I have to use Hamilton's method to find how many council members should represent each district. ​what is the standard divisor for Hamilton's method

Answer:

Angle A = 60º and Angle B = 30º

Step-by-step explanation:

We know that A and B are complementary angles, therefore ∠A + ∠B =90. (+)

On the other hand, ∠A is half as large as ∠B; this can be written algebraically as [tex]A=\frac{B}{2}[/tex]. (*)

If we substitute (*) in (+) we get:

[tex]\frac{B}{2}+B = 90\\\\ \frac{B+2B=180}{2} \\\\ B+2B=180\\\\ 3B=180\\\\ B=\frac{180}{3}\\\\ B=60[/tex]

And now we substitute the value of B in (+) and we get:

∠A+60 = 90

∠A = 90-60

∠A = 30

Answer:

Length = Width = 7 inches

Height = 14 inches

Step-by-step explanation:

Let's call  

x = length and width of the bottom and top

y = heigth

The bottom and top are squares, so their total area is

[tex]A_{bt}=x^2+x^2=2x^2[/tex]

the area of one side is xy. As we have 4 sides, the total area of the sides is

[tex]A_s=4xy[/tex]

The cost for the bottom and top would be

[tex]C_{bt}=\$4.2x^2=\$8x^2[/tex]

The cost for the sides would be

[tex]C_s=\$2.4xy=\$8xy[/tex]

The total cost to made the box is

[tex]C_t=\$(8x^2+8xy)[/tex]

But the volume of the box is  

[tex]V=x^2y=686in^3[/tex]

isolating

[tex]y=\frac{686}{x^2}[/tex]

Replacing this expression in the formula of the total cost

[tex]C_t(x)=8x^2+8x(\frac{686}{x^2})=8x^2+\frac{5488}{x}[/tex]

The minimum of the cost should be attained at the point x were the derivative of the cost is zero.

Taking the derivative

[tex]C'_t(x)=16x-\frac{5488}{x^2}[/tex]

If C' = 0 then

[tex]16x=\frac{5488}{x^2}\Rightarrow x^3=\frac{5488}{16}\Rightarrow x^3=343\Rightarrow x=\sqrt[3]{343}=7[/tex]

We have to check that this is actually a minimum.

To check this out, we take the second derivative

[tex]C''_t(x)=16+\frac{10976}{x^3}[/tex]

Evaluating this expression in x=7, we get C''>0, so x it is a minimum.

Now that we have x=7, we replace it on the equation of the volume to get

[tex]y=\frac{686}{49}=14in[/tex]

The dimensions of the most economical box are

Length = Width = 7 inches

Height = 14 inches

To find the dimensions of the container of least cost, assume the length of the base is x and the width of the base is also x. The height of the container is 686/(x^2). The dimensions of the container of least cost are approximately 10.33 inches by 10.33 inches by 6.68 inches.

To find the dimensions of the container of least cost, we need to minimize the cost function. Let's assume the length of the base is x, then the width of the base will also be x since it's a square. The height of the container will be 686/(x^2) since the volume is given as 686 in³. The total cost, C, is given by C = 4(2x^2) + 2(4x(686/(x^2))). Simplifying this expression gives C = 8x^2 + 5488/x. To find the minimum cost, we can take the derivative of C with respect to x and set it equal to zero:

dC/dx = 16x - 5488/x^2 = 0

Solving this equation gives x = √(343/2) ≈ 10.33. Therefore, the dimensions of the container of least cost are approximately 10.33 inches by 10.33 inches by 6.68 inches.

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Answer:

Y = [tex]e^{t}[/tex] +  [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex] + t - 18

Step-by-step explanation:

y''' − 3y'' + 3y' − y = ex − x + 21

Homogeneous solution:

First  we propose a solution:

Yh = [tex]e^{r*t}[/tex]

Y'h = [tex]r*e^{r*t}[/tex]

Y''h = [tex]r^{2}*e^{r*t}[/tex]

Y'''h = [tex]r^{3}*e^{r*t}[/tex]

Now we solve the following equation:

Y'''h - 3*Y''h + 3*Y'h - Yh = 0

[tex]r^{3}*e^{r*t}[/tex] - 3*[tex]r^{2}*e^{r*t}[/tex] + 3*[tex]r*e^{r*t}[/tex] - [tex]e^{r*t}[/tex] = 0

