape or a flat (2-D) surface occupy.

Create a square on paper by using a pencil. Two dimensions make it up. A shape's area on paper is the space it takes up.

Imagine that your square is made up of smaller unit squares.

The area of a figure is equal to the number of unit squares required to completely cover the surface area of a particular 2-D shape. Square cms, square feet, square inches, square meters, etc. are a few cmmon units for measuring area

To get the area of the square figures presented below, draw unit squares with 1-centimeter sides. Therefore, the shape will be measured.

According to our question-

723+685

1408-58

1350

Hence, Students will board the bus at 1350 even though it is not totally full.

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Answer:

a

Step-by-step explanation:

Answer:

A

........................

Answer:

13

Step-by-step explanation:

5-x+6=-2

X=13

First, we want to note two things:

We can describe this with an inequality:  x ≥ -10
Be sure you use ≥ and not >, since -10 is included.

We can describe this with interval notation: [ -10, infty )
Be sure you use [ and not ( on -10, since -10 is included.

You can also use set-builder notation: { x | x ≥ -10 }

The answer is 12 or option (d)

I took the test~ good luck!

The properties of Boolean Difference are: The Boolean difference of a variable and a function remains the same as the Boolean difference of the function. The Boolean difference of the product of two functions can be expressed as the XOR of two terms involving the Boolean differences of each function. The Boolean difference of the sum of two functions can be expressed as the XOR of two terms involving the Boolean differences and Boolean negations of the functions.

To prove the properties in Boolean Difference, we'll use the following definitions:

Boolean Difference: The Boolean difference of two variables x and y, denoted as dx dy, is the exclusive OR (XOR) of x and y.

Boolean Negation: The Boolean negation of a variable x, denoted as xˉ, is the complement (NOT) of x.

Now let's prove each property one by one:

dx dy df(x) = dx dy df(x)

This property states that taking the Boolean difference of a variable x and a function f(x) is equivalent to taking the Boolean difference of x and the derivative of f(x) with respect to xi.

Proof:

We know that the derivative of f(x) with respect to xi can be written as df(x)/dxi.

Using the definition of Boolean difference, we have:

dx dy df(x) = dx dy (df(x)/dxi)

= (dx dy df(x))/dxi

Since dx dy is a Boolean value and does not depend on xi, we can conclude that:

dx dy df(x) = dx dy (df(x)/dxi)

= dx dy df(x)

dx dy [f(x)⋅g(x)] = f(x)⋅dx dy g(x) ⊕ g(x)⋅dx dy f(x)

This property states that taking the Boolean difference of the product of two functions, f(x) and g(x), with respect to xi is equivalent to the XOR of two terms: f(x) multiplied by the Boolean difference of g(x) with respect to xi, and g(x) multiplied by the Boolean difference of f(x) with respect to xi.

Proof:

Using the definition of Boolean difference, we have:

dx dy [f(x)⋅g(x)] = dx dy [f(x)]⋅g(x) ⊕ f(x)⋅dx dy [g(x)]

= [dx dy f(x)]⋅g(x) ⊕ f(x)⋅[dx dy g(x)]

This follows from the distributive property of XOR over the Boolean product.

dx dy [f(x)+g(x)] = fˉ(x)⋅dx dy g(x) ⊕ gˉ(x)⋅dx dy f(x)

This property states that taking the Boolean difference of the sum of two functions, f(x) and g(x), with respect to xi is equivalent to the XOR of two terms: the Boolean negation of f(x) multiplied by the Boolean difference of g(x) with respect to xi, and the Boolean negation of g(x) multiplied by the Boolean difference of f(x) with respect to xi.

Proof:

Using the definition of Boolean difference and Boolean negation, we have:

dx dy [f(x)+g(x)] = dx dy [f(x)] ⊕ dx dy [g(x)]

= [dx dy f(x)]⋅[gˉ(x)] ⊕ [fˉ(x)]⋅[dx dy g(x)]

= fˉ(x)⋅dx dy g(x) ⊕ gˉ(x)⋅dx dy f(x)

dx dy [f(x)⊕g(x)] = dx dy f(x) ⊕ dx dy g(x)

This property states that taking the Boolean difference of the XOR (exclusive OR) of two functions, f(x) and g(x), with respect to xi is equivalent to the XOR of the Boolean differences of f(x) and g(x) with respect to xi.

Proof:

Using the definition of Boolean difference, we have:

dx dy [f(x)⊕g(x)] = dx dy [f(x)] ⊕ dx dy [g(x)]

= [dx dy f(x)] ⊕ [dx dy g(x)]

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Answer:

When radius is doubled:

Circumference becomes double.

