A survey of 1,107 tourists visiting Orlando was taken. Of those surveyed:

268 tourists had visited the Magic Kingdom

258 tourists had visited Universal Studios

68 tourists had visited both the Magic Kingdom and LEGOLAND

79 tourists had visited both the Magic Kingdom and Universal Studios

72 tourists had visited both LEGOLAND and Universal Studios

36 tourists had visited all three theme parks

58 tourists did not visit any of these theme parks

How many tourists only visited the LEGOLAND (of these three)?

Answer:

R={(10,2),(10,4),(10,6),(2,2),(2,4),(2,6),(4,2),(4,6)}

Step-by-step explanation:

We are given that

A={10,1,2,3,4,5,61}

B={0,1,2,3,4,5,6,7,81}

We are given that R be the relation from  A to B

R={gcd(a,b)=2,a[tex]\inA,b\inB[/tex]}

Gcd=Greatest common divisor  of a and b.

We have to find the elements in R

(10,2)=2,(10,4)=2,(10,6)=2

(2,2)=2,(2,4)=2,(2,6)=2

(4,2)=2,(4,6)=2

Therefore, R={(10,2),(10,4),(10,6),(2,2),(2,4),(2,6),(4,2),(4,6)}

Answer:

The solution for this system is [tex]x = 5, y = 3[/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]5x + 3y = 16[/tex]

[tex]-2x + y = -13[/tex]

This system has the following augmented matrix.

[tex]\left[\begin{array}{ccc}5&3&16\\-2&1&-13\end{array}\right][/tex]

The first step is dividing the first line by 5. So:

[tex]L_{1} = \frac{L_{1}}{5}[/tex]

We now have

[tex]\left[\begin{array}{ccc}1&\frac{3}{5}&\frac{16}{5}\\-2&1&-13\end{array}\right][/tex]

Now i want to reduce the first row, so I do:

[tex]L_{2} = L_{2} + 2L_{1}[/tex]

So we have

[tex]\left[\begin{array}{ccc}1&\frac{3}{5}&\frac{16}{5}\\0&\frac{11}{5}&-\frac{33}{5}\end{array}\right][\tex].

Now, the first step to reduce the second row is:

[tex]L_{2} = \frac{5L_{2}}{11}[/tex]

So we have:

[tex]\left[\begin{array}{ccc}1&\frac{3}{5}&\frac{16}{5}\\0&1&-3\end{array}\right][/tex].

Now, to reduce the second row, we do:

[tex]L_{1} = L_{1} - \frac{3L_{2}}{5}[/tex]

And the augmented matrix is:

[tex]\left[\begin{array}{ccc}1&0&5\\0&1&-3\end{array}\right][/tex]

The solution for this system is [tex]x = 5, y = 3[/tex].

Answer:

This is it:

Step-by-step explanation:

don't click any links

Answer: a. 1

Step-by-step explanation:

Given : Population mean :[tex]\mu=28\ cm[/tex]

Standard deviation : [tex]\sigma=1.3\ cm[/tex]

Let n be the number of times  a beach ball with a diameter of 26.7 cm differ from the mean.

Then, we have

[tex]28-n(1.3)=26.7\\\\\Rightarrow\ 1.3n=28-26.7\\\\\Rightarrow\ 1.3 n=1.3\\\\\Rightarrow\ n=1[/tex]

Hence, a beach ball with a diameter of 26.7 cm is 1 standard deviation differ from the mean.

Answer:

You can use the given hint as follows:

Step-by-step explanation:

Let [tex]A[/tex] be a square matrix that is a skew-symmetric matrix. Since the matrix [tex]R={\bf x}^{T}A{\bf x}[/tex] is matrix of size [tex]1\times 1[/tex] then it can be identified with an scalar. It is clear that [tex]R=R^{T}[/tex]. Then applying the properties of transposition we have

[tex]({\bf x}^{T}A{\bf x})^{T}=({\bf x}^{T})A^{T}({\bf x}^{T})^{T}={\bf x}^{T}(-A){\bf x}=-{\bf x}^{T}A{\bf x}[/tex]

Then,

[tex]{\bf x}^{T}A{\bf x}+{\bf x}^{T}A{\bf x}=0[/tex]

[tex]2{\bf x}^{T}A{\bf x}=0[/tex]

Then,

[tex]{\bf x}^{T}A{\bf x}=0[/tex]

For all column vector [tex]{\bf x}[/tex] of size [tex]n\times 1[/tex] .

