Jill has a swimming pool in her backyard. It is shaped like a rectangle and measures approximately 16.3 feet wide and 26.2 feet long. It is an average of 5 feet deep. During a few hot weeks during the summer, some water evaporates from the pool, and Jill needs to add 8 inches of water to the depth of the pool, using her garden hose. Although her water pressure varies, the water flows through Jill’s garden hose at an average rate of 10 gallons/minute.

How many liters will Jill need to add to her pool to return the water level to its original depth? How many gallons of water is this? Reminder: Volume = Length x Width x Depth

The answer for the sets corresponding to the given conditions is as follows:

a) A ∩ B = {4, 6, 8, 10, 12}.

b) A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

c)  B ∩ C = {}.

d) B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8...}.

e) Set A = {3, 5, 7}   and  Set B = {2,4, 6, 8, 10, 12}

Given:

Set A =  {x ∈ N : 3 ≤ x ≤ 13}

Set B =  {x ∈ N : x is even}

Set C  = {x ∈ N : x is odd}.

Solve each option:

(a) Find A ∩ B (the intersection of sets A and B):

Set A contains natural numbers from 3 to 13 (inclusive): A = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

Set B contains even natural numbers: B = {2, 4, 6, 8, 10, 12, ...}.

The intersection of A and B includes even numbers that are between 3 and 13: A ∩ B = {4, 6, 8, 10, 12}.

(b) Find A ∪ B (the union of sets A and B):

Set A contains natural numbers from 3 to 13 (inclusive): A = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

Set B contains even natural numbers: B = {2, 4, 6, 8, 10, 12, ...}.

The union of A and B includes all numbers from both sets: A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

(c) Find B ∩ C (the intersection of sets B and C):

Set B contains even natural numbers: B = {2, 4, 6, 8, 10, 12, ...}.

Set C contains odd natural numbers: C = {1, 3, 5, 7, 9, 11, ...}.

The intersection of B and C is the empty set, as there are no numbers that are both even and odd.

(d) Find B ∪ C (the union of sets B and C):

Set B contains even natural numbers: B = {2, 4, 6, 8, 10, 12, ...}.

Set C contains odd natural numbers: C = {1, 3, 5, 7, 9, 11, ...}.

The union of B and C includes all natural numbers: B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ...}.

(e) For the given example:

Set A = {3, 5, 7}

Set B = {2,4, 6, 8, 10, 12}

This example satisfies the conditions A ∩ B = {3, 5} and A ∪ B = {2, 3, 5, 7, 8}

The intersection and Unioun of all the sets is found from A and B.

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The intersection of sets A and B is {4, 6, 8, 10, 12}. The union of sets A and B is {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. The intersection of sets B and C is {}. The union of sets B and C is {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

(a) To find the intersection of sets A and B, we need to identify the elements that are common to both sets. In set A, we have: {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. In set B, we have: {4, 6, 8, 10, 12}. The elements that are common to both sets are: {4, 6, 8, 10, 12}. Therefore, A ∩ B = {4, 6, 8, 10, 12}.

(b) To find the union of sets A and B, we need to combine all the elements from both sets. In set A, we have: {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. In set B, we have: {4, 6, 8, 10, 12}. Combining these sets gives us: {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. Therefore, A ∪ B = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

(c) To find the intersection of sets B and C, we need to identify the elements that are common to both sets. In set B, we have: {4, 6, 8, 10, 12}. In set C, we have: {3, 5, 7, 9, 11, 13}. The elements that are common to both sets are: {}. Therefore, B ∩ C = {}.

(d) To find the union of sets B and C, we need to combine all the elements from both sets. In set B, we have: {4, 6, 8, 10, 12}. In set C, we have: {3, 5, 7, 9, 11, 13}. Combining these sets gives us: {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. Therefore, B ∪ C = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

Answer:  The required solution of the given IVP is

[tex]y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.[/tex]

Step-by-step explanation:  We are given to find the solution of the following initial value problem :

[tex]y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.[/tex]

Let [tex]y=e^{mx}[/tex] be an auxiliary solution of the given differential equation.

Then, we have

[tex]y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.[/tex]

Substituting these values in the given differential equation, we have

[tex]m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.[/tex]

So, the general solution of the given equation is

[tex]y(x)=Ae^x+Be^{-x},[/tex] where A and B are constants.

