MATHEMATICS COLLEGE

A survey of 300 parks showed the following. 15 had only camping. 20 had only hiking trails. 35 had only picnicking. 185 had camping. 140 had camping and hiking trails. 125 had camping and picnicking. 210 had hiking trails. Determine the number of parks that:

a. Had at least one of these features.
b. Had all three features.
c. Did not have any of these features.
d. Had exactly two of these features.

Answers

Answer 1

Answer:

Answer:

A. 290 parks dad at least one of these features.

b. 95 parks had all three features.

c. 10 parks did not have any of these features.

d. 125 parks had exactly two of these features.

Step-by-step explanation:

This will be solved using set notation according to the venn diagram attached.

Let n(U) be the total number of parks surveyed

n(C) be those that had camping = 185

n(H) be those that had hiking trails = 210

n(C∩H) be those that had camping and hiking trails = 140

n(C∩P) be those that had camping and picnicking = 125

n(C∩P'∩H') be those that had only camping = 15

n(C'∩P'∩H) be those that had only hiking trails = 20

n(C'∩P∩H') be those that had only picnicking = 35

Find the calculation in the attached file

Answer 2

Answer:

Final answer:

The number of parks that had at least one of the listed features was 135.

The number of parks that had all three features was 20.

The number of parks that did not have any of these features was 165.

Explanation:

To determine the number of parks that had at least one of the listed features, we can add up the numbers of parks that had only camping, only hiking trails, and only picnicking. Then we subtract the parks that had two or three of these features, as they were already counted in the previous step. Doing this calculation, we get:

Parks with at least one feature: 15 + 20 + 35 + (185 - 140 - 125) + (140 - 125) + (210 - 140 - 125) = 135

To find the number of parks that had all three features, we need to subtract the parks that had only camping, only hiking trails, only picnicking, or none of these features from the total number of parks (300). Doing this calculation, we get:

Parks with all three features: 300 - 15 - 20 - 35 - (185 - 140 - 125) - (140 - 125) - (210 - 140 - 125) - (300 - 135) = 20

To determine the number of parks that did not have any of these features, we subtract the parks with at least one feature from the total number of parks (300). Doing this calculation, we get:

Parks with no features: 300 - 135 = 165

To calculate the number of parks that had exactly two features, we add the intersections of each pair of features and subtract the parks that had all three features. Doing this calculation, we get:

Parks with exactly two features: (185 - 140 - 125) + (140 - 125) + (210 - 140 - 125) - (300 - 20) = 60

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Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie of this type contains at least two chocolate chips to be greater than 0.99. Find the smallest value of the mean that the distribution can take.

Answers

Answer:

$\lambda \geq 6.63835$

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that $X \sim Poisson(\lambda)$

The probability mass function for the random variable is given by:

$f(x)=(e^(-\lambda) \lambda^x)/(x!) , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda$

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

$P(X\geq 2)=1-P(X<2)=1-P(X\leq 1)=1-[P(X=0)+P(X=1)]$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=(e^(-\lambda) \lambda^0)/(0!)=e^(-\lambda)$

$P(X=1)=(e^(-\lambda) \lambda^1)/(1!)=\lambda e^(-\lambda)$

And replacing we have this:

$P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^(-\lambda) +\lambda e^(-\lambda)[]$

$P(X\geq 2)=1-e^(-\lambda)(1+\lambda)$

And we want this probability that at least of 99%, so we can set upt the following inequality:

$P(X\geq 2)=1-e^(-\lambda)(1+\lambda)\geq 0.99$

And now we can solve for $\lambda$

$0.01 \geq e^(-\lambda)(1+\lambda)$

Applying natural log on both sides we have:

$ln(0.01) \geq ln(e^(-\lambda)+ln(1+\lambda)$

$ln(0.01) \geq -\lambda+ln(1+\lambda)$

$\lambda-ln(1+\lambda)+ln(0.01) \geq 0$

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

$x_(n+1)=x_n -(f(x_n))/(f'(x_n))$

Where :

$f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)$

$f'(x_n)=1-(1)/(1+\lambda)$

Iterating as shown on the figure attached we find a final solution given by:

$\lambda \geq 6.63835$

Final answer:

The problem pertains to Poisson Distribution in probability theory, focusing on finding the smallest mean (λ) such that the probability of having at least two chocolate chips in a cookie is more than 0.99. This involves solving an inequality using the formula for Poisson Distribution.

Explanation:

This problem pertains to the Poisson Distribution, often used in probability theory. In particular, we're looking at the number of events (in this case, the number of chocolate chips) that occur within a fixed interval. Here, the interval under study is a single cookie. The question requires us to find the smallest value of λ (the mean value of the distribution) such that the probability of getting at least two chocolate chips in a cookie is more than 0.99.

Using the formula for Poisson Distribution, the probability of finding k copies of an event is given by:

P(X=k) = λ^k * exp(-λ) / k!

The condition here is that the probability of finding at least 2 copies is more than 0.99. Therefore, you formally need to solve the inequality:

P(X>=2) = 1 - P(X=0) - P(X=1) > 0.99

Substituting the values of P(X=0) and P(X=1) from our standard formula, you will need to calculate and find the smallest value of λ that satisfies this inequality.

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Chapter1
Review & Refresh
Simplify the expression.

