Sneha is inspecting the average cleanliness of the rooms in a hotel. There are 10 floors in the hotel, each with 20 rooms. From each floor, she randomly selects 2 rooms and inspects them and then takes the average of these scores. What type of sampling did Sneha use

Answer:

a) not independent

b) not mutually exclusive

Step-by-step explanation:

Given:

- A 6 sided die is rolled

- Event A is rolling an even number

- Event B is rolling a prime number

Find

- (a) Are A and B independent events?

- (b) Are A and B mutually exclusive events?

Solution:

- We will find the probability of each event:

                       set(Even number: A) = {2, 4, 6} = 3 outcomes

                       set(Prime number: B) = {2 , 3, 5} = 3 outcomes

- The probabilities are:

                        P(A) = 3/6 = 0.5

                        P(B) = 3/6 = 0.5

- For Event A and B to be independent then the following condition must match:

                        P ( A & B ) = P(A)*P(B)

                        set (A&B) = {2} = 1 outcome

                        P(A&B) = 1/6

                        1 / 6 = 0.5*0.5

                        1/6 = 0.25         ...... NOT INDEPENDENT

- For Event A and B to be mutually exclusive then the following condition must match:

                        P(A&B) = 0

                        P(A&B) = 1/6

Hence, we can say the events are NOT MUTUALLY EXCLUSIVE

No, events A and B are not independent events because the probability of both events occurring is not equal to the product of their individual probabilities. Furthermore, events A and B are not mutually exclusive events because they can both occur simultaneously if the number rolled is 2.

No, events A and B are not independent events. In order for two events to be independent, the probability of both events occurring should be equal to the product of the probabilities of each event occurring individually. However, in this case, the probability of rolling an even number (event A) is 3/6 (since there are three even numbers out of six total outcomes), and the probability of rolling a prime number (event B) is 2/6 (since there are two prime numbers out of six total outcomes). Therefore, P(A and B) is not equal to P(A) * P(B), indicating that events A and B are dependent.

Moreover, events A and B are not mutually exclusive events either. Mutually exclusive events are events that cannot occur at the same time, meaning they have no outcomes in common. In this case, event A (rolling an even number) and event B (rolling a prime number) can both occur simultaneously if the number rolled is 2, which is both even and prime. Therefore, P(A and B) is not equal to 0, indicating that events A and B are not mutually exclusive.

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Answer:

a) [tex] P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3}{0.45}= 0.667[/tex]

Represent the probability that the event B occurs given that the event A occurs first

b) [tex] P(B'|A) = \frac{0.15}{0.45}=0.333[/tex]

Represent the probability that the event B no occurs given that the event A occurs first

c) [tex] P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.3}{0.35}= 0.857[/tex]

Represent the probability that the event A occurs given that the event B occurs first

d) [tex] P(A'|B) = \frac{0.05}{0.35}=0.143[/tex]

Represent the probability that the event A no occurs given that the event B occurs first

Step-by-step explanation:

For this case we have the following probabilities given for the events defined A and B

[tex] P(A) = 0.45, P(B) = 0.35, P(A \cap B) =0.30[/tex]

For this case we can begin finding the probability for the complements:

[tex] P(B') =1-P(B) = 1-0.35= 0.65[/tex]

[tex] P(A') =1-P(A) = 1-0.45= 0.55[/tex]

For this case we are interested on the following probabilities:

Part a

[tex] P(B|A)[/tex]

For this case we can use the Bayes theorem and we can find this probability like this:

[tex] P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3}{0.45}= 0.667[/tex]

Represent the probability that the event B occurs given that the event A occurs first

Part b

[tex] P(B'|A) = \frac{P(B' \cap A)}{P(A}[/tex]

And for this case we can find [tex] P(B' \cap A) =P(A) -P(A\cap B)= 0.45-0.3=0.15[/tex]

And if we replace we got:

[tex] P(B'|A) = \frac{0.15}{0.45}=0.333[/tex]

Represent the probability that the event B no occurs given that the event A occurs first