This gives, after differentiating with respect to x that

[tex]y^\prime(x)=Ae^x-Be^{-x}.[/tex]

The given conditions implies that

[tex]y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

and

[tex]y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]

Adding equations (i) and (ii), we get

[tex]2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.[/tex]

From equation (i), we get

[tex]\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.[/tex]

Substituting the values of A and B in the general solution, we get

[tex]y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.[/tex]

Thus, the required solution of the given IVP is

[tex]y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.[/tex]

Step-by-step explanation:

Assuming that the 350 taking history and the 300 taking math each includes the 270 taking both history and math, then:

N(H or M) = N(H) + N(M) − N(H and M)

N = 350 + 300 − 270

N = 380

There are 380 first-year students taking history or mathematics.

[tex]f(t)=\begin{cases}9&\text{for }0\le t<2\\(t-5)^2&\text{for }2\le t<5\\2te^{6t}&\text{for }t\ge5\end{cases}[/tex]

and presumably 0 for [tex]t<0[/tex]. We can express [tex]f(t)[/tex] in terms of the unit step function,

[tex]u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t<c\end{cases}[/tex]

[tex]f(t)=9(u(t)-u(t-2))+(t-5)^2(u(t-2)-u(t-5))+2te^{6t}u(t-5)[/tex]

Quick explanation: [tex]9u(t)=9[/tex] for [tex]t\ge0[/tex], and [tex]9u(t-2)=9[/tex] for [tex]t\ge2[/tex]. So subtracting these will cancel the value of 9 for all [tex]t\ge2[/tex] and leave us with the value of 9 over the interval we want, [tex]0\le t<2[/tex]. The same reasoning applies for the other 3 terms.

Recall the time displacement theorem:

[tex]\mathcal L_s\{f(t-c)u(t-c)\}=e^{-sc}\mathcal L_s\{f(t)\}[/tex]

By this property, we have

[tex]\mathcal L_s\{9u(t)\}=\mathcal L_s\{9\}=\dfrac9s[/tex]

[tex]\mathcal L_s\{9u(t-2)\}=e^{-2s}\mathcal L_s\{9\}=\dfrac{9e^{-2s}}s[/tex]

[tex]\mathcal L_s\{(t-5)^2u(t-2)\}=\mathcal L_s\{((t-2)-3)^2u(t-2)\}[/tex]

[tex]=e^{-2s}\mathcal L_s\{(t-3)^2\}=\left(\dfrac2{s^3}-\dfrac6{s^2}+\dfrac9s\right)e^{-2s}[/tex]

[tex]\mathcal L_s\{(t-5)^2u(t-5)\}=e^{-5s}\mathcal L_s\{t^2\}=\dfrac{2e^{-5s}}{s^3}[/tex]

[tex]\mathcal L_s\{2te^{6t}u(t-5)\}=\mathcal L_s\{2e^{30}(t-5)e^{6(t-5)}+10e^{30}e^{6(t-5)}\}[/tex]

[tex]=2e^{30-5s}\mathcal L_s\{te^{6t}+5e^{6t}\}=2e^{30-5s}\left(\dfrac1{(s-6)^2}+\dfrac5{s-6}\right)[/tex]

Putting everything together, we end up with

[tex]\boxed{\mathcal L_s\{f(t)\}=\dfrac{(2-6s)e^{-2s}-2e^{-5s}}{s^3}+\dfrac9s-\dfrac{2e^{30-5s}(29-5s)}{(s-6)^2}}[/tex]

Explanation:

"Any normal distribution" may have arbitrary mean and standard deviation. The "standard normal distribution" has a mean of zero and a standard deviation of 1.

Answer:

495

Step-by-step explanation:

To find this you have to find the LCM of the two times which in this case is 33 and 45. The LCM of those two is 495.

The minutes it will be before this happens again is 495.

The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value.

Assume that an airline’s flights for Miami leave every 33 minutes and flights from Dallas leave every 45 minutes.

To find the LCM of the two times which in this case is 33 and 45.

Factor;

33 = 3 x 11

45 = 5 x 3 x 3

Thus, The LCM of those two is 495.

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Answer:

d

Step-by-step explanation:

13 cant be divided equally nor cubed  because its not an even number u can try to give all thirteen of then

The simpler version of the initial problem is arranging three people in a line. There are three choices for the first spot, two for the second, and one for the third, which results in a total of six possible arrangements. This involves the principle of permutation in combinatorics.

The subject of the given problem can be defined as permutations. If we're looking for a simpler version of it, we should choose a problem which still involves line-up or arrangement of a smaller number of people. Hence, the best option is: 'In how many ways can three people be lined up for a picture?'

To solve this simpler problem, we consider the number of available spots for each person in the line. For the first spot, there are 3 people that could be selected. After the first person is chosen, there are only 2 people left for the second spot. Lastly, there is only 1 person left for the third spot. So, the total number of ways we can line up 3 people for a picture is 3*2*1 = 6 ways.