[tex]r^{3} - 3r^{2} + 3r - 1 = 0[/tex]

To solve the equation we must propose a solution to the  polynomial :

r = 1

To find the other r we divide the polynomial by (r-1) as you can see  

attached:

solving the equation:

(r-1)([tex]r^{2} - 2r + 1[/tex]) = 0

[tex]r^{2} - 2r + 1[/tex] = 0

r = 1

So we have 3 solution [tex]r_{1} = r_{2} =r_{3}[/tex] = 1

replacing in the main solution

Yh =  [tex]e^{t}[/tex] +  [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex]

The t and [tex]t^{2}[/tex] is used because we must have 3 solution  linearly independent

Particular solution:

We must propose a Yp solution:

Yp = [tex]c_{1} (t^{3} + t^{2} + t + c_{4} )e^{t} + c_{2} t + c_{3}[/tex]

Y'p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + c_{1}( 3t^{2} + 2t + 1 )e^{t} + c_{2}[/tex]

Y''p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + c_{1}(6t + 2)e^{t}[/tex]

Y'''p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + 6c_{1}e^{t}[/tex]

Y'''p - 3*Y''p + 3*Y'p - Yp = [tex]e^{t} - t + 21[/tex]

[tex]6c_{1}e^{t} - 18c_{1} te^{t} - 6c_{1} e^{t} + 6c_{1} te^{t} + 9c_{1} t^{2} e^{t} + 3c_{1}e^{t} + 3c_{2} - c_{2} t -  c_{3}[/tex] = [tex]e^{t} - t + 21[/tex]

equalizing coefficients of the same function:

- 12c_{1} = 0

9c_{1} = 0

3c_{1} = 0

c_{1} = 0

3c_{2} - c_{3} = 21 => c_{5} = [tex]\frac{1}{3}[/tex]

-c_{2} = -1

c_{2} = 1

c_{3} = -18

Then we have:

Y = [tex]e^{t}[/tex] +  [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex] + t - 18

Answer:

Part 1.

In 12 hours 30 units of  Humulin R insulin in 300 mL is to be infused.

So, per hour = [tex]\frac{30}{12}= 2.5[/tex] units are to be infused.

Part 2.

Now 30 units Humulin R insulin in 300 mL of normal saline (NS).

So, 2.5 units will be in : [tex]\frac{300\times2.5}{30}= 25[/tex] mL

Hence. 2.5 units Humulin R insulin in 25 mL of normal saline (NS).

Answer:

You must pick at least 13 pennies to be sure of getting at least five from the same year.

Step-by-step explanation:

You have:

12 1967 pennies

7 1868 pennies

11 1971 pennies.

how many must you pick to be sure of getting at least five pennies from the same year?

This value is the multiplication of the number of different pennies by the antecessor of the number of pennies you want, added by 1.

So

You have 3 differennt pennies

You want to get at least five from the same year.

[tex]3*4 + 1 = 13[/tex]

You must pick at least 13 pennies to be sure of getting at least five from the same year.

For example, if you pick 12 pennies, you can have four from each year.

Adding three values, 13 is the smallest number that you need at least one term of the addition being equal or bigger than 5.

To be sure of getting at least five pennies from the same year, you need to pick 35 pennies in total.

To ensure that you get at least five pennies from the same year, you need to consider the worst-case scenario. In this case, the worst-case scenario is where you pick pennies from each of the three different years first before getting five from the same year. So, you need to pick the maximum number of pennies from each year first before reaching the desired goal. The maximum number of pennies you need to pick is:

12 + 7 + 11 + 5 = 35 pennies

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Answer:

a) The patient weighs 75.15kg = 75.1kg, rounded to the nearest tenth.

b) The patient should 75.1mg a day of the drug.

c) The patient should receive 0.37mL per dose, rounded to the nearest hundreth.

Step-by-step explanation:

These problems can be solved by direct rule of three, in which we have cross multiplication.

a. How many kilograms does the patient weigh?

The problem states that patient weighs 167lb. Each lb has 0.45kg. So:

1 lb - 0.45kg

167 lb - xkg

[tex]x = 167*0.45[/tex]

[tex]x = 75.15[/tex]kg

The patient weighs 75.15kg = 75.1kg, rounded to the nearest tenth.

b. How many milligrams should the patient receive per day?

