To reduce laboratory​ costs, water samples from two public swimming pools are combined for one test for the presence of bacteria. Further testing is done only if the combined sample tests positive. Based on past​ results, there is a 0.005 probability of finding bacteria in a public swimming area. Find the probability that a combined sample from two public swimming areas will reveal the presence of bacteria. Is the probability low enough so that further testing of the individual samples is rarely​ necessary?

Answer: Probability that students who did not attend the class on Fridays given that they passed the course is 0.043.

Step-by-step explanation:

Since we have given that

Probability that students attend class on Fridays = 70% = 0.7

Probability that who went to class on Fridays would pass the course = 95% = 0.95

Probability that who did not go to class on Fridays would passed the course = 10% = 0.10

Let A be the event students passed the course.

Let E be the event that students attend the class on Fridays.

Let F be the event that students who did not attend the class on Fridays.

Here, P(E) = 0.70 and P(F) = 1-0.70 = 0.30

P(A|E) = 0.95,  P(A|F) = 0.10

We need to find the probability that they did not attend on Fridays.

We would use "Bayes theorem":

[tex]P(F\mid A)=\dfrac{P(F).P(A\mid F)}{P(E).P(A\mid E)+P(F).P(A\mid F)}\\\\P(F\mid A)=\dfrac{0.30\times 0.10}{0.70\times 0.95+0.30\times 0.10}\\\\P(F\mid A)=\dfrac{0.03}{0.695}=0.043[/tex]

Hence, probability that students who did not attend the class on Fridays given that they passed the course is 0.043.

Answer:

  $5036.34

Step-by-step explanation:

Each year, 8% of the existing balance is added to the existing balance, effectively multiplying the amount by 1.08. If that is done for 12 years, the effective multiplier is 1.08^12 ≈ 2.51817. The the amount in the bank at the end of that time is ...

  $2000×2.51817 = $5036.34

The amount in the bank after 12 years with an annual interest rate of 8% on a principal amount of 2000 dollars, compounded annually, will be approximately $5025.90.

This is a compound interest problem. The formula used to solve this type of problem is A = P(1 + r/n)^(nt), where:

In this case, P = $2000, r = 8% or 0.08, t = 12 years and n = 1 (as interest is compounded annually). Substituting these values in the equation, we get:

A = 2000(1 + 0.08/1)^(1*12)

.

The resulting Amount A after 12 years will be approximately $5025.90.

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Answer:

[tex]\large\boxed{x=6\pm3\sqrt{10}}[/tex]

Step-by-step explanation:

[tex]x^2-12x+36=90\\\\x^2-2(x)(6)+6^2=90\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\(x-6)^2=90\iff x-6=\pm\sqrt{90}\\\\x-6=\pm\sqrt{9\cdot10}\\\\x-6=\pm\sqrt9\cdot\sqrt{10}\\\\x-6=\pm3\sqrt{10}\qquad\text{add 6 to both sides}\\\\x=6\pm3\sqrt{10}[/tex]

Answer:

x² + 9x + 8 = (x + 1)(x + 8)

x² + 9x + 8 = (x + 8)(x + 1)

Step-by-step explanation:

* Lets explain how to factorize the polynomial using the pattern

- The form of the quadratic polynomial is x² + px + q, where p is the

  coefficient of x and q is the numerical term

∵ x² + (a + b)x + (ab) = (x + a)(x + b)

- From the formula above the coefficient of x is the sum of the two

 factors a and b

∴ p = a + b and q = ab

- That means p is the sum of two numbers and q is the product of

  the same numbers

* Lets solve the problem

∵ x² + 9x + 8 is a quadratic polynomial

∵ x² + px + q is the form of quadratic polynomial

∴ p = 9 and q = 8

∵ p = a + b and q = ab

∴ a + b = 9 ⇒ (1)

∴ ab = 8 ⇒ (2)

- We must to find two numbers their product is 8 and their sum is 9

∵ The possibility of 8 as a product of two numbers is:

   2 × 4 OR 1 × 8

∵ The sum of 1 + 8 = 9

∴ The value of a and b are 1 and 8

- It does't matter which of them = 1 or which of them = 8

∴ x² + (a + b)x + ab = x² + (1 + 8)x + (1)(9)

