Factor completely 3x4 − 48. 3(x2 − 4)(x2 + 4) 3(x − 2)(x + 2)(x + 2)(x + 2) 3(x − 2)(x + 2)(x2 + 4) 3(x − 2)(x + 2)(x2 − 4)

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.

Answer:

The Final amount in the account after 5 years will be $108.26

Step-by-step explanation:

Hello, great question. These types are questions are the beginning steps for learning more advanced Equations.

Since we are talking about compounding interest we can use the Exponential Growth Formula to calculate the interest over the next couple of years. The formula is the following,

[tex]y = a*(1+r)^{t}[/tex]

Where:

We first need to calculate the 2% interest for the first year,

[tex]y = 100*(1+0.02)^{1}[/tex]

[tex]y = 100*1.02[/tex]

[tex]y = 102[/tex]

So after the first year the account will have $102. Now we can use the $102 to calculate the next 4 years of interest using the new interest rate of 1.5%

[tex]y = 102*(1+0.015)^{4}[/tex]

[tex]y = 102*(1.015)^{4}[/tex]

[tex]y = 102*1.0614[/tex]

[tex]y = 108.26[/tex]

The Final amount in the account after 5 years will be $108.26

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

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:

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 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]

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:

The probability that we shall make exactly n selections is [tex]P(X = n)=(\frac{17}{35})^{n-1}\frac{18}{35}[/tex].

Step-by-step explanation:

It is given that an urn contains 4 white and 4 black balls and we randomly choose 4 balls. If 2 of them are white and 2 are black, we stop.

The total number of ways to select exactly 2 white and 2 black balls.

[tex]^4C_2\times ^4C_2=\frac{4!}{2!(4-2)!}\times \frac{4!}{2!(4-2)!}=6\times 6=36[/tex]

The total number of ways to select 4 balls from 8 balls is

[tex]^8C_4=\frac{8!}{4!(8-4)!}=\frac{8\times 7\times 6\times 5\times 4!}{4\times 3\times 2\times 1\times !4!}=70[/tex]

The probability of selecting exactly 2 white and 2 black balls is

[tex]p=\frac{36}{70}=\frac{18}{35}[/tex]

The probability of not selecting exactly 2 white and 2 black balls is

[tex]q=1-p=1-\frac{18}{35}=\frac{17}{35}[/tex]

If we not get exactly 2 white and 2 black balls, then we replace the balls in the urn and again randomly select 4 balls.

The probability that we shall make exactly n selections is

[tex]P(X = n)=(q)^{n-1}p[/tex]

[tex]P(X = n)=(\frac{17}{35})^{n-1}\frac{18}{35}[/tex]

Therefore the probability that we shall make exactly n selections is [tex]P(X = n)=(\frac{17}{35})^{n-1}\frac{18}{35}[/tex].

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

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:

<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:

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

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.

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:

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:

Step-by-step explanation:

P\left ( defective item\right )=0.035

Using binomial distribution

Where p= probability of success

q=probability of failure

Here p=0.035

q=1-0.035=0.965

[tex]^nC_{r}P^{r}q^{n-r}[/tex]

(i)for exactly 3 defective items i.e. r=3

P[tex]\left ( r=3\right )[/tex]=[tex]^{44}C_{3}[/tex][tex]\left ( 0.035\right )^{3}\left ( 0.965\right )^{44-3}[/tex]

P=[tex]\frac{44!}{41!3!}\times \left ( 0.035\right )^3\left ( 0.965\right )^{41}[/tex]

P=0.1317

(ii)No defective item i.e. r=0

P[tex]\left ( r=0\right )[/tex]=[tex]^{44}C_{0}[/tex][tex]\left ( 0.035\right )^{0}[/tex][tex]\left ( 0.965\right )^{44-0}[/tex]

P=[tex]\frac{44!}{44!0!}\times \left ( 0.035\right )^0\left ( 0.965\right )^{44}[/tex]

P=0.2085

(iii)At least 1 defective item

P=1-P(zero defective item)

P=1-[tex]^{44}C_{1}\left ( 0.035\right )^{1}\left ( 0.965\right )^{44-1}[/tex]

P=1-[tex]\frac{44!}{43!1!}\times \left ( 0.035\right )^1[/tex][tex]\left ( 0.965\right )^{43}[/tex]

P=0.6671

(a) The probability of exactly 3 defective items: approximately 0.1318
(b) The probability of no defective items: approximately 0.2085
(c) The probability of at least 1 defective item: approximately 0.7915

(a) Probability of Exactly 3 Defective Items
To find the probability of getting exactly 3 defective items in a run of 44, we will use the binomial probability formula:

[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]

where
- [tex]\( X \)[/tex] is the random variable representing the number of defective items,
- [tex]\( k \)[/tex] is the number of defective items we want to find the probability for (in this case, 3),
- [tex]\( \binom{n}{k} \)[/tex] is the number of combinations of [tex]\( n \)[/tex] items taken [tex]\( k \)[/tex] at a time.