The drug has 1mg/kg. The patient weighs 75.1kg. So

1 mg - 1 kg

x mg - 75.1kg

[tex]x = 75.1[/tex]mg

The patient should 75.1mg a day of the drug.

c. How many milliliters should the patient receive per dose?

The drug is SubQ, q12h. This means that the drug is administered twice a day, so there are 2 doses. 75.1mg of the drug are administered a day. so:

2 doses - 75.1mg

1 dose - xmg

[tex]2x = 75.1[/tex]

[tex]x = \frac{75.1}{2}[/tex]

[tex]x = 37.5[/tex] SubQ, q12h

For each dose, the patient should receive 37.5mg. Each 40mg of the drug has 0.4mL. So:

40mg - 0.4ml

37.5mg - xmL

[tex]40x = 0.4*37.4[/tex]

[tex]x = \frac{0.4*37.4}{40}[/tex]

[tex]x = 0.374mL[/tex]

The patient should receive 0.37mL per dose, rounded to the nearest hundreth.

Answer:

Proved

Step-by-step explanation:

To prove that if n is a perfect square, then n+1 can never be a perfect square

Let n be a perfect square

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

Let [tex]n+1 = y^2[/tex]

Subtract to get

[tex]1 = y^2-x^2 =(y+x)(y-x)[/tex]

Solution is y+x=y-x=1

This gives x=0

So only 0 and 1 are consecutive integers which are perfect squares

No other integer satisfies y+x=y-x=1

Answer:

K=6

Step-by-step explanation:

we know that

A function is: a relation from a set of inputs to a set of possible outputs where each input is related to exactly one output

so

In this problem

If the x-coordinate of the first point is equal to the x-coordinate of the second point, then the relation will not be a function

Remember that the x-coordinate is the input value and the y-coordinate is the output value

Equate the x-coordinates and solve for k

[tex]1.5k-4=-0.5k+8[/tex]

[tex]1.5k+0.5k=8+4[/tex]

[tex]2k=12[/tex]

[tex]k=6[/tex]

therefore

If the value of k is 6 the, the relation will not be a function

Final answer:

To determine when the relation A is not a function, set the first elements of the ordered pairs equal to each other. Solving for k, we find that k must equal 6 to make the relation not a function, as this value of k would result in the same input for both pairs, violating the definition of a function.

Explanation:

For a relation to be considered a function, each input value must be associated with exactly one output value. Looking at the provided relation A = {(1.5k-4, 7), (-0.5k+8, 15)}, we can see that there are two ordered pairs. For A to not be a function, the first element of both ordered pairs must be the same because this would mean that a single input is associated with two different outputs, which violates the definition of a function.

We set the first elements equal to each other to find the value of k that would make the relation not a function: 1.5k - 4 = -0.5k + 8. Solving this equation, we add 0.5k to both sides and add 4 to both sides to obtain 2k = 12, and then divide both sides by 2 to find that k = 6.

Therefore, when k equals 6, the relation A will not be a function because both ordered pairs will have the first element equal to 5, which corresponds to two different output values (7 and 15).

Answer:

MNP is a Congruent Triangle according to Side-Side-Side Congruence.

Step-by-step explanation:

Whenever we talk about Euclidean Geometry and Congruence of Triangles. We are taking into account that in any given plane, three given points, in this case, M, N, and P is and its segments between two points make up a Triangle.

In this case MNP and an identical one and PNM

To be called congruent, it's necessary to have the same length each side and when it comes to angles, congruent angles have the same measure.

A postulate, cannot be proven since it's self-evident. And there's one that fits for this case which says

"Every SSS (Side-Side-Side) correspondence is a congruence"

So this is why we can classify MNP as Side-Side-Side congruence since its segments are the same size MN, NP, and MP for both of them. The order of the letters does not matter.