∵ x² + (a + b)x + (ab) = (x + a)(x + b)

∴ x² + (1 + 8)x + (1)(9) = (x + 1)(x + 8)

∴ x² + 9x + 8 = (x + 1)(x + 8)

- OR

∴ x² + 9x + 8 = (x + 8)(x + 1)

Answer:

The volume of the prism is 27√3/2

Step-by-step explanation:

* Lets revise the triangular prism properties

- The triangular prism has five faces

- Two bases and three side faces

- The two bases are triangles

- The three side faces are rectangles

- The rule of its volume is Area of its base × its height

* Lets solve the problem

- The triangular prism has two bases which are equilateral triangles

- The length of each side of the triangular base is 3"

- The height of the prism is 6"

∵ The volume of the prism = area of the base × its height

∵ The base is equilateral triangle of side length 3"

- The area of any equilateral triangle is √3/4 s²

∴ The area of the base of the prism = √3/4 × (3)² = 9√3/4

∵ The length of the height of the prism is 6"

∴ The volume of the prism = 9√3/4 × 6 = 27√3/2

* The volume of the prism is 27√3/2

The angle between radius and tangent to circle is 90 degrees.

The quadrilateral formed by the two tangents and the two rays has two angles of 90 degrees, an angle of 40 degrees and an unknown angle.

The sum of the angles of a quadrilateral is 360 degrees.

⇒  x = 360° - 90 - 90 - 40 = 140°

Step-by-step explanation:

Let n = number of years since 1995

Let C = cost of gym membership in any particular year

Initial cost in 1995 = $25

Additional cost each year = $6

We can say the following:

Cost at any given year = cost in 1995 + ($6 x number of years after 1995)

Or expressed as the following :  C = 25 + 6n   (Ans)

In 2009, the number of years since 1995,

= 2009 - 1995

= 14 years

Hence, cost in 2009,

= $25 + ($6 x 14 years)

= $109 (Ans)

 $109   is the Cost of A GYM Membership in 2009.

What does "cost" mean to you?

given,

Let n = number of years since 1995

Let C = cost of gym membership in any particular year

Initial cost in 1995 = $25

Additional cost each year = $6

formula ,

Cost at any given year = cost in 1995 + ($6 x number of years after 1995)

put value in formula

             C = 25 + 6n  

now,

In 2009, the no of member in  years since 1995,

= 2009 - 1995

= 14 years

Hence, cost of gym membership in 2009,

= $25 + ($6 x 14 years)

= $109

Therefore, cost of gym membership in 2009 = $109

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Answer:

The solution is 22.98.

Step-by-step explanation:

Here, the given equation,

[tex]5 + 69\times 0.96^t = 32[/tex],

Let [tex]f(t) = 5 + 69\times 0.96^t[/tex]

And, [tex]f(t) = 32[/tex]

Where, t represents x-axis and f(t) represents y-axis,

Since, [tex]f(t) = 5 + 69\times 0.96^t[/tex] is an exponential decay function having y-intercept (0,74).

Also, f(t) = 32 is the line, parallel to x-axis,

Thus, after plotting the graph of the above functions,

We found that they are intersecting at (22.984, 32)

Hence, the solution of the given equation = x-coordinate of the intersecting point = 22.984 ≈ 22.98

To solve the given equation, 5 + 69 × 0.96t = 32, you start by subtracting 5 from both sides, then divide by 69. Then, divide both sides by 0.96 to solve for t. The solution is t ≈ 0.41 (rounded to two decimal places).

To solve the equation 5 + 69 × 0.96t = 32 using the crossing-graphs method, we first simplify the equation:

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Answer:

14.6328% , $5836.03

Step-by-step explanation:

Here we are going to use the formula

[tex]A_{0}(1-r)^n = A_{n}[/tex]

[tex]A_{0}[/tex] = 39000

r=?