Answer:

The term algorithm derives from the title of the Persian and Muslim mathematician of the 9th century Abu Abdullah Muhammad ibn Musa Al-Khwarizmi.

Step-by-step explanation:

The term algorithm derives from the title of the Persian and Muslim mathematician of the 9th century Abu Abdullah Muhammad ibn Musa Al-Khwarizmi. He was a mathematician, astronomer, and geographer during  Abbasid Caliphate and he was a researcher at the House of Wisdom in Baghdad.

His systematic method of analyzing linear and quadratic problems resulted in algebra. It is a title derived from all his  collection of 830 book title on the topic, "The Compendious Book on Completion and Balancing Calculation."

Answer:

The correct option is b.

Step-by-step explanation:

The given function is

[tex]f(x)=-5x+1[/tex]

We need to find the value of f(-2). It means we have to find the value of given function at x=-2.

Substitute x=-2 in the given function to find the value of f(-2).

[tex]f(-2)=-5(-2)+1[/tex]

On simplification we get

[tex]f(-2)=10+1[/tex]

[tex]f(-2)=11[/tex]

The value of f(-2) is 11. Therefore the correct option is b.

Answer:

462 000mg

Step-by-step explanation:

1gram = 1000milligrams

Hence...462 grams...,

; 462 × 1000 = 462 000mg

462 grams is equal to 462,000 milligrams .

To convert grams ( g ) to milligrams ( mg ), you need to use the following conversion factor:

1 gram ( g ) = 1000 milligrams ( mg )

milligram (mg) is equal to 1/1000 grams (g).

1 mg = (1/1000) g = 0.001 g

The mass m in grams (g) is equal to the mass m in milligrams (mg) divided by 1000:

m(g) = m(mg) / 1000

This means that there are 1000 milligrams in 1 gram. Now, let's use this conversion factor to calculate 462 grams in milligrams:

462 grams * ( 1000 mg / 1 g ) = 462,000 milligrams

So, 462 grams is equal to 462,000 milligrams.

To know more about milligrams click here :

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Answer:

Maximum safe height can be reached by ladder = 15.03. ft

Step-by-step explanation:

Given,

Let's assume the maximum safe height of wall = h

angle formed between ladder and ground = 70°

length of ladder = 16 ft

From the given data, it can be seen that ladder will form a right angle triangle structure with the wall

So,from the concept of trigonometry,

[tex]Sin70^o\ =\ \dfrac{\textrm{maximum safe height of wall}}{\textrm{length of ladder}}[/tex]

[tex]=>Sin70^o\ =\ \dfrac{h}{16\ ft}[/tex]

[tex]=>\ h\ =\ 16\times Sin70^o[/tex]

=> h = 16 x 0.9396

=> h = 15.03 ft

So, the maximum safe height that can be reached by the ladder will be 15.03 ft.

Answer:

20 L of 22% solution and 8 L of 36% solution

Step-by-step explanation:

Volume of 22% solution + volume of 36% solution = volume of 26% solution

x + y = 28

Acid in 22% solution + acid in 36% solution = acid in 26% solution

0.22x + 0.36y = 0.26(28)

0.22x + 0.36y = 7.28

Solve the system of equations using either elimination or substitution.  I'll use substitution:

x = 28 − y

0.22(28 − y) + 0.36y = 7.28

6.16 − 0.22y + 0.36y = 7.28

0.14y = 1.12

y = 8

x = 28 − y

x = 20

The chemist should use 20 L of 22% solution and 8 L of 36% solution.

Answer:

There should be mixed 20 L of the 22% acid solution with 8L of the 36% acid solution

Step-by-step explanation:

We are mixing two acids.

x = liters of 22% acid solution

y = liters of 36% acid solution

x + y = 28    (total liters)

0.22x +0.36y = 0.26* 28  

Since x+y=28 means y = 28-x

Now we will use substitution to find x

0.22x + 0.36(28-x) = 0.26 * 28

0.22x + 10.08 - 0.36x = 7.28

0.14x = 2.8

x = 20

y = 28 - 20 = 8  

⇒ We use 20 liters of the 22% solution to be mixed with 8 liters of the 36% solution to form  28l of a  26% acid solution.

Answer: (1,0)

Step-by-step explanation: What is the x-intercept of the linear equation y = 4x – 4?

y = 4x - 4

x-intercept ⇒ y = 0

which means that we need to substitute the y by 0.