Answer:

[tex]m\angle B=m\angle C=120^{\circ}[/tex]

[tex]m\angle A=m\angle D=60^{\circ}[/tex]

Step-by-step explanation:

Trapezoid ABCD is isosceles trapezoid, because AB = CD (given). In isosceles trapezoid, angles adjacent to the bases are congruent, then

Since BK ⊥ AD, the triangle ABK is right triangle. In this triangle,  AB = 8, AK = 4. Note that the hypotenuse AB is twice the leg AK:

[tex]AB=2AK.[/tex]

If in the right triangle the hypotenuse is twice the leg, then the angle opposite to this leg is 30°, so,

[tex]m\angle ABK=30^{\circ}[/tex]

Since BK ⊥ AD, then BK ⊥ BC and

[tex]m\angle KBC=90^{\circ}[/tex]

Thus,

[tex]m\angle B=30^{\circ}+90^{\circ}=120^{\circ}\\ \\m\angle B=m\angle C=120^{\circ}[/tex]

Now,

[tex]m\angle A=m\angle D=180^{\circ}-120^{\circ}=60^{\circ}[/tex]

Answer:

y = -0.9x+10.1

Step-by-step explanation:

The equation of the line is:

[tex]y=mx+b[/tex]

You have been asked to stimate m and b. To do so, first find the product between each pair of x and y and the value of x squared:

[tex]\left[\begin{array}{cccc}x&y&x*y&x^2\\1&9&9&1\\3&7&21&9\\5&7&35&25\\7&3&21&49\\9&2&18&81\end{array}\right][/tex]

Then calculate the total sum of all columns:

[tex]\left[\begin{array}{cccc}x&y&x*y&x^2\\1&9&9&1\\3&7&21&9\\5&7&35&25\\7&3&21&49\\9&2&18&81\\\bold{25}&\bold{28}&\bold{104}&\bold{165}\end{array}\right][/tex]

m can be calculated following the next equation:

[tex]m=\frac{\frac{\sum{xy}-\sum{y}}{n}}{\sum{x^2}-\frac{(\sum{x})^2}{n}}[/tex]

where n is the number of (x, y) couples (5 in our case).

Replacing the values calculated previously:

[tex]m=\frac{104-\frac{25*28}{5} }{165-\frac{25^2}{5} }=\frac{104-\frac{700}{5} }{165-\frac{625}{5} } = \frac{104-140}{165-125 } = \frac{-36}{40} = -0.9[/tex]

For b:

[tex]b=\bar{y}- m\bar{x}=\frac{\sum{y}}{n}-m\frac{\sum{x}}{n}=\frac{28}{5}-(-0.9)\frac{25}{5}= \frac{28}{5}+\frac{22.5}{5}=\frac{50.5}{5}=10.1[/tex]

In the figure attached you can see the points given and the stimated line.

Answer:

[tex]y(x) = e^{3x} [Acos x+Bsin x][/tex]

Step-by-step explanation:

Given is a differential equation

[tex]y" - 6y' +10y=0.[/tex]

We have characteristic equation as

[tex]m^2-6m+10 =0[/tex]

The above quadratic cannot be factorised hence use formula

[tex]m=\frac{6+/-\sqrt{36-40} }{2} \\=3+i, 3-i[/tex]

Hence general solution would be

[tex]y(x) = e^{3x} [Acos x+Bsin x][/tex]

Answer:

2 mL

Step-by-step explanation:

Given:

Volume of water for injection = 23 mL

Resulting concentration = 200,000 units per mL

Amount of drug in the vial = 5,000,000 units

Now,

Let the final volume of the solution be 'x' mL

Now, concentration = [tex]\frac{\textup{units of the powder}}{\textup{Total volume of the soltuion}}[/tex]

thus,

200,000 = [tex]\frac{\textup{5,000,000}}{\textup{x}}[/tex]

or

x = 25 mL

also,

Total volume 'x' = volume of water + volume of powder

or

25 mL = 23 + volume of powder

or

Volume of powder = 2 mL

Answer: The relation is a function.

Step-by-step explanation: In this situation, we are given a relation in the form of a graph and we are asked if it represents a function. In this situation, we would you something called the vertical line test. In other words, if we can draw a vertical line that passes through more than one point on the graph, then the relation is not a function. Notice that in this problem, it's impossible to draw a vertical line that passes through more than one point on the graph so the relation is a function.

Therefore, this relation must be a function.

Answer: There were 469,532 American prison for drug related offenses in 2015.

Step-by-step explanation:

Since we have given that

Number of Americans in prison for drug related offenses in 1980 = 40,900

Rate of increment in Americans in prison from 1980 to 2015 = 1048%

So, Number of Americans who were in prison for drug related offenses in 2015 is given by

[tex]\dfrac{100+1048}{100}\times 40900\\\\\dfrac{1148}{100}\times 40900\\\\=1148\times 409\\\\=469,532[/tex]

Hence, there were 469,532 American prison for drug related offenses in 2015.