[tex]A_{8}[/tex] = 11000

n=8

Hence

[tex]39000(1-r)^8 = 11000[/tex]

[tex](1-r)^8 = \frac{11000}{39000}[/tex]

[tex](1-r)^8 = 0.2820[/tex]

[tex](1-r) = 0.2820^{\frac{1}{8}[/tex]

[tex](1-r) = 0.2820^{0.125}[/tex]

[tex](1-r) = 0.8536[/tex]

[tex](1-0.8536=r[/tex]

[tex]r = 0.1463[/tex]

Hence r= 0.1463

In percentage form r = 14.63%

Now let us see calculate the value of car in 2003 that is after 12 years

we use the main formula again

[tex]A_{0}(1-r)^n = A_{n}[/tex]

[tex]A_{0}[/tex] = 39000

r=0.1463

[tex]A_{12}[/tex] = ?

n=12

[tex]39000(1-0.14634)^{(12} = A_{12}[/tex]

[tex]39000(0.8536)^{12} = A_{12}[/tex]

[tex]39000*0.1497 = A_{12}[/tex]

[tex]A_{12}=5840.34[/tex]

Hence the car's value will be depreciated to $5840.34 (approx) by 2003.

The annual rate of change between 1995 and 2003 is -0.1463

The annual rate of change between 1995 and 2003 is -14.63%

The value of the car in 2007 would be $5,844.24

The value of the car decreases as the years go by. This is referred to as depreciation. Depreciation is the decline in value of an asset as a result of wear and tear.

In order to determine the annual rate of change, use this formula:

g = [tex](FV / PV) ^{\frac{1}{n} } - 1[/tex]

Where:

g = depreciation rate

FV = value of the car in 2003 = $11,000

PV = value of the car in 1995 = $39,000

n = number of years = 2003 - 1995 = 8

[tex](11,000 / 39,00)^{\frac{1}{8} } - 1[/tex] = -0.1463 = -14.63%

The value of car in 7 years can be determined using this formula:

FV = P (1 + g)^n

$39,000 x (1 - 0.1463)^12

$39,000 x 0.8537^12 = $5,844.24

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Answer:

[tex]\boxed{\text{197 m}}[/tex]

Step-by-step explanation:

The formula relating distance (d), speed (s), and time (t) is

d = st

1. Calculate the distance

d = 269 s × 4.6 m·s⁻¹ = 1240 m

2.Calculate the track radius

The distance travelled is the circumference of a circle

[tex]\begin{array}{rcl}C & = & 2 \pi r\\1240 & = & 2 \pi r\\\\r & = & \dfrac{1240}{2 \pi }\\\\& = & 197\\\end{array}\\\text{The radius of the track is }\boxed{\textbf{197 m}}[/tex]

The radius in meters is 196.9 meters.

The runner ran around the track in 269 seconds at a speed of 4.6 m/s. This will enable us to find the distance around the track which is the circumference of the track.

Distance = Speed × time

= 4.6 × 269

= 1,237.4 meters

The distance here is the circumference which can also be found by the formula:

Circumference = π × diameter

1,237.4 = 22/7 × Diameter

Diameter = 1,237.4 ÷ 22/7

= 393.7 meters

Now that we have the diameter, the radius is:

= Diameter / 2

= 196.9 meters

In conclusion, the radius is 196.3 meters

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Answer:

Step-by-step explanation:

The answer above this answer is correct

The correct answer is:

                   Option: E

   E.   None of the above

[tex]Ax=0[/tex] has infinite many solutions if det(A)=0

and is non-singular or invertible otherwise i.e. when det(A)≠0

The matrix that will be formed by the given set of vectors is:

[tex]A=\begin{bmatrix}-1 &2 &0 &-1 \\ -1&-1 &0 &-1 \\ 0 &0 &1 &0 \\ 0 &0 &0 &1 \end{bmatrix}[/tex]

Also, determinant i.e. det of matrix A is calculated by:

[tex]\begin{vmatrix}-1 &-2 &0 &-1 \\ -1&-1 &0 &-1 \\ 0 &0 &1 &0 \\ 0 &0 &0 &1 \end{vmatrix}=1(1(1+2))=3[/tex]

Hence, determinant is not equal to zero.

This means that the matrix is invertible and non-singular.

Answer:

  450 km

Step-by-step explanation:

Equations

We can define 3 variables: a, b, d. Let "a" and "b" represent the speeds of the cars leaving cities A and B, respectively. Let "d" represent the distance between the two cities. We can write three equations in these three variables:

1. The relation between "a" and "b":

  a = b -10 . . . . . . . the speed of car A is 10 kph less than that of car B

2. The relation between speed and distance when the cars leave at the same time:

  d = (a +b)·5 . . . . . . distance = speed × time

3. Note that the time it takes car B to travel 150 km to the meeting point is (150/b). (time = distance/speed) The total distance covered is ...

  distance covered by car A in 4 1/2 hours + distance covered by both cars (after car B leaves) = total distance

  4.5a + (150/b)(a +b) = d

__

Solution

Substituting for d, we have ...