0 = 4x-4

4x = 4

x = 1

As it is a linear equation, 1st degree, there is only one point.

This way, the linear y = 4x - 4 intercept x on point (1,0)

Answer:

the answer is D

Step-by-step explanation:

Answer with Step-by-step explanation:

Part 1)

we know that

[tex]e^{i\theta }=cos(\theta )+isin(\theta )[/tex]

thus [tex]-i=e^{\frac{-i\times (4n-1)\pi }{2}}[/tex]

thus [tex](-i)^i=(e^{\frac{-i\times (4n-1)\pi }{2}})^i\\\\(-i)^i=e^{\frac{-i^2\times (4n-1)\pi }{2}}=e^{\frac{(4n-1)\pi }{2}}\\\\\therefore (-i)^i=e^{\frac{(4n-1)\pi }{2}}[/tex] where 'n' is any integer

Part 2)

We have [tex]-1=e^{(2n+1)\pi }\\\\\therefore (-1)^{i}=(e^{i(2n+1)\pi })^{i}\\\\(-1)^i=(e^{i^2(2n+1)\pi })\\\\(-1)^i=e^{-(2n+1)\pi }[/tex] where 'n' is any integer

Answer:

The house with an area 3,600 square feet is more unusual

Step-by-step explanation:

Given:

Number of houses surveyed = 583

Mean price = $176,678

Standard deviation = $61,029

Mean house size = 1,676 square ft

standard deviation = 582 square ft

Now,

the as z score = [tex]\frac{\textup{(X - mean )}}{\textup{standard deviation}}[/tex]

thus,

for selling value of $357,000

z score = [tex]\frac{\textup{(357,000 - 176,678 )}}{\textup{61,029}}[/tex]

or

z score = 2.95

and for house with an area 3,600 square feet

z score =  [tex]\frac{\textup{(3600 - 1676)}}{\textup{582}}[/tex]

or

z score = 3.30

Hence, the house with an area 3,600 square feet is more unusual

Final answer:

To determine the more unusual house in the market, we calculate the z-scores. A house priced at $357,000 has a z-score of 2.95, while a house of 3,600 square feet has a z-score of 3.31. Therefore, the larger house size is more unusual.

Explanation:

To determine which house is more unusual in the given market, we need to calculate the number of standard deviations each value is from the mean, also known as the z-score. The z-score is calculated by taking the difference between the value and the mean, and then dividing by the standard deviation. For the price of the house, the z-score is calculated as follows:

Z = (Value - Mean) / Standard Deviation

For the $357,000 house price:

Z = ($357,000 - $176,678) / $61,029 = 2.95

For the 3,600 square ft house:

Z = (3,600 - 1,676) / 582 = 3.31

The house with an area of 3,600 square ft is 3.31 standard deviations away from the mean, whereas the $357,000 house price is 2.95 standard deviations away from the mean. Hence, the house with an area of 3,600 square ft is more unusual compared to the market's average.

Answer:

Sum = 10,232

Step-by-step explanation:

The given sequence is Arithmetic Progression.

Arithmetic Progression is a sequence in which every two neighbor digits have equal distances.

For finding the sum of given series firstly we find the number of terms in given series.

For finding the nth term, we use formula

aₙ = a + (n - 1) d

where, aₙ = value of nth term

a = First term

n = number of term

d = difference

Now, In given sequence: 3+11+19+27+.....+395+403

a = 3, d = 8, aₙ = 403

∴ 403 = 3 + (n - 1) × 8

⇒ n = 51

Now, the sum of series is determined by formula,

Sₙ = n ÷ 2 [ a + l]

where l = last term

⇒ Sₙ = 51 ÷ 2 [ 3 + 403]

⇒ Sₙ = 51 × 203

⇒ Sₙ = 10,232

Final answer:

The series 3+11+19+27+.....+395+403 is an arithmetic sequence with a common difference of 8. It has 51 terms, and the sum can be found using the formula Sn = n(a1 + an) / 2, which gives us a sum of 10353.

Explanation:

To add the numbers in the series 3+11+19+27+.....+395+403, we need to recognize that this series is an arithmetic sequence, where each term increases by a common difference. In this case, the common difference is 8 (since 11 - 3 = 8, 19 - 11 = 8, and so on).