In this type of problems what we have to do is unit conversion. In order to do so we need all the equivalences which we will be mentioning during the explanation of the problem:

First of all the answer is asked to be in acre-feet and we can see the data we are getting from the rain is in [tex]in*Km^{2}[/tex] not even a volume unit.

To calculate the volume of poured rain we need to have both numbers in the same units, we will convert [tex]Km^{2}[/tex] to [tex]in^{2}[/tex] using the equivalence [tex]1 Km^{2}=1550001600in^{2}[/tex] like this:

[tex]101Km^{2}*\frac{1550001600in^{2} }{1Km^{2}}=156550161600in^{2}[/tex]

it is possible now to calculate the volume ([tex]Volume_{cuboid}=Area*Height[/tex]) like this:

[tex]Volume_{cuboid}=156550161600in^{2}*2.5in=391375404000in^{3}[/tex]

Now we just need to convert this volume to acre-feet and we will do so using the equivalence [tex]1acre-foot=751271680in^{3}[/tex] like this:

[tex]391375404000in^{3}*\frac{1acre-foot}{751271680in^{3}}=5199.50403658 acre-feet[/tex]

5199.50403658 acre-feet would be the answer to our problem

Answer: The total purchase price is $ 29,032

Step-by-step explanation:

Hi, to solve this problem you have to solve the percentages of each taxes first.

So :

The next step is adding the taxes results to the speedboat cost.

so:

speedboat = $24,500

speedboat + taxes = $24,500 + $1,347.5 +$3,185 = $29,032

The total purchse price for the speedboat is $29,032.

Answer:

I think that what you are trying to show is:  If [tex]s[/tex] is irrational and [tex]r[/tex] is rational, then [tex]r+s[/tex]  is rational. If so, a proof can be as follows:

Step-by-step explanation:

Suppose that [tex]r+s[/tex] is a rational number. Then [tex]r[/tex] and [tex]r+s[/tex] can be written as follows

[tex]r=\frac{p_{1}}{q_{1}}, \,p_{1}\in \mathbb{Z}, q_{1}\in \mathbb{Z}, q_{1}\neq 0[/tex]

[tex]r+s=\frac{p_{2}}{q_{2}}, \,p_{2}\in \mathbb{Z}, q_{2}\in \mathbb{Z}, q_{2}\neq 0[/tex]

Hence we have that

[tex]r+s=\frac{p_{1}}{q_{1}}+s=\frac{p_{2}}{q_{2}}[/tex]

Then

[tex]s=\frac{p_{2}}{q_{2}}-\frac{p_{1}}{q_{1}}=\frac{p_{2}q_{1}-p_{1}q_{2}}{q_{1}q_{2}}\in \mathbb{Q}[/tex]

This is a contradiction because we assumed that [tex]s[/tex] is an irrational number.

Then [tex]r+s[/tex] must be an irrational number.

Final answer:

Explaining the impact of selling additional sandwiches on MTA's profits.

Explanation:

To prepare the schedule showing the impact on MTA's profits:

Calculate the total income from selling 800 additional sandwiches: $2.85 x 800 = $2,280.

Calculate the total cost of producing 800 sandwiches: $5 x 800 = $4,000.

Subtract the cost from the income to find the profit impact: $2,280 - $4,000 = lower profit.

Explanation (150 words): Adding 800 sandwiches at a cost of $5 each while selling at $2.85 results in a loss, impacting MTA's profits negatively. This shift reduces the overall profitability due to the increased production costs outweighing the revenue generated from the additional sandwich sales.

Answer:

The dosage has 1.2ml

Step-by-step explanation:

This problem can be solved as a rule of three problem.

In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.

When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too.

When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease.

Unit conversion problems, like this one, is an example of a direct relationship between measures.

First step: The first step is the conversion from mcg to mg.

The bottle has 60 mcg. How much is this in ml? 1 mcg has 0.001mg. So:

1 mcg - 0.001mg

60mcg - x mg

x = 60*0.001

x = 0.06mg

Final step: The bottle of elixir contains 0.05 mg per 1 ml. Calculate the dosage in ml.