  4.5a + 150/b(a +b) = 5(a +b)

  4.5ab +150a +150b = 5ab +5b^2 . . . . . . multiply by b, eliminate parentheses

  5b^2 +0.5ab -150(a +b) = 0 . . . . . . . . . . subtract the left side

Now, we can substitute for "a" and solve for b.

  5b^2 + 0.5b(b-10) -150(b -10 +b) = 0

  5.5b^2 -5b -300b +1500 = 0 . . . . . . . . eliminate parentheses

  11b^2 -610b +3000 = 0 . . . . . . . . . . . . . multiply by 2

  (11b -60)(b -50) = 0 . . . . . . . . . . . . . . . . factor

The solutions to this equation are ...

  b = 60/11 = 5 5/11 . . . and . . . b = 50

Since b must be greater than 10, the first solution is extraneous, and the values of the variables are ...

The distance between A and B is 450 km.

_____

Check

When the cars leave at the same time, their speed of closure is the sum of their speeds. They will cover 450 km in ...

  (450 km)/(40 km/h +50 km/h) = 450/90 h = 5 h

__

When car A leaves 4 1/2 hours early, it covers a distance of ...

  (4.5 h)(40 km/h) = 180 km

before car B leaves. The distance remaining to be covered is ...

  450 km - 180 km = 270 km

When car B leaves, the two cars are closing at (40 +50) km/h = 90 km/h, so will cover that 270 km in ...

  (270 km)/(90 km/h) = 3 h

In that time, car B has traveled (3 h)(50 km/h) = 150 km away from city B, as required.

Answer:

450km

Step-by-step explanation:

Take it that each automobile travels at 30 km an hour, for 150 km, meaning it will be 450 km apart.

Final answer:

Explanation of quartiles, deciles, and percentiles in a dataset of gas-powered generator invoice payment times. Lastly, the 67th percentile in the data set is approximately 46 days.

Explanation:

The first quartile (Q1) is found by locating the median of the lower half of the data set, which results in a value of 34 days. The third quartile (Q3) is the median of the upper half of the data set, yielding a value of 47 days. The second decile corresponds to the 20th percentile, which is approximately 31 days. The eighth decile corresponds to the 80th percentile, which is approximately 47 days. Lastly, the 67th percentile in the data set is approximately 46 days.

Final answer:

The problem requires formulating a linear programming model to maximize the profit function Z = 9x + 7y with constraints on machine time for product A and B (2x + 3y ≤ 6 for machine M and 4y ≤ 8 for machine L) and the non-negativity restrictions (x, y ≥ 0).

Explanation:

The linear programming problem can be stated in mathematical terms with an objective function and constraints for a firm making products A and B. The objective is to maximize profit, which is the sum of 9 dollars per unit of product A and 7 dollars per unit of product B. Let the number of products A and B produced be represented by variables x and y, respectively.

The objective function to maximize is Z = 9x + 7y.

Constraints:

Answer:

The probability of the sample mean foot length less than 23 cm is 0.120

Step-by-step explanation:

* Lets explain the information in the problem

- The eighth-graders asked to measure the length of their right foot at

  the beginning of the school year, as part of a science project

- The foot length is approximately Normally distributed, with a mean of

 23.4 cm

∴ μ = 23.4 cm

- The standard deviation of 1.7

∴ σ = 1.7 cm

- 25 eighth-graders from this population are randomly selected

∴ n  = 25

- To find the probability of the sample mean foot length less than 23

∴ The sample mean x = 23, find the standard deviation σx

- The rule to find σx is σx = σ/√n

∵ σ = 1.7 and n = 25

∴ σx = 1.7/√25 = 1.7/5 = 0.34

- Now lets find the z-score using the rule z-score = (x - μ)/σx

∵ x = 23 , μ = 23.4 , σx = 0.34

∴ z-score = (23 - 23.4)/0.34 = -1.17647 ≅ -1.18

- Use the table of the normal distribution to find P(x < 23)

- We will search in the raw of -1.1 and look to the column of 0.08

∴ P(X < 23) = 0.119 ≅ 0.120

* The probability of the sample mean foot length less than 23 cm is 0.120

ooops, i made a mistake.  deleted. Give the other guy brainly

sorry

Check the picture below.