This is a basic principle called permutation in combinatorics which is a fundamental concept in mathematics that deals with counting, both as a means and an end in obtaining results.

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The probability that has been chosen Bag 1 is 0.2087.

Given that, bag 1 contains 10 blue marbles, while bag 2 contains 15 blue marbles.

Here we have;

Bag 1 contains 10 blue marbles

Bag 2 contains 15 blue marbles

Chosen a bag at random and throw 5 red marbles in it.

[tex]Required Probability = P(\frac{Bag 1}{2 red and 3 blue marbles})[/tex]

= [tex]\frac{P(bag 1)\cap(2 Red \ and \ 3 blue)}{P(2 \ red \ and \ 3 \ blue \ marbles)}[/tex]

= [tex]\frac{\frac{1}{2}\times ^6C_2\times^{10}C_3}{\frac{1}{2}\times^6C_2\times^{10}C_3+\frac{1}{2}\times^6C_2\times^{15}C_3}[/tex]

= 0.2087

Therefore, the probability that has been chosen Bag 1 is 0.2087.

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To find the probability of choosing Bag 1 given there are 2 red marbles among the 5 chosen marbles, we can use Bayes' theorem to calculate the probability.

To solve this problem, we can use Bayes' theorem to find the probability that Bag 1 was chosen given there are exactly 2 red marbles among the 5 chosen marbles. Let's denote Bag 1 as event A and Bag 2 as event B.

The answer to the question is the probability of choosing Bag 1 given there are exactly 2 red marbles among the 5 chosen marbles.

The average value of [tex]\(f(x)\) over \([-1, 3]\)[/tex] is 13. The function equals its average value at certain [tex]\(x\)[/tex] values.

To find the average value of the function [tex]\( f(x) = 4x^3 - 3x^2 \)[/tex] over the interval [tex]\([-1, 3]\)[/tex], we'll first calculate the definite integral of the function over that interval and then divide it by the length of the interval.

The formula for the average value of a function [tex]\( f(x) \)[/tex] over an interval [tex]\([a, b]\)[/tex] is given by:

[tex]\[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \][/tex]

For the function [tex]\( f(x) = 4x^3 - 3x^2 \)[/tex] over the interval [tex]\([-1, 3]\)[/tex], we have:

[tex]\[ \text{Average value} = \frac{1}{3-(-1)} \int_{-1}^{3} (4x^3 - 3x^2) \, dx \][/tex]

First, let's find the integral:

[tex]\[ \int (4x^3 - 3x^2) \, dx = \frac{4}{4}x^4 - \frac{3}{3}x^3 + C \]\[ = x^4 - x^3 + C \][/tex]

Now, we'll evaluate this integral from -1 to 3:

[tex]\[ \left[ x^4 - x^3 \right]_{-1}^{3} = (3^4 - 3^3) - ((-1)^4 - (-1)^3) \]\[ = (81 - 27) - (1 - (-1)) \]\[ = 54 - 2 \]\[ = 52 \][/tex]

So, the definite integral is 52.

Now, we'll find the average value:

[tex]\[ \text{Average value} = \frac{1}{3-(-1)} \times 52 \]\[ = \frac{52}{4} \]\[ = 13 \][/tex]

The average value of the function [tex]\( f(x) = 4x^3 - 3x^2 \)[/tex] over the interval [tex]\([-1, 3]\) is 13.[/tex]

To find the values of [tex]\( x \)[/tex] in the interval for which the function equals its average value, we set [tex]\( f(x) \)[/tex] equal to 13 and solve for [tex]\( x \):[/tex]

[tex]\[ 4x^3 - 3x^2 = 13 \][/tex]

This equation can be solved numerically. By using methods like graphing, Newton's method, or a numerical solver, we can find the roots of this equation within the interval [tex]\([-1, 3]\).[/tex] These roots will be the [tex]\( x \)[/tex] values where the function equals its average value.

The average value of [tex]\(f(x) = 4x^3 - 3x^2\)[/tex]over [tex]\([-1, 3]\) is 13.[/tex]

The values of x that make [tex]\(f(x) = 13\)[/tex] are approximately -0.771, 1.979, and 2.792.