Area becomes four times.

When diameter is doubled:

Circumference becomes double.

Area becomes four times.

Step-by-step explanation:

Given that

Radius of a circle is doubled.

Diameter of circle is doubled.

To study:

The effect on circumference and area on doubling the radius and diameter.

Solution/explanation:

Let us discuss about the formula for circumference and area.

Formula for Circumference of a circle in form of radius:

\(C =2\pi r\)

It is a linear equation in 'r'. So by doubling the radius will double the circumference.

Formula for Area of a circle in form of radius:

\(A =\pi r^2\)

It is a quadratic equation in 'r'. So by doubling the radius will make the area as four times the earlier area.

Testing using example:

Let the initial radius of a circle = 7 cm

Initial circumference = \(2 \times \frac{22}{7} \times 7 = 44 cm\)

Initial area = \(\frac{22}{7} \times 7 \times 7 =154 cm^2\)

After doubling:

Radius = 14 cm

circumference = \(2 \times \frac{22}{7} \times 14 = 88 cm\)   (Twice the initial circumference)

area = \(\frac{22}{7} \times 14 \times 14 =616 cm^2\) (4 times the initial area)

------------------------------------

Formula for Circumference of a circle in form of Diameter:

\(C =\pi D\)

It is a linear equation in 'D'. So by doubling the diameter will double the circumference.

Formula for Area of a circle in form of diameter:

\(A =\dfrac{1}{4}\pi D^2\)

It is a quadratic equation in 'D'. So by doubling the Diameter will make the area as four times the earlier area.

Testing using example:

Let the initial diameter of a circle = 28 cm

Initial circumference = \(\frac{22}{7} \times 28 = 88 cm\)

Initial area = \(\frac{1}{4}\times \frac{22}{7} \times 28 \times 28 =616cm^2\)

After doubling:

Diameter = 56 cm

circumference = \(\frac{22}{7} \times 56= 176 cm\)   (Twice the initial circumference)

area = \(\frac{1}{4}\times \frac{22}{7} \times 56 \times 56 =2464cm^2\) (4 times the initial area)

So, the answer is justified:

When radius is doubled:

Circumference becomes double.

Area becomes four times.

When diameter is doubled:

Circumference becomes double.

Area becomes four times.

Option a, Yes players with future transfer plans do experience significant points variance if those plans have an impact on their point performance.

There are two probable conclusions: (a) players with a future transfer plan do indeed have a greater point variance, and (b) players with a future transfer plan do indeed have a higher point median.

The presentation of players with and without move expectations ought to be analyzed, jumbling variables ought to be considered, and factual tests ought to be rushed to survey the legitimacy of these speculations.

The outcome of this research would ultimately establish whether or not the possibility of a future transfer plan had an impact on a player's point production.

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Answer:

T

Step-by-step explanation:

Just did it

Answer:

y=1/4x+6

Step-by-step explanation:

Answer:

sorry do not know the answer, but this is all i got for you. hope it helps.

Step-by-step explanation:

One of the simplest and most common transformations in geometry is the 180-degree rotation, both clockwise and counterclockwise. If P (x, y) is a point that must be rotated 180 degrees about the origin, the coordinates of this point after the rotation will only be of the opposite signs of the original coordinates.

The fraction \(\frac{20}{2}\) is 10 as a whole number.

A fraction is a mathematical expression that represents a part or a division of a whole. It is used to represent numbers that are not whole numbers or integers. A fraction consists of two components:

1. Numerator: The numerator is the number on the top of the fraction. It represents the quantity or part of the whole being considered.

2. Denominator: The denominator is the number at the bottom of the fraction. It represents the total number of equal parts into which the whole is divided.

Given the fraction \(\frac{20}{2}\),

divide numerator and denominator by 2,

\(\frac{20\div 2}{2\div 2}\) = \(\frac{10}{1}\) = 10.

So, \(\frac{20}{2}\) = 10

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Answer:

why is the link is blurry I can see it good

The rate at which the copy machine prints is 10 copies per minute.

The copy machine prints 10 copies per minute.

This means that the rate at which the copy machine prints is 10 copies per minute.

Rate is a measure of how fast something happens over a specific time interval.

In this case, the rate of printing is the number of copies produced per minute.

Since the machine prints 10 copies in 1 minute, we can say that its printing rate is 10 copies per minute.

This indicates that every minute, the machine is capable of producing 10 copies.