1. 18x - 6x + 2x

2. 4b – 7 – 15b + 3

3. 15(6-g)

4. –24 + 2(y – 9)

Please help me with these 4 problems!!:))

Answers

Answers:

1) 18x - 6x + 2x = 14x

Combine like terms by subtracting 6 from 18 and adding 2

2) 4b - 7 - 15b + 3 = - 11b - 4

Combine like terms add 4b and - 15b ; add -7 and 3

3) 15(6-g) = 90 - 15g

Distribute the 15 to both terms

4) -24 + 2(y - 9) = 2y - 42

Distribute the 2 to (y-9) and combine like terms

What is 783,264 rounded to the nearest ten thousand

Answers

Answer:

780,000 if it was at least 785k then you round it to 790k

Let S be the sphere of radius 1 centered at (2, 4, 6). Find the distance from S to the plane x + y + z = 0.

Answers

Answer:

5.928

Step-by-step explanation:

Given that:

The relation of the plane x+y+z= 0

Suppose (x,y,z) is any point on the plane.

Then the difference between (2,4,6) to (x,y,z) is:

$d^2 = (x-2)^2 + (y -4)^2 + ( z -6)^2 \n \n d^2 = (x^2 -4x+4) + ( y^2-8y +16) +(z^2 -12z + 36)$

$d^2 = x^2 + y^2 +z^2 -4x -8y -12z +4 +16 +36$

$d^2 = x^2 +y^2 + z^2 -4x -8y -12z +56$

∴

$f(x,y,z) =d^2 = x^2 + y^2 + z^2 - 4x -8y - 12 z +56 - - - (1)$

To estimate the maximum and minimum values of the function f(x,y,z) subject to the constraint g(x,y,z) = x+y+z =0

By applying Lagrane multipliers;

If we differentiate equation (1) with respect to x; we have:

f(x,y,z) = 2x -4

If we differentiate equation (1) with respect to y; we have:

f(x,y,z) = 2y - 8

If we differentiate equation (1) with respect to z; we have:

f(x,y,z) = 2z - 12

Differentiating g(x,y,z) with respect to x, we have:

$g_x(x,y,z) = 1$

Differentiating g(x,y,z) with respect to y, we have:

$g_y(x,y,z) = 1$

Differentiating g(x,y,z) with respect to z, we have:

$g_z(x,y,z) = 1$

Calculating the equations $\bigtriangledown f = \lambda \bigtriangleup g \ \ \ \& \ \ \ g(x,y,z) =0$

$f_x = \lambda g_x\n$

$2x - 4 = \lambda (1)$

$2x= 4 + \lambda$

$x= 2 + (\lambda )/(2) --- (2)$

$f_y = \lambda g_y$

$2x -8 = \lambda(1)$

$2x = 8+ \lambda$

$x = 4+(\lambda)/(2) --- (3)$

$f_z = \lambda g_z$

$2x -12 = \lambda (1)$

$x = 6 + (\lambda )/(2) --- (4)$

x+y+z = 0 - - - (5)

replacing x, y, z values in the given constraint

x + y + z = 0

$2+(\lambda)/(2)+4+(\lambda)/(2)+6+(\lambda)/(2)=0$

$12 + (3 \lambda )/(2)=0$

$(3 \lambda )/(2)=-12$

$3 \lambda=-12 * 2$

$3 \lambda=-24$

$\lambda=(-24)/(3)$

$\lambda=-8$

Therefore, from equation (2)

$x=2 +( \lambda )/(2)$

$x=2 +( -8 )/(2)$

x = 2 - 4

x = - 2

From equation (3)

$x=4 +( \lambda )/(2)$

$x=4 +( -8 )/(2)$

x = 4 - 4

x = 0

From equation (3)

$x=6 +( \lambda )/(2)$

$x=6 +( -8 )/(2)$

x = 6 -4

x = 2

i.e (x,y,z) = (-2, 0, 2)

∴

$d^2 = (x-2)^2 +(y-4)^2 + (z -6)^2$

$d^2 = (-2-2)^2 +(0-4)^2 + (2 -6)^2$

$d^2 = 16 +16 + 16$

$d^2 =48$

$d =√(48)$

$d= \pm 6.928$

since we are taking only the positive integer because distance cannot be negative, then:

The distance from the center of the sphere to the plane is 6.928.

However, the distance from the surface S to the plane is:

6.928 - radius of the sphere.

where;

the radius of the sphere is given as 1

Then:

the distance from the surface S to the plane is:

6.928 - 1

= 5.928

a bag contains 5 blue sticks,4 red sticks, and 3 orange sticks and you ask a friend to pick one without looking. what is probability that the stick will be blue?

Answers

the probability of getting a blue stick will be 5/12.

What is probability?

A probability is a numerical representation of the likelihood or chance that a specific event will take place. Both proportions ranging from 0 to 1 and percentages ranging from 0% to 100% can be used to describe probabilities.

Given, a bag containing 5 blue sticks,4 red sticks, and 3 orange sticks and you ask a friend to pick one without looking.

probability = desired outcome / maximum possible outcomes

For the probability of getting a blue stick

since there are 5 blue pen

desired outcome = 5

maximum possible outcomes = 5 + 4 + 3

maximum possible outcomes = 12

thus, probability = 5/12

therefore, 5/12 will be the probability of getting a blue stick.

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The probability is 5/12

At a computer store, a customer is considering 9 different computers,10 different monitors,8 different printers and 2different scanners. Assuming that each of the components is compatible with one another and that one of each is to be selected. Determine the number of different computer systems possible.

Answers

Answer:

The number of different computer systems possible is 1440.

Step-by-step explanation:

For each computer, there are 10 options of monitor.

For each monitor, there are 8 printers.

For each printer, there are 2 scanners.

There are 9 computers.

Determine the number of different computer systems possible.

9*10*8*2 = 1440

The number of different computer systems possible is 1440.

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The table shows the price for different numbers of notebooks:Number of Notebooks Price (in dollars)2 123 184 5 30

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