To find the average value of the function [tex]\(f(x) = 4x^3 - 3x^2\)[/tex]over the interval [-1, 3], we can use the formula for the average value of a function over an interval [a, b] :

[tex]\[ A = \frac{1}{b - a} \int_{a}^{b} f(x) dx. \][/tex]

Determine the integral of f(x) :

  To find the integral of f(x), we first compute the antiderivative of[tex]\(4x^3 - 3x^2\):[/tex]

  [tex]\[ \int (4x^3 - 3x^2) dx = x^4 - x^3. \][/tex]

Evaluate the integral over [tex]\([-1, 3]\):[/tex]

  Now, let's find [tex]\(\int_{-1}^{3} (4x^3 - 3x^2) dx\):[/tex]

[tex]\[ \int_{-1}^{3} (4x^3 - 3x^2) dx = (x^4 - x^3) \Big|_{-1}^{3}. \][/tex]

  Evaluate the antiderivative at 3 and -1 :

  - When [tex]\(x = 3\), \(3^4 - 3^3 = 81 - 27 = 54\),[/tex]

  - When [tex]\(x = -1\), \((-1)^4 - (-1)^3 = 1 + 1 = 2\).[/tex]

  Thus, [tex]\[ \int_{-1}^{3} (4x^3 - 3x^2) dx = 54 - 2 = 52. \][/tex]

Find the average value over [-1, 3]:

  Using the result from step 2, the average value over [tex]\([-1, 3]\)[/tex]  is:

 [tex]\[ A = \frac{1}{3 - (-1)} \cdot 52 = \frac{1}{4} \cdot 52 = 13. \][/tex]

Therefore, the average value of [tex]\(f(x) = 4x^3 - 3x^2\)[/tex] over the interval [tex]\([-1, 3]\) is \(13\).[/tex]

Now, let's find the values of x in [-1, 3] for which f(x) = 13):

[tex]\[ 4x^3 - 3x^2 = 13. \][/tex]

Rearrange the equation:

[tex]\[ 4x^3 - 3x^2 - 13 = 0.[/tex]

This cubic equation is more complex to solve algebraically. The approximate solutions can be obtained numerically, using a graphing calculator, computational software, or by iterative methods.

Using an approximation method, we get the following solutions (rounded to three decimal places):

1. [tex]\( x \approx -0.771 \),[/tex]

2. [tex]\( x \approx 1.979 \),[/tex]

3. [tex]\( x \approx 2.792 .[/tex]

These are the values of x in the interval [-1, 3] for which f(x) = 13).

Question :

Find the average value of the function over the given interval and all values of x in the interval for which the function equals its average value. (Round your answer to three decimal places.) f(x) = 4x3 − 3x2, [−1, 3]

Answer: 0.1841

Step-by-step explanation:

Given: Mean : [tex]\mu=8.2[/tex]

Standard deviation : [tex]\sigma = 1.34[/tex]

The formula to calculate z-score is given by :_

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x= 9, we have

[tex]z=\dfrac{9-8.2}{1.34}\approx0.60[/tex]

For x= 10, we have

[tex]z=\dfrac{10-8.2}{1.34}\approx1.34[/tex]

The P-value = [tex]P(0.6<z<1.34)=P(z<1.34)-P(z<0.6)[/tex]

[tex]=0.9098773-0.7257469=0.1841304\approx0.1841[/tex]

Hence, the probability that at your next exam, you will have a stress level between 9 and 10 = 0.1841

Answer:

that would be 4253.28 feet

that would be 1296.4 m

Step-by-step explanation:

I think that's the answer and I hope it helps :)

Answer:

v.w = 22

Step-by-step explanation:

We are given

r = <8, 8, -6>; v = <3, -8, -3>; w = <-4, -2, -6>

and we need to evaluate v.w

Using the formula:  v.w = vxwx+vywy+vzwz

Putting values and solving:

v.w = 3(-4)+(-8)(-2)+(-3)(-6)

v.w = -12+16+18

v.w = 22

So, v.w = 22

Answer:

JL=2 units

Step-by-step explanation:

we know that

If k is the midpoint in the line JL

then

JL=JK+KL

JK=KL

substitute the given values

2x-5=3x-8

Solve for x

3x-2x=-5+8

x=3

so

JK=2x-5=2(3)-5=1 units

KL=3x-8=3(3)-8=1 units

therefore

JL=JK+KL=1+1=2 units

Let [tex]A(t)[/tex] denote the amount of salt in the tank at time [tex]t[/tex]. We're told that [tex]A(0)=0[/tex].