To further understand the concept, we can think of it in terms of a ratio. The ratio of copies to time is 10 copies per 1 minutes.

This ratio represents the rate at which the copy machine operates.

It's important to note that the rate of printing remains constant as long as the machine operates under the same conditions.

In this scenario, where 10 copies are printed per minute, the rate remains steady unless any changes are made to the machine's functionality or settings.

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The amount that will have accumulated in 10 years after the last deposit is approximately $13,299.25.

To calculate the accumulated amount, we can use the formula for compound interest:

A = P(1 + r/n)^(nt)

Where:

A = Accumulated amount

P = Principal amount (initial deposit)

r = Annual interest rate (as a decimal)

n = Number of times interest is compounded per year

t = Number of years

In this case, Sklyer has made deposits of $680 at the end of every quarter for 13 years, so the principal amount (P) is $680. The annual interest rate (r) is 5%, which is 0.05 as a decimal. The interest is compounded annually, so the number of times interest is compounded per year (n) is 1. And the number of years (t) for which we need to calculate the accumulated amount is 10.

Plugging these values into the formula, we have:

A = $680(1 + 0.05/1)^(1*10)

  = $680(1 + 0.05)^10

  = $680(1.05)^10

  ≈ $13,299.25

Therefore, the amount that will have accumulated in 10 years after the last deposit is approximately $13,299.25.

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Answer: 30

Step-by-step explanation:

i don’t know, but i know it’s 30

Answer: 28

Step-by-step explanation:

Answer:

4

Step-by-step explanation:

The formula for the surface area of a cube is 6s²

You set the equation: 6s² = 96

-First, divide both sides by 6

→s²=16

-Then, square root both sides and solve

→√s²=√16

→s=4

Finally, you get the answer of 4 for the length of the edge of the cube with the surface area of 96

When equation f(x) = 3x + 4 is transformed to h(x) = f(x + 3) and plotted on the graph, both the lines of the equations are parallel to each other.

So, the equations are graphs are:

Now, graph both equations as follows:

Therefore, when equation f(x) = 3x + 4 is transformed to h(x) = f(x + 3) and plotted on the graph, both the lines of the equations are parallel to each other.

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Answer:

45.4782609 miles per hour

Step-by-step explanation:

523 miles

11 1/2 hours

523 / 11.5

Therefore, the series 4/5 - 4/7 + 4/9 - 4/11 + 4/13 - ... is convergence in nature.

To test the convergence or divergence of the given series: 4/5 - 4/7 + 4/9 - 4/11 + 4/13 - ..., we can use the alternating series test. The alternating series test states that if a series has the form (-1)^n b_n, where b_n is a positive sequence that is decreasing and approaches zero as n approaches infinity, then the series converges.

In the given series, we have b_n = 4/(2n+3), which is a positive sequence that approaches zero as n approaches infinity. To see that b_n is decreasing, we can note that b_{n+1}/b_n = 2(2n+3)/(2n+5) < 1 for all n, since the numerator is always less than the denominator. Therefore, b_n is a positive, decreasing sequence that approaches zero as n approaches infinity.

Furthermore, the series has the form (-1)^n b_n, where the sign alternates between positive and negative. Therefore, we can apply the alternating series test, which tells us that the series converges.

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Answer:

y= -5/4x+

Step-by-step explanation:

find the gradient first ∆y/∆5x

-4-1/ 8-4

=-5/4

assume one of the points to be x, y

y-1/ x-4 =-5/4

cross multiply

4y-4=-5x+20

4y= -5x+20+4

divide both sides by 4

y= -5/4x+

The radius of the oatmeal container is 0.2 inches.

The surface area of the plastic is 0.08π square inches

The base and lid are made of plastic and have a radius of R.

The base and lid have a combined surface area of:

πr² + πr² = 2πr² square inches

The total surface area of the oatmeal container is 40 square inches.

The surface area of the cardboard is 16πr square inches and the surface area of the plastic is 2πr² square inches.

Thus, we have the equation:

40 = 16πr + 2πr²

We can solve for R as follows:

16πr + 2πr² = 40

2πr^2 + 16πr - 40 = 0

2πr^2 + 20πr - 4πr - 40 = 0

20r(πr + 2) - 4(πr + 2) = 0

(20r - 4)(πr + 2) = 0

20r - 4 = 0 or πr + 2 = 0

20r = 4 or πr = -2

r = 0.2

Thus, the radius of the oatmeal container is 0.2 inches.

The surface area of the plastic is 2πr² = 2π(0.2)² = 0.08π square inches

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