Answer:

218.75 miles / 12.5 hours

437.5 miles / 25 hours

656.25 miles / 37.5 hours

Step-by-step explanation:

35 miles / 2 hours = 87.5 miles / 5 hours

This is the  ratio of 2.5. So, the other proportions are

87.5 x 2.5 miles / 5 x 2.5 hours = 218.75 miles / 12.5 hours

87.5 x 5 miles / 5 x 5 hours = 437.5 miles / 25 hours

87.5 x 7.5 miles / 5 x 7.5 hours = 656.25 miles / 37.5 hours

Answer:

[tex]\large\boxed{A=153\ cm^2}[/tex]

Step-by-step explanation:

Look at the picture.

We have

square with side length a = 9

trapezoid with base lengths b₁ = 9 and b₂ = 6 and the height length h = 6

right triangle with legs lengths l₁ = 3 + 6 = 9 and l₂ = 6

The formula of an area of a square

[tex]A=a^2[/tex]

Substitute:

[tex]A_I=9^2=81\ cm^2[/tex]

The formula of an area of a trapezoid:

[tex]A=\dfrac{b_1+b_2}{2}\cdot h[/tex]

Substitute:

[tex]A_{II}=\dfrac{9+6}{2}\cdot6=\dfrac{15}{2\!\!\!\!\diagup_1}\cdot6\!\!\!\!\diagup^3=(15)(3)=45\ cm^2[/tex]

The formula of an area of a right triangle:

[tex]A=\dfrac{l_1l_2}{2}[/tex]

Substitute:

[tex]A_{III}=\dfrac{(9)(6)}{2}=\dfrac{54}{2}=27\ cm^2[/tex]

The area of the shape:

[tex]A=A_I+A_{II}+A_{III}\\\\A=81+45+27=153\ cm^2[/tex]

I think to solve this question, subtract the rate of feet per second disappearing from the distance of feet between the 2 cars.

100-85=15fps

The car is traveling at 15 feet per second.

Hope this helps!

The Laplace Transform of a function f(t) is a tool used to solve differential equations by converting them into simpler algebraic equations. The transform itself is given by the integral ℒ{f(t)} = ∫∞₀ e−stf(t) dt, but without knowing the specific form of f(t), the exact transformation cannot be computed.

The Laplace Transform, as defined by Definition 7.1.1, is a mathematical tool often used to handle differential equations, especially in the fields of Physics and Engineering. The main idea behind the transform is to convert the differential equations, which are difficult to solve, into simple algebraic equations. These simple equations are relatively easy to solve. Once solved, the inverse Laplace Transform is employed to obtain the solution to the original differential equation.

The question asks to compute the Laplace Transform ℒ{f(t)} of a function. By definition, the Laplace transform of a function f(t), defined for t ≥ 0 is given by the integral ℒ{f(t)} = ∫∞₀ e−stf(t) dt. The question doesn't give an explicit form of the function f(t). In general, if f(t) = e^(-αt), where α is a constant, then its Laplace Transform is given by ℒ{e^(-αt)} = 1/(s + α) for s > α.

If you could kindly provide me with the specific function f(t), I can better determine the Laplace Transform ℒ{f(t)}

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Answer:

A

Step-by-step explanation:

Volume varies directly with the square of the radius, so:

V = k r²

When V = 80, r = 4.

80 = k (4)²

k = 5

V = 5r²

When r = 3:

V = 5 (3)²

V = 45

The flow is 45 L/min.

Answer:

  sin(2θ) = 0.96

Step-by-step explanation:

In the third quadrant, both sin(θ) and cos(θ) are negative. Then the double-angle trig identity tells us ...

  sin(2θ) = 2·sin(θ)·cos(θ) = -2cos(θ)√(1 -cos(θ)²) . . . . using the negative root

Filling in the given value, we have

  sin(2θ) = -2·(-0.6)(√(1-(-0.6)²) = 2·0.6·0.8 = 0.96

Answer: Hence, the probability that the whole shipment would be accepted is 0.371.