Part c

[tex] P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.3}{0.35}= 0.857[/tex]

Represent the probability that the event A occurs given that the event B occurs first

Part d

[tex] P(A'|B) = \frac{P(A' \cap B)}{P(B}[/tex]

And for this case we can find [tex] P(A' \cap B) =P(B) -P(A\cap B)= 0.35-0.3=0.05[/tex]

And if we replace we got:

[tex] P(A'|B) = \frac{0.05}{0.35}=0.143[/tex]

Represent the probability that the event A no occurs given that the event B occurs first

Conditional probabilities are calculated as the ratio of the intersection of two events to the probability of the given event. P(B|A) is approximately 0.6667, P(B'|A) is approximately 0.3333, P(A|B) is approximately 0.8571, and P(A'|B) is approximately 0.1429.

When considering the probability of randomly selecting a student at a university who has a Visa (Event A) or a MasterCard (Event B), we can use the given probabilities to calculate the following:

P(B|A) is the probability that a selected individual has a MasterCard given they have a Visa. It is calculated as P(A [tex]\cap[/tex] B) / P(A). Given P([tex]A \cap B[/tex]) = 0.30 and P(A) = 0.45, P(B|A) = 0.30 / 0.45 which rounds to 0.6667.

P(B'|A) is the probability that a selected individual does not have a MasterCard given they have a Visa. It is 1 - P(B|A), which is 1 - 0.6667, rounding to 0.3333.

P(A|B) is the probability that a selected individual has a Visa given they have a MasterCard. It is calculated as P(A \\cap B) / P(B). Given P(A [tex]\cap[/tex] B) = 0.30 and P(B) = 0.35, P(A|B) = 0.30 / 0.35 which rounds to 0.8571.

P(A'|B) is the probability that a selected individual does not have a Visa given they have a MasterCard. It is 1 - P(A|B), which is 1 - 0.8571, rounding to 0.1429.

Answer:

50% probability that a randomly selected respondent voted for Obama.

Step-by-step explanation:

We have these following probabilities:

60% probability that an Ohio resident does not have a college degree.

If an Ohio resident does not have a college degree, a 52% probability that he voted for Obama.

40% probability that an Ohio resident has a college degree.

If an Ohio resident has a college degree, a 47% probability that he voted for Obama.

What is the probability that a randomly selected respondent voted for Obama?

This is the sum of 52% of 60%(non college degree) and 47% of 40%(college degree).

So

[tex]P = 0.52*0.6 + 0.47*0.4 = 0.5[/tex]

50% probability that a randomly selected respondent voted for Obama.

Answer:

V=15.44

Step-by-step explanation:

We have a formula

V=\int^{π/3}_{-π/3} A(x) dx ,

where A(x) calculate as  cross sectional.

We have:

Inner radius: 5 + sec(x) - 5= sec(x)  

Outer radius: 7 - 5=2, we get

A(x)=π 2²- π· sec²(x)

A(x)=π(4-sec²(x))

Therefore, we calculate the volume V, and we get

V=\int^{π/3}_{-π/3} A(x) dx

V=\int^{π/3}_{-π/3} π(4-sec²(x)) dx

V=[ π(4x-tan(x)]^{π/3}_{-π/3}

V=π·(8π/3-2√3)

V=15.44

We use a site geogebra.org to plot the graph.

The volume of the solid obtained by rotating the region bounded by y = 5 + sec(x), -π/3 ≤ x ≤ π/3, y = 7 about y = 5 is calculated using the disc method formula in calculus, resulting in a volume of 8π²/3 cubic units.

The region that we need to rotate around the line y=5 is bounded by the curves y=7 and y=5+sec(x) over the interval -π/3 ≤ x ≤ π/3. The resulting solid is a disc shaped object. We can find the volume of this solid using the disc method formula in calculus:

V = π ∫ [R(x)]² dx

, where R(x) = radius function. The radius function is the absolute difference between y=7 (the upper curve) and y=5 (the line of rotation), which equals 2. Therefore, the integral becomes:

V = π ∫ from -π/3 to π/3 [2]² dx

, which simplifies to

V = 4π [x] from -π/3 to π/3

. Finally, evaluate this expression by subtracting the lower limit from the upper limit, giving

V = 4π(2π/3) = 8π²/3

cubic units.