First, we need to find the number of terms in the series. We know that an arithmetic series can be expressed as ann = a1 + (n - 1)d, where an is the nth term, a1 is the first term, d is the common difference, and n is the number of terms. Plugging in the values we have:

403 = 3 + (n - 1)8
400 = (n - 1)8
n - 1 = 50
n = 51

So there are 51 terms in the series. The sum of an arithmetic series is given by Sn = n(a1 + an) / 2. Plugging in the values we found:

S51 = 51(3 + 403) / 2
S51 = 51(406) / 2
S51 = 10353

Therefore, the sum of the series is 10353.

Answer:

0.024

Step-by-step explanation:

Given,

Red marbles = 4,

Green marbles = 7,

Total marbles = 4 +7 = 11,

Ways of choosing 3 marbles =[tex]^{11}C_3[/tex]

Ways of choosing 3 red marble = [tex]^4C_3[/tex]

Hence, the probability of 3 red marble = [tex]\frac{^4C_3}{^{11}C_3}[/tex]

[tex]=\frac{\frac{4!}{3!1!}}{\frac{11!}{3!8!}}[/tex]

[tex]=\frac{4}{165}[/tex]

≈ 0.024

Answer:

The answer is [tex]x=a^{-1}cb^{-1}[/tex].

Step-by-step explanation:

First, it is important to recall that the group law is not commutative in general, so we cannot assume it here. In order to solve the exercise we need to remember the axioms of group, specially the existence of the inverse element, i.e., for each element [tex]g\in G[/tex] there exist another element, denoted by [tex]g^{-1}[/tex] such that [tex]gg^{-1}=e[/tex], where [tex]e[/tex] stands for the identity element of G.

So, given the equality [tex] axb=c [/tex] we make a left multiplication by [tex]a^{-1}[/tex] and we obtain:

[tex]a^{-1}axb =a^{-1}c. [/tex]

But, [tex]a^{-1}axb = exb = xb[/tex]. Hence, [tex]xb = a^{-1}c[/tex].

Now, in the equality [tex]xb = a^{-1}c[/tex] we make a right multiplication by [tex]b^{-1}[/tex], and we obtain

[tex] xbb^{-1} = a^{-1}cb^{-1}[/tex].

Recall that [tex]bb^{-1}=e[/tex] and [tex]xe=x[/tex]. Therefore,

[tex]x=a^{-1}cb^{-1}[/tex].

Answer:

132 games played.

Step-by-step explanation:

Let x be the

We have been given that in a baseball league consisting of 12 teams, each team plays each of the other teams twice.

Since there are 12 teams, so each team will play with [tex](12-1)[/tex] teams except itself.

The total number of games played would be 12 times [tex](12-1)[/tex]:

[tex]\text{The total number of games played}=12(12-1)[/tex]

[tex]\text{The total number of games played}=12(11)[/tex]

[tex]\text{The total number of games played}=132[/tex]

Therefore, there will be 132 games played.

The total number of games that will be played in the baseball league with 12 teams is 66.

To find the number of games that will be played in the baseball league, we need to consider that each team plays each of the other teams twice. Since there are 12 teams, each team will play 11 other teams. However, this counts each game twice (once for each team).

So, to find the total number of games, we can use the formula: Total number of games = (Number of teams * Number of teams - Number of teams) / 2.

Substituting the values, we get: Total number of games = (12 * 12 - 12) / 2 = 66.

Answer: 1000 hot dogs and and 1600 sodas were sold.

Step-by-step explanation:

Let x be the number of hot dogs and y be the number of sodas.

Given : The concession stand at an ice hockey rink had receipts of $6200 from selling a total of 2600 sodas and hot dogs.

Each soda sold for $2 and each hot dog sold for $3 .

Then, we have the following system of two linear equations:-

[tex]x+y=2600-----------(1)\\\\3x+2y=6200-----------(2)[/tex]

Multiplying 2 on both sides of (1), we get

[tex]2x+2y=5200------------(3)[/tex]

Now, Eliminate equation (3) from equation (2), we get

[tex]x=1000[/tex]

Put x=1000 in (1), we get

[tex]1000+y=2600\\\\\Rightarrow\ y=2600-1000=1600[/tex]

Hence, 1000 hot dogs and and 1600 sodas were sold.

Answer and Explanation:

To find : Convert the given units ?