The dosage has 0.06 mg, so:

0.05mg - 1 ml

0.06mg - xml

0.05x = 0.06

[tex]x = \frac{0.06}{0.05}[/tex]

x = 1.2ml

The dosage has 1.2ml

Answer:

OBSERVATIONAL

Step-by-step explanation:

Let`s see the two kind of experiments mentioned:

In this case, the patient's parent only answer a questionnairewith no intervention from the doctors into the pacients at all, so it's an observational  experiment

Answer:

The concentration of the drug is 50mg/mL.

Step-by-step explanation:

This problem can be solved by a rule of three.

In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.

When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too.

When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease.

In this problem, as the number of mg increases, so does the number of mL. It means the we have a direct rule of three.

Tha problem states that the antibody is available in a prefilled syringe containing 40 mg/0.8 mL. Calculate the concentration of drug on a mg/mL basis.

The problem wants to know how many mg are there in a mL of the drug.

40mg - 0.8mL

x mg - 1 mL

0.8x = 40

[tex]x = \frac{40}{0.8}[/tex]

x = 50mg.

The concentration of the drug is 50mg/mL.

Answer:

[tex] (x-4)^2+(y+5)^2=90[/tex]

Step-by-step explanation:

The equation of a circle of radius r, centered at the point (a,b) is

[tex](x-a)^2+(y-b)^2=r^2[/tex]

We already know the center is at [tex](4,-5)[/tex], we are just missing the radius. To find the radius, we can use the fact that the circle passes through the point (7,4), and so the radius is just the distance from the center to this point (see attached image). So we find the distance by using distance formula between the points (7,4) and (4,-5):

radius[tex]=\sqrt{(7-4)^2+(4-(-5))^2}=\sqrt{3^2+9^2}=\sqrt{90}[/tex]

And now that we know the radius, we can write the equation of the circle:

[tex] (x-4)^2+(y-(-5))^2=\sqrt{90}^2[/tex]

[tex] (x-4)^2+(y+5)^2=90[/tex]

Answer:

number of 1 bedroom units are 96

number of 1 bedroom units are 64

number of 1 bedroom units are 32

Step-by-step explanation:

Let the number of 1 bedroom units be 'a'

number of 2 bedroom units be 'b'

and,

number of 3 bedroom units be 'c'

now,

according to the question

a + b + c = 192 ................. (1)

also,

b + c = a ..............(2)

and,

a = 3c ...................(3)

now,

substituting value of 'a' from 3  into 2, we get

b + c = 3c

or

b = 2c ...................(4)

also,

from 3 and 1

3c + b + c = 192

or

4c + b = 192  ................(5)

now from 4 and 5,

4c + 2c = 192

or

6c = 192

or

c = 32 units

now, substituting c in equation 4, we get

b = 2 × 32 = 64 units

and, substituting c in equation (3), we get

a = 3 × 32 = 96 units

Therefore,

number of 1 bedroom units are 96

number of 1 bedroom units are 64

number of 1 bedroom units are 32

Answer:

The solution is: [tex](x_{1}, x_{2}, x_{3}) = (1,0,-4)[/tex]

Step-by-step explanation:

The Gauss-Jordan elimination method is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.

We have the following system:

[tex]2x_{1} - x_{2} + 3x_{3} = -10[/tex]

[tex]x_{1} - 2x_{2} + x_{3} = -3[/tex]

[tex]x_{1} - 5x_{2} + 2x_{3} = -7[/tex]

This system has the following augmented matrix:

[tex]\left[\begin{array}{ccc}2&-1&3|-10\\1&-2&1|-3\\1&-5&2| -7\end{array}\right][/tex]

To make the reductions easier, i am going to swap the first two lines. So

[tex]L1 <-> L2[/tex]

Now the matrix is:

[tex]\left[\begin{array}{ccc}1&-2&1|-3\\2&-1&3|-10\\1&-5&2| -7\end{array}\right][/tex]

Now we reduce the first row, doing the following operations

[tex]L2 = L2 - 2L1[/tex]

[tex]L3 = L3 - L1[/tex]

So, the matrix is:

[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&3&1|-4\\0&-3&1| -4\end{array}\right][/tex]

Now we divide L2 by 3

[tex]L2 = \frac{L2}{3}[/tex]

So we have

[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&1&\frac{1}{3}|\frac{-4}{3}\\0&-3&1| -4\end{array}\right][/tex]

Now we have:

[tex]L3 = 3L2 + L3[/tex]

So, now we have our row reduced matrix:

[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&1&\frac{1}{3}|\frac{-4}{3}\\0&0&2| -8\end{array}\right][/tex]