So, substituting into the formula:
[tex]\[ P(X = 3) = \binom{44}{3} (0.035)^3 (1 - 0.035)^{41} \][/tex]

After calculating, we find:
[tex]\[ P(X = 3) \approx 0.13177807290504395 \][/tex]

Thus, the probability of getting exactly 3 defective items is approximately 0.1318.

(b) Probability of No Defective Items
To determine the probability of having no defective items, we calculate:

[tex]\[ P(X = 0) = \binom{44}{0} (0.035)^0 (1 - 0.035)^{44} \][/tex]

Here:
[tex]\[ \binom{44}{0} = 1 \\ (0.035)^0 = 1 \\ (1 - 0.035)^{44} \approx 0.20854596293662794[/tex]

Thus, the probability of having no defective items is approximately 0.2085.

(c) Probability of At Least 1 Defective Item
To find the probability of at least 1 defective item, it is easier to calculate the complement—the probability of having no defective items—and subtract it from 1:

[tex]\[ P(X \geq 1) = 1 - P(X = 0) \][/tex]

From part (b), we know [tex]\( P(X = 0) \)[/tex]:
[tex]\[ P(X \geq 1) = 1 - 0.20854596293662794 \approx 0.791454037063372 \][/tex]

Therefore, the probability of having at least 1 defective item is approximately 0.7915.

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:

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.

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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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.

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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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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:163.54$

Step-by-step explanation:

Given data

Volume of Storage(V)=[tex]{10m^3}[/tex]

Length=2breadth

Let Length be L,Breadth be & height be H

therefore

10=LBH

Now substitutes the values

10=2[tex]{B^2}H[/tex]

5=[tex]{B^2}H[/tex]

Now cost for base is [tex]{C_1}=2{B^2}\times10[/tex]

Cost for side walls is[tex]{C_2}={2LH}\times6+2BH}\times6[/tex]

Now total cost(C)=[tex]{C_1}+{C_2}[/tex]

C=20[tex]{B^2}H[/tex]+[tex]{2LH}\times6[/tex]+[tex]2BH}\times6[/tex]

C=20[tex]{B^2}H[/tex]+24BH+[tex]12BH[/tex]

C=[tex]20{B^2}+36B\times\frac{5}{B^{2}}[/tex]

Now Differentiating With respect to Breadth to get minimum cost

[tex]\frac{\mathrm{d} C}{\mathrm{d} B}=0[/tex]

[tex]we\ get\ B=\sqrt[3]{4.5}=1.65m[/tex]

[tex]L=3.30m[/tex]

[tex]H=1.836m[/tex]

and mimimum cost C

[tex]{C=163.54\$}[/tex]

To find the cost of materials for the cheapest container, we need to minimize the total cost. The total cost function is C(x) = 20x^2 + 180x. However, there is no minimum cost for the container since it cannot have a negative width, resulting in a cost of $0.

Let's denote the width of the rectangular storage container as x. According to the given information, the length of the base is twice the width, so the length would be 2x. The height can be calculated by dividing the volume of the container by the area of the base. Therefore, the height would be 10/(x * 2x) = 5/(2x).

The cost of the base would be the area of the base multiplied by the cost per square meter, which is 10 * (x * 2x) = 20x^2. The cost of the sides would be the sum of each side multiplied by the cost per square meter, which is 6 * (2x * 5) + 6 * (x * 5) + 6 * (2x * 5) = 180x.

To find the cost of materials for the cheapest container, we need to minimize the total cost, which is the sum of the cost of the base and the cost of the sides. Therefore, the total cost function is C(x) = 20x^2 + 180x.

To find the minimum cost, we can take the derivative of C(x) with respect to x, set it equal to 0, and solve for x. The value of x that satisfies this equation will give us the width of the container that minimizes the cost.

C'(x) = 40x + 180 = 0

40x = -180

x = -4.5

Since the width cannot be negative, we disregard this solution.

Therefore, there is no minimum cost for the container since it cannot have a negative width. In this case, the cost of materials for the cheapest container would be $0.

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[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]

See more about Divergence Theorem at brainly.com/question/6960786

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 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

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

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]

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:

  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%.