Solution :  

a) 2.5 km to mm

[tex]1\ km = 1000000\ mm[/tex]

[tex]2.5\ km = 2.5\times 1000000\ mm[/tex]

[tex]2.5\ km = 2500000\ mm[/tex]

b) 0.05 cm to mm

[tex]1\ cm = 10\ mm[/tex]

[tex]0.05\ cm =0.05\times 10\ mm[/tex]

[tex]0.05\ cm =0.5\ mm[/tex]

c) 200.5 g to kg

[tex]1\ g = 0.001\ kg[/tex]

[tex]200.5\ g =200.5\times 0.001\ kg[/tex]

[tex]200.5\ g =0.2005\ kg[/tex]

d) 0.03 tone into g

[tex]1\ t =1000000\ g[/tex]

[tex]0.03\ t =0.03\times 1000000\ g[/tex]

[tex]0.03\ t =30000\ g[/tex]

e) 3.0412 sec into hour

[tex]1\ sec =\frac{1}{3600}\ hr[/tex]

[tex]3.0412\ sec =3.0412\times \frac{1}{3600}\ hr[/tex]

[tex]3.0412\ sec =0.000844\ hr[/tex]

Final answer:

To find the quadratic function whose graph contains the points (0, -2), (-5, -17), and (3, -17), we can substitute the x and y values of each point into the standard quadratic function equation y = ax² + bx + c...Therefore, the quadratic function whose graph contains the points (0, -2), (-5, -17), and (3, -17) is y = -x² - 2x - 2.

Explanation:

To find the quadratic function whose graph contains the points (0, -2), (-5, -17), and (3, -17), we can substitute the x and y values of each point into the standard quadratic function equation y = ax^2 + bx + c. This will give us a system of three equations with three variables (a, b, c) that we can solve to find the values of a, b, and c.

Substituting the coordinates (0, -2), we get -2 = a(0)²+ b(0) + c, which simplifies to -2 = c.

Substituting the coordinates (-5, -17), we get -17 = a(-5)² + b(-5) + c, which simplifies to -17 = 25a - 5b + c.

Substituting the coordinates (3, -17), we get -17 = a(3)² + b(3) + c, which simplifies to -17 = 9a + 3b + c.

Since we know that c = -2, we can substitute this value into the other two equations to get -17 = 25a - 5b - 2 and -17 = 9a + 3b - 2.

Simplifying these equations, we get:

25a - 5b = -15 (equation 1)

9a + 3b = -15 (equation 2)

Now we can solve this system of equations using any method such as substitution or elimination to find the values of a and b.

Multiplying equation 1 by 3 and equation 2 by 5, we get:

75a - 15b = -45 (equation 3)

45a + 15b = -75 (equation 4)

Adding equation 3 and equation 4 together, we eliminate the variable b and get:

120a = -120

Dividing both sides by 120, we get a = -1.

Substituting this value of a back into equation 1, we get:

25(-1) - 5b = -15

-25 - 5b = -15

Adding 25 to both sides, we get -5b = 10.

Dividing both sides by -5, we get b = -2.

Therefore, the quadratic function whose graph contains the points (0, -2), (-5, -17), and (3, -17) is y = -x² - 2x - 2.

Answer:

The remainder is 3.

Step-by-step explanation:

We have to find out,

[tex]10^z-1(mod 4)=?\text{ where }z\geq 2[/tex]

If z = 2,

[tex]10^{2}-1=100-1=99[/tex]

∵ 99 ( mod 4 ) = 3,

Suppose,

[tex](10^{k}-1)(mod 4)=3\forall \text{ k is an integer greater than 2,}[/tex]

Now,

[tex](10^{k+1}-1) ( mod 4)[/tex]

[tex]= (10^k.10 - 10+9)(mod 4)[/tex]

[tex] = 10(mod 4)\times (10^k-1)(mod 4 ) + 9 ( mod 4)[/tex]

[tex]= (2\times 3)(mod 4) + 1[/tex]

[tex]=2+1[/tex]

[tex]=3[/tex]

Hence, our assumption is correct.

The remainder of [tex]10^z -1[/tex] divided by 4 is 3 where, z ≥ 2.

Answer:

If the intersection is finite the statement  is true, but if the intersection is infinite the statement is false.

Step-by-step explanation:

From the statement of the problem I am not sure if the intersection is finite or infinite. Then, I will study both cases.

Let us consider first the finite case: [tex]A = \cap_{i=1}^{n}A_i[/tex]. Because the condition A1 ⊇ A2 ⊇ A3 ⊇ A4 ... we can deduce that the set [tex]A_n[/tex] is a subset of each set [tex]A_i[/tex] with [tex] i\leq n[/tex]. Thus,

[tex]\cap_{i=1}^{n}A_i = A_n[/tex].