For [tex]0\le t\le15[/tex], the salt flows in at a rate of (1/5 lb/gal)*(5 gal/min) = 1 lb/min. When the regulating mechanism fails, 20 lbs of salt is dumped and no more salt flows for [tex]t>15[/tex]. We can capture this in terms of the unit step function [tex]u(t)[/tex] and Dirac delta function [tex]\delta(t)[/tex] as

[tex]\text{rate in}=u(t)-u(t-15)+20\delta(t-15)[/tex]

(in lb/min)

The salt from the mixed solution flows out at a rate of

[tex]\text{rate out}=\left(\dfrac{A(t)\,\mathrm{lb}}{50+(5-5)t\,\mathrm{gal}}\right)\left(5\dfrac{\rm gal}{\rm min}\right)=\dfrac A{10}\dfrac{\rm lb}{\rm min}[/tex]

Then the amount of salt in the tank at time [tex]t[/tex] changes according to

[tex]\dfrac{\mathrm dA}{\mathrm dt}=u(t)-u(t-15)+20\delta(t-15)-\dfrac A{10}[/tex]

Let [tex]\hat A(s)[/tex] denote the Laplace transform of [tex]A(t)[/tex], [tex]\hat A(s)=\mathcal L_s\{A(t)\}[/tex]. Take the transform of both sides to get

[tex]s\hat A(s)-A(0)=\dfrac1s-\dfrac{e^{-15s}}s+20e^{-15s}-\dfrac1{10}\hat A(s)[/tex]

Solve for [tex]\hat A(s)[/tex], then take the inverse of both sides.

[tex]\hat A(s)=\dfrac{\frac{10-10e^{-15s}}{s^2}+\frac{200e^{-15s}}s}{10s+1}[/tex]

[tex]\implies\boxed{A(t)=10-10e^{-t/10}+\left(30e^{3/2-t/10}-10\right)u(t-15)}[/tex]

Answer:

The distance is 9.17 feet.

Step-by-step explanation:

The ramp, vertical distance it is lifted, and the ground form a right triangle, whose hypotenuse the ramp, and whose base and perpendicular are the ground and the lifted distance respectively.

Thus we have a triangle whose hypotenuse [tex]H[/tex] is 10 feet, the perpendicular [tex]P[/tex] is 4 feet, and a base [tex]B[/tex] feet.

The Pythagorean theorem gives:

[tex]H^2=P^2+B^2[/tex]

We substitute the values [tex]H=10[/tex], [tex]P =4[/tex] and solve for B:

[tex]B=\sqrt{H^2-P^2} =\sqrt{10^2-4^2} =9.17.[/tex]

Thus the distance is 9.17 feet.

Answer:

the Answer is ≈ 9.17 feet

Step-by-step explanation:

it is correct on edge  2020

Answer:

<CBA

Step-by-step explanation:

The angle name could be either

<ABC or <CBA

The vertex must be in the middle

Answer:

Yes, the answer is CBA

Step-by-step explanation:

The only way to get a term of degree 25 is by taking 3 copies of [tex]x^8[/tex], 1 copy of [tex]x[/tex], and 6 copies of 1. Then the coefficient of [tex]x^{25}[/tex] is

[tex]\dbinom{10}3\dbinom71\dbinom66=\dbinom{10}{3,1,6}=\dfrac{10!}{3!6!}=\boxed{840}[/tex]

Answer:

239,580 ways of seating

Step-by-step explanation:

11 students will be divided into 2 groups. One group of 7 people and one group of 4 people. So first we need to find the number of ways of dividing 11 students into these 2 groups.

First group is of 7 people. We have to select 7 people out of 11. The order of selection does not matter so this is a combination problem. Selecting 7 people from 11 can be expressed as 11C7.

Formula for combination is:

[tex]^{n}C_{r}=\frac{n!}{r!(n-r)!}[/tex]

For the given case this would be:

[tex]^{11}C_{7}=\frac{11!}{7! \times 4!}=330[/tex]

So, there are 330 ways of selecting a group of 7 from 11 students. When these 7 students are selected the remaining 4 will go to the other group. So, we can say there are 330 ways to divide the 11 students in groups of 7 and 4. Note that if you start with group of 4 students, the answer will still the same because 11C4 is also equal to 330.

Next we have to arrange 7 students on a round table. The number of possible arrangements would be = (7 - 1)! = 6! = 720

Similarly, to arrange 4 people on a round table, the number of possible arrangements would be = (4 - 1)! = 3! = 6

Since, for each selection of the 330 groups, there are 720 + 6 possible seating arrangements, so the total number of possible seating arrangements would be:

330 ( 720 + 6) = 239,580 ways

Thus, there are 239,580 ways of seating 11 students.

There are 86400 ways the students can be seated in the dining hall.

There are two circular tables in the dining hall, one can seat 7 people and the other can seat 4 people. The students need to be seated in a way that they can be accommodated on these two tables.

The number of ways the students can be seated is:

1) Assign the even-numbered students to the table that can seat 7 people. There are 6 even-numbered students.