Many would be rejected.

Step-by-step explanation:

Since we have given that

Number of tablets to be tested = 42

Probability of getting a defect = 5% = 0.05

We need to find the probability that this whole shipment will be accepted.

As we have mentioned that if there is only one or none defect, then the whole shipment would be accepted.

P(accepted) = P(either none or one defect) =  P(X=0)+P(X=1)

[tex]P(X=0)=(1-0.05)^{42}=(0.95)^{42}=0.115\\\\and\\\\P(X=1)=42\times (0.05)(0.95)^{41}=0.006\times 42=0.256[/tex]

So, P(Accepted) = 0.115+0.256=0.371

Hence, the probability that the whole shipment would be accepted is 0.371.

Many would be rejected.

Answer:

720

Step-by-step explanation:

Given : The word  ABSENT

To Find: How many different 6-letter arrangements can be formed using the letters in the word ABSENT, if each letter is used only once?

Solution:

Number of letters in ABSENT = 6

So, No. of arrangements can be formed using the letters in the word ABSENT, if each letter is used only once = 6!

                                                   = [tex]6 \times 5 \times 4\times 3 \times 2 \times 1[/tex]

                                                   = [tex]720[/tex]

So, Option C is true

Hence there are 720 different 6-letter arrangements can be formed using the letters in the word ABSENT.

Final answer:

There are 720 different 6-letter arrangements that can be formed from the word ABSENT, by applying the permutation formula 6! = 720. The correct option is c.

Explanation:

The question asks: How many different 6-letter arrangements can be formed using the letters in the word ABSENT, if each letter is used only once? To answer this, we need to calculate the number of permutations of 6 letters taken from 6. This is a simple permutation problem where we use the formula for permutations which is n!, where n is the total number of items to choose from and ! denotes a factorial, meaning the product of all positive integers up to n.

Given that the word ABSENT has 6 letters and we are arranging all 6, we have 6! = 6×5×4×3×2×1 = 720. Therefore, there are 720 different 6-letter arrangements that can be formed using the letters in ABSENT, with each letter used only once.

Answer:

  b = 1.098

Step-by-step explanation:

Each year, the GDP is 9.8% higher than the year before, so the multiplier each year is 1 + 9.8% = 1.098. This is the value of b.

  b = 1.098

Answer:

x = -1 and y = -4

Step-by-step explanation:

It is given that,

3x - 2y = 5     ----(1)

-2x - 3y = 14 ------(2)

To find the solution of equations

(1) * 2 ⇒

6x - 4y = 10  -----(3)

(2) * 3 ⇒

-6x - 9y =  42   ----(4)

eq(3) + eq(4) ⇒

6x - 4y = 10  -----(3)

-6x - 9y =  42  ----(4)

 0   - 13y = 52

y = 52/(-13) = -4

Substitute the value of y in eq(1)

3x - 2y = 5     ----(1)

3x - (2 * -4) = 5

3x  +8 = 5

3x = 5 - 8 = -3

x = -3/3 = -1

Therefore x = -1 and y = -4

Answer:

The solution is:

[tex](-1, -4)[/tex]

Step-by-step explanation:

We have the following equations

[tex]3x - 2y =5[/tex]

[tex]-2x - 3y = 14[/tex]

To solve the system multiply by [tex]\frac{3}{2}[/tex] the second equation and add it to the first equation

[tex]-2*\frac{3}{2}x - 3\frac{3}{2}y = 14\frac{3}{2}[/tex]

[tex]-3x - \frac{9}{2}y = 21[/tex]

[tex]3x - 2y =5[/tex]

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

[tex]-\frac{13}{2}y=26[/tex]

[tex]y=-26*\frac{2}{13}[/tex]

[tex]y=-4[/tex]

Now substitute the value of y in any of the two equations and solve for x

[tex]-2x - 3(-4) = 14[/tex]

[tex]-2x +12 = 14[/tex]

[tex]-2x= 14-12[/tex]

[tex]-2x=2[/tex]

[tex]x=-1[/tex]

The solution is:

[tex](-1, -4)[/tex]

Answer:

The point of intersection [tex]P\left(\dfrac{1133}{122},\dfrac{149}{122},\dfrac{699}{122}\right)[/tex]

Step-by-step explanation:

Equation of line:

[tex]x=7+9t[/tex]

[tex]y=3-7t[/tex]

[tex]z=7-5t[/tex]

Equation of plane:

[tex]5x-6y-7z=-1[/tex]

We need to find the point of intersection of line and plane.