Therefore, as [tex]A_n[/tex] is infinite, the intersection is infinite.

Now, if we consider the infinite intersection, i.e. [tex]A = \cap_{k=1}^{\infty}A_k[/tex] the reasoning is slightly different. Take the sets

[tex]A_k = (0,1/k)[/tex] (this is, the open interval between 0 and [tex]1/k[/tex].)

Notice that (0,1) ⊇ (0,1/2) ⊇ (0, 1/3) ⊇(0,1/4) ⊇...So, the hypothesis of the problem are fulfilled. But,

[tex]\cap_{k=1}^{\infty}(0,1/k) = \empyset[/tex]

In order to prove the above statement, choose a real number [tex]x[/tex] between 0 and 1. Notice that, no matter how small [tex]x[/tex] is, there is a natural number [tex]K[/tex] such that [tex]1/K<x[/tex]. Then, the number [tex]x[/tex] is not in any interval [tex](0,1/k)[/tex] with [tex]k>K[/tex]. Therefore, [tex]x[/tex] is not in the set [tex]\cap_{k=1}^{\infty}(0,1/k)[\tex].

Answer:

There is a 44.12% probability that the defective product came from C.

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

[tex]P = \frac{P(B).P(A/B)}{P(A)}[/tex]

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

-In your problem, we have:

P(A) is the probability of the customer receiving a defective product. For this probability, we have:

[tex]P(A) = P_{1} + P_{2} + P_{3}[/tex]

In which [tex]P_{1}[/tex] is the probability that the defective product was chosen from plant A(we have to consider the probability of plant A being chosen). So:

[tex]P_{1} = 0.35*0.25 = 0.0875[/tex]

[tex]P_{2}[/tex] is the probability that the defective product was chosen from plant B(we have to consider the probability of plant B being chosen). So:

[tex]P_{2} = 0.15*0.05 = 0.0075[/tex]

[tex]P_{3}[/tex] is the probability that the defective product was chosen from plant B(we have to consider the probability of plant B being chosen). So:

[tex]P_{3} = 0.50*0.15 = 0.075[/tex]

So

[tex]P(A) = 0.0875 + 0.0075 + 0.075 = 0.17[/tex]

P(B) is the probability the product chosen being C, that is 50% = 0.5.

P(A/B) is the probability of the product being defective, knowing that the plant chosen was C. So P(A/B) = 0.15.

So, the probability that the defective piece came from C is:

[tex]P = \frac{0.5*0.15}{0.17} = 0.4412[/tex]

There is a 44.12% probability that the defective product came from C.

Find the difference per row:

10 seats in the first row

30 seats in the sixth row:

30 -10 = 20 seats difference.

6-1 = 5 rows difference.

20 seats /  5 rows = 4 seats per row.

This means for every additional row, there are 4 more seats per row.

The equation would be:

Sn = S +(n-1)*d

Where d is the difference = 4

S = number of seats from starting row = 10

n = the number of rows wanted

S(11) = 10 + (11-1)*4

S(11) = 10 + 10*4

S(11) = 10 + 40

S(11) = 50

Check:

Row 6 = 30 seats

Row 7 = 30 + 4 = 34 seats

Row 8 = 34 + 4 = 38 seats

Row 9 = 38 + 4 = 42 seats

Row 10 = 42 + 4 = 46 seats

Row 11 = 46 + 4 = 50 seats.

Answer:

150 students take physics.

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of this set.

I am going to say that:

-The set A represents the students that take calculus.

-The set B represents the students that take physics

-The set C represents the students that take chemistry.

-The set D represents the students that do not take any of them.

We have that:

[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]

In which a is the number of students that take only calculus, [tex]A \cap B[/tex] is the number of students that take both calculus and physics, [tex]A \cap C[/tex] is the number of students that take both calculus and chemistry and [tex]A \cap B \cap C[/tex] is the number of students that take calculus, physics and chemistry.