2) Assign the odd-numbered students to the table that can seat 4 people. There are 5 odd-numbered students.

3) Calculate the number of ways these students can be arranged on their respective tables. For the table with 7 seats, there are 6 students to be seated, so the number of ways is 6!. For the table with 4 seats, there are 5 students to be seated, so the number of ways is 5!.

4) Multiply the number of ways for each table to get the total number of ways to seat the students: 6! * 5! = 720 * 120 = 86400.

Therefore, there are 86400 ways the students can be seated.

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Answer:

The length of a side of the square is 22 inches.

Step-by-step explanation:

Let each side of square be = s

Let each side of triangle be = s - 8

Perimeter of square, p₁ = 4s

Perimeter of triangle = p₂ = 3s

                                          = 3(s-8)

                                          = 3s - 24

Therefore, according to the question

p₁ - p₂ = 46

4s - (3s - 24) = 46

4s - 3s + 24 = 46

s = 46 - 24

s = 22

The length of a side of the square is 22 inches.

Answer:

The required answer is [tex]x+C=\frac{1}{2}\ln|\frac{2+x-y}{y-x}|[/tex].

Step-by-step explanation:

The given differential equation is

[tex]\frac{dy}{dx}=(x-y+1)^2[/tex]

Substitute u=x-y+1 in the above equation.

[tex]\frac{du}{dx}=1-\frac{dy}{dx}[/tex]

[tex]\frac{dy}{dx}=1-\frac{du}{dx}[/tex]

[tex]1-\frac{du}{dx}=u^2[/tex]

[tex]1-u^2=\frac{du}{dx}[/tex]

Using variable separable method, we get

[tex]dx=\frac{du}{1-u^2}[/tex]

Integrate both the sides.

[tex]\int dx=\int \frac{du}{1-u^2}[/tex]

[tex]x+C=\frac{1}{2}\ln|\frac{1+u}{1-u}|[/tex]      [tex][\because \int \frac{dx}{a^2-x^2}=\frac{1}{2a}\n|\frac{a+x}{a-x}|+C][/tex]

Substitute u=x-y+1 in the above equation.

[tex]x+C=\frac{1}{2}\ln|\frac{1+x-y+1}{1-(x-y+1)}|[/tex]

[tex]x+C=\frac{1}{2}\ln|\frac{2+x-y}{y-x}|[/tex]

Therefore the required answer is [tex]x+C=\frac{1}{2}\ln|\frac{2+x-y}{y-x}|[/tex].

Answer:

190578024 ways.

Step-by-step explanation:

We are asked to find the number of ways in which a committee of 5 be chosen from 120 employees to interview prospective applicants.

We will use combinations to solve our given problem.

[tex]_{r}^{n}\textrm{C}=\frac{n!}{(n-r)!r!}[/tex], where,

n = Total number of items,

r = Number of items being chosen at a time.

Upon substituting our given values in above formula, we will get:

[tex]_{5}^{120}\textrm{C}=\frac{120!}{(120-5)!5!}[/tex]

[tex]_{5}^{120}\textrm{C}=\frac{120!}{115!*5!}[/tex]

[tex]_{5}^{120}\textrm{C}=\frac{120*119*118*117*116*115!}{115!*5*4*3*2*1}[/tex]

[tex]_{5}^{120}\textrm{C}=\frac{120*119*118*117*116}{5*4*3*2*1}[/tex]

[tex]_{5}^{120}\textrm{C}=\frac{120*119*118*117*116}{120*1}[/tex]

[tex]_{5}^{120}\textrm{C}=\frac{119*118*117*116}{1}[/tex]

[tex]_{5}^{120}\textrm{C}=\frac{190578024}{1}[/tex]

Therefore, the committee of five can be chosen from 120 employees in 190578024 ways.

Answer:

$1850.

Step-by-step explanation:

We are asked to find amount of interest earned in one year on an amount of $5000.

We will use simple interest formula to solve our given problem.

[tex]I=Prt[/tex], where,

I = Amount of interest earned,

r = Annual interest rate in decimal form.

t = Time in years.

Let us convert our given interest rate in decimal form.

[tex]37\%=\frac{37}{100}=0.37[/tex]

Upon substituting our given value in simple interest formula, we will get:

[tex]I=\$5000\times 0.37\times 1[/tex]

[tex]I=\$5000\times 0.37[/tex]

[tex]I=\$1850[/tex]

Therefore, an amount of $1850 is earned as interest in one year.