Point of intersection: When both line and plane meet at single point.

So, put the value of x, y and z into plane.

[tex]5(7+9t)-6(3-7t)-7(7-5t)=-1[/tex]

[tex]35+45t-18+42t-49+35t=-1[/tex]

[tex]122t=-1+32[/tex]

[tex]t=\dfrac{31}{122}[/tex]

Substitute the value of t into x, y and z

[tex]x=7+9\cdot \dfrac{31}{122}=\dfrac{1133}{122}[/tex]

[tex]y=3-7\cdot \dfrac{31}{122}=\dfrac{149}{122}[/tex]

[tex]z=7-5\cdot \dfrac{31}{122}=\dfrac{699}{122}[/tex]

Point of intersection:

[tex]\left(\dfrac{1133}{122},\dfrac{149}{122},\dfrac{699}{122}\right)[/tex]

Hence, The point of intersection [tex]P\left(\dfrac{1133}{122},\dfrac{149}{122},\dfrac{699}{122}\right)[/tex]

Answer: The probability that 3 or more belong to the ruling clique is 0.34.

Step-by-step explanation:

Since we have given that

Number of total members = 50

Number of belonging to ruling clique = 10

Number of belonging to second class member = 40

We need to find the probability that 3 or more belong to the ruling clique.

Let X be the number of outcomes belong to ruling clique.

So, it becomes,

P(X≥3)=1-P(X<3)

[tex]P(X\geq 3)=1-P(X=1)-P(X=2)\\\\P(X\geq 3)=1-\dfrac{^{10}C_1\times ^{40}C_5}{^{50}C_6}-\dfrac{^{10}C_2\times ^{40}C_4}{^{50}C_6}\\\\P(X\geq 3)=1-0.41-0.25\\\\P(X\geq 3)=0.34[/tex]

Hence, the probability that 3 or more belong to the ruling clique is 0.34.

The probability of selecting 3 or more members from the ruling clique when choosing 6 members randomly from a club of 50 members (10 in ruling clique, 40 second-class) is 8.56%.

This probability problem can be solved using the concepts of Combinations and Binomial Theorem. You need to determine the number of ways to choose 3, 4, 5, or 6 members from the ruling clique (10 members) and the remaining from the second-class members (40 members). For each case, divide by the total number of ways to choose 6 members from all 50 members to get the probability. Sum up all the probabilities for each case to get the total probability of having 3 or more from the ruling clique.

1. Number of ways of choosing 3 from the ruling clique and 3 from the second class: C(10,3)*C(40,3) = 120*9880 = 1,185,600 ways

2. Number of ways of choosing 4 from the ruling clique and 2 from the second class: C(10,4)*C(40,2) = 210*780 = 163,800 ways

3. Number of ways of choosing 5 from the ruling clique and 1 from the second class: C(10,5)*C(40,1) = 252*40 = 10,080 ways

4. Number of ways of choosing 6 from the ruling clique and 0 from the second class: C(10,6)*C(40,0) = 210*1 = 210 ways

Total ways to choose 3 or more from the ruling clique: 1,185,600 + 163,800 + 10,080 +210 = 1,359,690 ways

From 50 members, the total ways to choose 6: C(50,6) = 15,890,700 ways

The Probability of 3 or more from the ruling clique = 1,359,690 / 15,890,700 = 0.0856 or 8.56%

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Answer: [tex]\dfrac{4}{165}[/tex]

Step-by-step explanation:

Given : The number of yellow balls in the box = 4

The total number of balls = [tex]4+2+5=11[/tex]

Since the given situation has dependent events.

Then, the probability of that all three balls are yellow is given by :-

[tex]\text{P(YYY)}==\dfrac{^4P_3}{^{11}P_3}=\dfrac{4\times3\times2}{11\times10\times9}\\\\\Rightarrow\ \text{P(YYY)}=\dfrac{4}{165}[/tex]

Hence, the probability of winning =[tex]\dfrac{4}{165}[/tex]