By the same logic, we have:

[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]

[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]

This diagram has the following subsets:

[tex]a,b,c,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C), D[/tex]

There are 360 people in my school. This means that:

[tex]a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) + D = 360[/tex]

The problem states that:

15 take calculus, physics, and chemistry, so:

[tex]A \cap B \cap C = 15[/tex]

15 don't take any of them, so:

[tex]D = 15[/tex]

75 take both calculus and chemistry, so:

[tex]A \cap C = 75[/tex]

75 take both physics and chemistry, so:

[tex]B \cap C = 75[/tex]

30 take both physics and calculus, so:

[tex]A \cap B = 30[/tex]

Solution:

The problem states that 180 take calculus. So

[tex]a + (A \cap B) + (A \cap C) + (A \cap B \cap C) = 180[/tex]

[tex]a + 30 + 75 + 15 = 180[/tex]

[tex]a = 180 - 120[/tex]

[tex]a = 60[/tex]

Twice as many students take chemistry as take physics:

It means that: [tex]C = 2B[/tex]

[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]

[tex]B = b + 75 + 30 + 15[/tex]

[tex]B = b + 120[/tex]

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

[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]

[tex]C = c + 75 + 75 + 15[/tex]

[tex]C = c + 165[/tex]

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

Our interest is the number of student that take physics. We have to find B. For this we need to find b. We can write c as a function o b, and then replacing it in the equations that sums all the subsets.

[tex]C = 2B[/tex]

[tex]c + 165 = 2(b+120)[/tex]

[tex]c = 2b + 240 - 165[/tex]

[tex]c = 2b + 75[/tex]

The equation that sums all the subsets is:

[tex]a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) + D = 360[/tex]

[tex]60 + b + 2b + 75 + 30 + 75 + 15 + 15 = 360[/tex]

[tex]3b + 270 = 360[/tex]

[tex]3b = 90[/tex]

[tex]b = \frac{90}{3}[/tex]

[tex]b = 30[/tex]

30 students take only physics.

The number of student that take physics is:

[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]

[tex]B = b + 75 + 30 + 15[/tex]

[tex]B = 30 + 120[/tex]

[tex]B = 150[/tex]

150 students take physics.

Final answer:

Using a Venn Diagram approach and the information given, we find that 45 students take physics at the school.

Explanation:

To find out how many students take physics at the school, we can use the Venn Diagram principle and the given data. We know that 15 students take calculus, physics, and chemistry together. Additionally, 180 students take calculus, and twice as many students take chemistry as take physics. With 75 students taking both calculus and chemistry, and another 75 taking both physics and chemistry, while only 30 take both physics and calculus, we can establish relationships and solve for the number of students taking each subject.

Let's denote the number of students taking physics as P. Then, the number of students taking chemistry would be 2P.

To avoid double counting, we must subtract those taking all three subjects once for each combination:

With twice as many students in chemistry as in physics, we can write the equation:

180 + 2P + P - (15 + 60 + 15 + 60) + 15 = 360

Solving for P:

Therefore, 45 students take physics at the school.

Answer:

tatho you get the aswer

Step-by-step explanation:

all you have to do is add

Answer:

noice

Step-by-step explanation:

Answer:

The solution is [tex]f(t)=-\ln \left|\cos \left(t\right)\right|+C[/tex]

Step-by-step explanation:

We know that this ordinary differential equation (ODE) is separable if we can write F(x,y) = f(x)g(y) for some function f(x), g(x).

We can write this ODE in this way

[tex]cos(t) \cdot f'(t)=sin(t)\\f'(t)=\frac{sin(t)}{cos(t)}[/tex]

[tex]\mathrm{If\quad }f^{'} \left(x\right)=g\left(x\right)\mathrm{\quad then\quad }f\left(x\right)=\int g\left(x\right)dx[/tex]

[tex]f(t) =\int\limits{\frac{sin(t)}{cos(t)}} \, dt[/tex]

To solve this integral we need to follow this steps

[tex]\int \frac{\sin \left(t\right)}{\cos \left(t\right)}dt = \\\mathrm{Apply\:u-substitution:}\:u=\cos \left(t\right)\\\int \frac{\sin \left(t\right)}{u}dt \\\mathrm{And \:du=-sin(t)\cdot dt}\\\mathrm{so \>dt=\frac{du}{-sin(t)}}\\\int \frac{\sin \left(t\right)}{u}dt = -\int \frac{1}{u}du[/tex]

[tex]\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)\\-ln|u|\\\mathrm{Substitute\:back}\:u=\cos \left(t\right)\\-\ln \left|\cos \left(t\right)\right|\\[/tex]

Add the constant of integration

[tex]f(t)=-\ln \left|\cos \left(t\right)\right|+C[/tex]

The given Statement which we have to prove using mathematical induction is

   [tex]5^n\geq 2*2^{n+1}+100[/tex]

for , n≥4.