[tex]\vec F(x,y,z)=\langle x^2,y^2,z^2\rangle\implies\mathrm{div}\vec F(x,y,z)=2x+2y+2z[/tex]

By the divergence theorem,

[tex]\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\iiint_R\mathrm{div}\vec F(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz[/tex]

where [tex]R[/tex] the region with [tex]S[/tex] as its boundary. Convert to spherical coordinates, taking

[tex]\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi[/tex]

Then the volume integral is

[tex]\displaystyle\iiint_R\mathrm{div}\vec F(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz[/tex]

[tex]=2\displaystyle\int_0^{2\pi}\int_0^\pi\int_0^5(x+y+z)\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta[/tex]

[tex]=2\displaystyle\int_0^{2\pi}\int_0^\pi\int_0^5(\cos\theta\sin\varphi+\sin\theta\sin\varphi+\cos\varphi)\rho^3\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta=\boxed{0}[/tex]

In this exercise we have to use the divergent theorem to calculate the flow of the given equation, so we will find that:

[tex]\int\limits \int\limits \int\limits_R {divF(x,y,z)} \, dx dy dz= 0[/tex]

So from the given equation, we will find that:

[tex]\int\limits \int\limits_S {F} \, ds = \int\limits \int\limits \int\limits_R {div F(x, y, z) } \, dx dy dz[/tex]

where [tex]R[/tex] the region with [tex]S[/tex] as its boundary. Convert to spherical coordinates, taking:

[tex]\left[\begin{array}{c}x= \rho cos(\theta) sin(\phi) \\y= \rho sin(\theta) sin(\phi) \\z= \rho cos (\phi) \end{array}\right[/tex]

Then the volume integral is:

[tex]\int\limits \int\limits \int\limits_R {divF(x,y,z)} \, dxdydz\\= 2 \int\limits^{2\pi}_0 \int\limits^{\pi}_0 \int\limits^{5}_0 {(x+y+z)\rho ^2 sin(\phi) d(\rho) d(\phi) d(\theta)= 0[/tex]

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Answer:

  about 34.9%

Step-by-step explanation:

The probability of not making a marking error is 0.9. The probability of doing that 10 times independently is 0.9^10 ≈ 0.34868 ≈ 34.9%.

I'm going to guess that you meant to include parentheses somewhere, so that the ODE is supposed to be

[tex]y'=\dfrac{y^2x}{y^3+x^3}+\dfrac yx[/tex]

Then substitute [tex]y(x)=xv(x)[/tex] so that [tex]y'(x)=xv'(x)+v(x)[/tex]. Then

[tex]xv'+v=\dfrac{x^3v^2}{x^3v^3+x^3}+v[/tex]

[tex]xv'=\dfrac{v^2}{v^3+1}[/tex]

which is separable as

[tex]\dfrac{v^3+1}{v^2}\,\mathrm dv=\dfrac{\mathrm dx}x[/tex]

Integrate both sides: on the left,

[tex]\displaystyle\int\frac{v^3+1}{v^2}\,\mathrm dv=\int\left(v+\frac1{v^2}\right)\,\mathrm dv=\dfrac12v^2-\dfrac1v[/tex]

The other side is trivial. We end up with

[tex]\dfrac12v^2-\dfrac1v=\ln|x|+C[/tex]

Solve in terms of [tex]y(x)[/tex]:

[tex]\boxed{\dfrac{y^2}{2x^2}-\dfrac xy=\ln|x|+C}[/tex]

Answer:  Each muffin tin contains 3 oz of batter.

Step-by-step explanation:  Given that a baker pours  108 oz of batter into 36 muffin tins such that each tin has same amount of batter.

We are to calculate the quantity of batter in each tin.

We will be using the UNITARY method to solve the given problem.

Quantity of batter in 36 muffin tins = 108 oz.

Therefore, the quantity of batter in 1 muffin tin is given by

[tex]Q_t=\dfrac{108}{36}=3~\textup{oz}.[/tex]

Thus, each muffin tin contains 3 oz of batter.

By dividing 108 oz of batter by 36 muffin tins, you find that each tin contains 3 oz of batter. This simple division problem helps distribute the batter evenly. Each tin thus gets exactly 3 oz.

To find out how much batter is in each muffin tin, you need to divide the total amount of batter by the number of muffin tins.

Here are the steps:

So, there are 3 oz of batter in each muffin tin.

Non zero vector x perpendicular to u and v : x = [tex][ \frac{-39}{14} x_2 - \frac{26}{14} x_3 , x_2 , x_3, \frac{17}{14}x_2 + \frac{23}{14} x_3 ][/tex]

Given, v= [-2,-8,-7,2] u= [6,7,-2,8]

Let the vector be x = [[tex]x_1 , x_2 , x_3, x_4[/tex]]

Now x is non xero vector perpendicular to vector 'v' and 'u' .