We start from the bottom line, where we have:

[tex]2x_{3} = -8[/tex]

[tex]x_{3} = \frac{-8}{2}[/tex]

[tex]x_{3} = -4[/tex]

At second line:

[tex]x_{2} + \frac{x_{3}}{3} = \frac{-4}{3}[/tex]

[tex]x_{2} - \frac{4}{3} = -\frac{4}{3}[/tex]

[tex]x_{2} = 0[/tex]

At the first line

[tex]x_{1} -2x_{2} + x_{3} = -3[/tex]

[tex]x_{1} - 4 = -3[/tex]

[tex]x_{1} = 1[/tex]

The solution is: [tex](x_{1}, x_{2}, x_{3}) = (1,0,-4)[/tex]

Answer:

0001 0110 1110 1011 0010

Step-by-step explanation:

First decimal number = 11540

second decimal number = 5972

BCD of 0 to 9 decimal number

Decimal number                               BCD

       0                                                  0000

       1                                                    0001

       2                                                   0010

       3                                                   0011

       4                                                   0100

       5                                                   0101

       6                                                   0110

       7                                                    0111

       8                                                    1000

       9                                                    1001

Hence, we can write above two numbers in BCD form as

11540 => 0001 0001 0101 0100 0000

5972 => 0101 1001 0111 0010

So, sum of 11540 and 5972 in BCD form can be given by

       0001 0001 0101 0100 0000

        +       01 01 1001 01 11  001 0

      0001  01 10 11 10 10 11  001 0

So, the sum of 11540 and 5972 in BCD form is

                  0001 0110 1110 1011 0010

Answer:

The probability that exactly 5 are unable to complete the  race is 0.1047

Step-by-step explanation:

We are given that 25% of all who enters a race do not complete.

30 have entered.

what is the probability that exactly 5 are unable to complete the  race?

So, We will use binomial

Formula : [tex]P(X=r) =^nC_r p^r q^{n-r}[/tex]

p is the probability of success i.e. 25% = 0.25

q is the probability of failure =  1- p  = 1-0.25 = 0.75

We are supposed to find the probability that exactly 5 are unable to complete the  race

n = 30

r = 5

[tex]P(X=5) =^{30}C_5 (0.25)^5 (0.75)^{30-5}[/tex]

[tex]P(X=5) =\frac{30!}{5!(30-5)!} \times(0.25)^5 (0.75)^{30-5}[/tex]

[tex]P(X=5) =0.1047[/tex]

Hence the probability that exactly 5 are unable to complete the  race is 0.1047

Final answer:

The height of the pile after 3 weeks is 43 181/192 inches.

Explanation:

To find the height of the pile after 3 weeks, we need to add the increase in height for each week. The initial height of the pile is 17 3/4 inches. For each week, the height increases by 8 7/12 inches.

So after 1 week, the height is (17 3/4 + 8 7/12) inches.

After 2 weeks, the height is [(17 3/4 + 8 7/12) + 8 7/12] inches.

And after 3 weeks, the height is [((17 3/4 + 8 7/12) + 8 7/12) + 8 7/12] inches.

Let's calculate:

So, after 3 weeks, the height of the pile of newspapers is 43 181/192 inches.

Final answer:

The number 4320 is an abundant number because the sum of its proper divisors exceeds 4320. All positive multiples of 6 greater than 6 are abundant because their smallest divisor, 6, is the sum of its proper divisors, and they have additional divisors that increase the sum.

Explanation:

To determine if the number 4320 is perfect, abundant, or deficient, we must compare the sum of its proper divisors (excluding the number itself) with the number. A perfect number is equal to the sum of its divisors. A number is abundant if the sum of its divisors is greater than the number, and it is deficient if the sum is less.

For 4320, the divisors are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 27, 30, 32, 36, 40, 45, 48, 54, 60, 64, 72, 80, 90, 96, 108, 120, 135, 144, 160, 180, 192, 216, 240, 270, 288, 320, 360, 432, 480, 540, 576, 720, 864, 1080, and 1440. Their sum is greater than 4320, so it is an abundant number.

Regarding positive multiples of 6 greater than 6 being abundant, the smallest divisor of such a multiple is always 6, which is already the sum of its divisors (1, 2, and 3). Since there are additional divisors beyond 1, 2, and 3, the sum of divisors must exceed the number, making it abundant.