⇒For, n=4

LHS

[tex]=5^4\\\\5*5*5*5\\\\=625\\\\\text{RHS}=2.2^{4+1}+100\\\\=64+100\\\\=164[/tex]

 LHS >RHS

Hence this statement is true for, n=4.

⇒Suppose this statement is true for, n=k.

 [tex]5^k\geq 2*2^{k+1}+100[/tex]

                      -------------------------------------------(1)

Now, we will prove that , this statement is true for, n=k+1.

[tex]5^{k+1}\geq 2*2^{k+1+1}+100\\\\5^{k+1}\geq 2^{k+3}+100[/tex]

LHS

[tex]5^{k+1}=5^k*5\\\\5^k*5\geq 5 \times(2*2^{k+1}+100)----\text{Using 1}\\\\5^k*5\geq (3+2) \times(2*2^{k+1}+100)\\\\ 5^k*5\geq 3\times (2^{k+2}+100)+2 \times(2*2^{k+1}+100)\\\\5^k*5\geq 3\times(2^{k+2}+100)+(2^{k+3}+200)\\\\5^{k+1}\geq (2^{k+3}+100)+3\times2^{k+2}+400\\\\5^{k+1}\geq (2^{k+3}+100)+\text{Any number}\\\\5^{k+1}\geq (2^{k+3}+100)[/tex]

Hence this Statement is true for , n=k+1, whenever it is true for, n=k.

Hence Proved.

Answer:

(a) 1:2

(b) 6 cups

(c) [tex]\dfrac{2}{3}[/tex]

Step-by-step explanation:

(a) Given,

amount of rice mixture contains= 1 cup

amount of water mixture contains= 2 cups

[tex]\textrm{So, the ratio of rice to water}\ =\ \dfrac{\textrm{amount of rice in mixturte}}{\textrm{amount of water in mixture}}[/tex]

                                                              [tex]=\ \dfrac{1}{2}[/tex]

So, the ratio of rice to water is 1:2.

(b) Amount of rice in mixture = 3 cups

[tex]\textrm{So, the ratio of rice to water}\ =\ \dfrac{\textrm{amount of rice in mixturte}}{\textrm{amount of water in mixture}}[/tex]

                      [tex]=>\ \dfrac{1}{2}\ =\ \dfrac{3}{\textrm{amount of water in mixture}}[/tex]

                       => amount of water in mixture = 3 x 2

                                                                          = 6 cups

(c) [tex]\textrm{amount of rice in mixture}\ =\dfrac{1}{3}[/tex]

 [tex]\textrm{So, the ratio of rice to water}\ =\ \dfrac{\textrm{amount of rice in mixturte}}{\textrm{amount of water in mixture}}[/tex]

   [tex]=>\ \dfrac{1}{2}\ =\ \dfrac{\dfrac{1}{3}}{\textrm{amount of water in mixture}}[/tex]

  [tex]=>\textrm{amount of water in mixture}\ =\ \dfrac{2}{3}[/tex]

So, the amount of water in the mixture will be [tex]\dfrac{2}{3}[/tex] cup.

Answer:

The improper integral converges and [tex]\int_0^{-\infty} e^{7x}dx = -\frac{1}{7}[/tex].

Step-by-step explanation:

First, I assume that the integral in question is

[tex]\int_0^{-\infty} e^{7x}dx[/tex].

Now, the integral is improper because, at least, one of the limits is [tex]\pm\infty[/tex]. We need to recall that an improper integral

[tex]\int_0^{-\infty} f(x)dx[/tex]

converges, by definition, if the following limit exist:

[tex]\lim_{A\rightarrow -\infty} \int_0^A f(x)dx = \int_0^{-\infty} f(x)dx[/tex].

In this particular case we need to study the limit

[tex]\lim_{A\rightarrow -\infty} \int_0^A e^{7x}dx[/tex].

In order to complete this task we calculate the integral [tex]\int_0^A e^{7x}dx[/tex]. Then,

[tex]\int_0^A e^{7x}dx = \frac{e^{7x}}{7}\Big|_0^A = \frac{e^{7A}}{7} - \frac{1}{7}[/tex].

Substituting the above expression into the limit we have

[tex]\lim_{A\rightarrow -\infty} \frac{e^{7A}}{7} - \frac{1}{7} = - \frac{1}{7}[/tex]

because

[tex]\lim_{A\rightarrow -\infty} \frac{e^{7A}}{7}=0[/tex].