So,

x . v = 0

[tex]-2x_1 - 8x_2 - 7x_3 + 2x_4 = 0[/tex] .........1

x . u = 0

[tex]6x_1 + 7x_2 -2x_3 + 8x_4 = 0[/tex] .........2

Solve 1 and 2 to eliminate [tex]x_4[/tex] .

Multiply 1 with 4 to make the coefficients of [tex]x_4[/tex] same .

[tex]-8x_1 - 32x_2 - 28x_3 + 8x_4 = 0[/tex]

[tex]6x_1 + 7x_2 -2x_3 + 8x_4 = 0[/tex]

Subtract two equations,

[tex]-14x_1 -39x_2 -26x_3 = 0[/tex]

[tex]-14x_1 = 39x_2 + 26x_3[/tex]

[tex]x_1 = \frac{-39}{14} x_2 - \frac{26}{14} x_3[/tex]

From equation 1,

[tex]x_4 = x_1 + 4x_2 + \frac{7}{2} x_3[/tex]

[tex]x_4 = \frac{-39}{14} x_2 - \frac{26}{14} x_3+ 4x_2 + \frac{7}{2} x_3\\\\x_4 = \frac{17}{14}x_2 + \frac{23}{14} x_3[/tex]

Thus x = [tex][ \frac{-39}{14} x_2 - \frac{26}{14} x_3 , x_2 , x_3, \frac{17}{14}x_2 + \frac{23}{14} x_3 ][/tex]

[tex]x_2 = [-39/14 , 1 , 0 , 17/14] + x_3[-26/14, 0 , 1 , 23/14 ][/tex]

[tex]x_1 , x_3[/tex] are arbitrary .

For every value of [tex]x_2 , x_3[/tex] vector x is obtained.

Know more about non zero vector,

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To find a vector x that is perpendicular to vectors v and u, we can use the cross product.

To find a vector x that is perpendicular to vectors v and u, we can use the cross product. The cross product of two vectors is a vector that is perpendicular to both of them. To find the cross-product, we can use the formula:

x = (v2u3 - v3u2, v3u1 - v1u3, v1u2 - v2u1)

Plugging in the values, we get:

x = (-8(-2) - (-7)(7), (-7)(6) - (-2)(-2), (-2)(8) - (-8)(6)) = (1, 52, -32)

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Answer:

1577.20 ft³

Step-by-step explanation:

Cube of length = 12 ft = a

Hole diameter which is cutout = 4 ft = d

Hole radius which is cutout = 4/2 =2 ft = r

Volume of the cube = a³

⇒Volume of the cube = a×a×a

⇒Volume of the cube = 12×12×12

⇒Volume of the cube = 1728 ft³

The hole cut out will be in the shape of a cylinder

Volume of cylinder = πr²h

⇒Volume of cylinder = π×2²×12

⇒Volume of cylinder = 150.79 ft³

Now volume of the solid figure with hole cut out is

Volume of the cube - Volume of cylinder

=1728 - 150.79

=1577.20 ft³

∴ Volume of solid figure not including hole cutout is 1577.20 ft³

Answer:

26

Step-by-step explanation:

Converting 11010 to base 10.

1*24=16

1*23=8

0*22=0

1*21=2

0*20=0

Adding all to get Ans=26_10

Step2 converting 26_10 to 10

The equation calculation formula for 26_10 number to 10 is like this below.

10|26  

10|2|6

10|2|2

Ans:26_10

Assuming the given number is in base 2, we have

[tex]11010_2=2^4+2^3+2^1=16+8+2=26_{10}[/tex]

Answer with explanation:

(A)

It is given that, A is invertible, That is inverse of matrix exist.

    [tex]|A|=|A^{-1}|\neq 0[/tex]

That is,  [tex]|A|=|A^{-1}|=1[/tex], is incorrect Statement.

False

(B)

If a Matrix has , either any row or column has all entry equal to Zero, then value of Determinant is equal to 0.

Any matrix with a row of all zeros has a determinant of 1 ,is incorrect Statement.

False

(C)

The Meaning of Singular matrix is that , then Determinant of Singular Matrix is equal to Zero.

For, a n×n , matrix, whether n is Odd or even

  [tex]A^{T}= -A\\\\|A^{T}|=|-A|=(-1)^n|A|[/tex]

So, the statement, If A is a skew symmetric matrix,  [tex]A^{T}= -A[/tex],and A has size n x n then A must be singular if n is odd ,is incorrect Statement.

False