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Answer:

8 weeks of production from A and 4 weeks of production of B required to produce 28,000 monitors and 132,000 televisions.

Step-by-step explanation:

Let the number of weeks of production for team A be x and for team B be y.

In a week, facility A produces 2,000 computer monitors.

In a week ,facility A produces 3,000 computer monitors.

Total monitors to be produced = 28,000

[tex]2000x+3000y=28,000[/tex]

[tex]2x+3y=28[/tex]..[1]

In a week, facility A produces 10,000 flat panel televisions.

In a week ,facility A produces 13,000 flat panel televisions.

Total flat panel televisions to be produced = 132,000

[tex]10,000x+13,000y=132,000[/tex]

[tex]10x+13y=132[/tex]..[2]

Solving equation [1] and [2] by eliminationg method.

5 × [1] + (-1) × [2]

[tex]10x+15y=140[/tex]

[tex]-10x-13y=-132[/tex]

y = 4

[tex]x = \frac{28-3\times 4}{2}=\frac{16}{2}=8[/tex]

8 weeks of production from A and 4 weeks of production of B required to produce 28,000 monitors and 132,000 televisions.

To produce 28,000 monitors and 132,000 televisions, it would take approximately 3.5 weeks for monitors and 3.67 weeks for televisions.

To determine the number of weeks of production required, we need to divide the total number of monitors and televisions needed by the production rate of each facility.

For monitors: Facility A produces a total of 2000 + 3000 = 5000 monitors per week, and Facility B produces 3000 monitors per week. Therefore, the total production rate for monitors is 5000 + 3000 = 8000 monitors per week.

For televisions: Facility A produces a total of 10,000 + 13,000 = 23,000 televisions per week, and Facility B produces 13,000 televisions per week. Therefore, the total production rate for televisions is 23,000 + 13,000 = 36,000 televisions per week.

To produce 28,000 monitors and 132,000 televisions, it would take:

Monitors: 28,000 / 8000 = 3.5 weeks

Televisions: 132,000 / 36,000 = 3.67 weeks

Therefore, the number of weeks required to produce the desired quantities is approximately 3.5 weeks for monitors and 3.67 weeks for televisions.

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Answer:

[tex]y(t) = \$1900*1.0375^t[/tex]

Step-by-step explanation:

If y is the future value of the $1900 investment, in dollars, after t years at a rate of 3.75% per year, compounded annually. The exponential function that describes the relationship between the variables y and t is:

[tex]y(t) = \$1900*(1.0375)^t[/tex]

This relationship means that for every year t, the amount y will increase by a factor of 1.0375.

To represent the value of an investment of $1900 at a 3.75% annual interest rate compounded annually, the exponential function y = 1900(1 + 0.0375)^t is used, where t is the number of years.

The question asks for an exponential function that shows the relationship between the value of an investment, y, and time, t, when the principal is $1900 and the interest rate is 3.75% compounded annually. The standard formula for an investment compounded annually is y = P(1 + r)t, where P is principal, r is the annual interest rate as a decimal, and t is the number of years.

For this specific case, the function would be y = 1900(1 + 0.0375)t. This represents the value of the investment after t years, given the initial investment and interest rate provided.

Answer:

The balance after 5 years will be $2,187.91.

Step-by-step explanation:

This problem can be solved by the following formula:

[tex]A = P(1 + r)^{t}[/tex]

In which A is the final amount(balance), P is the principal(the deposit), r is the interest rate and t is the time, in years.

In this problem, we have that:

[tex]P = 1800, r = 0.05, t = 4[/tex]

We want to find A

So

[tex]A = P(1 + r)^{t}[/tex]

[tex]A = 1800(1+0.05)^{5}[/tex]

[tex]A = 2,187.91[/tex]

The balance after 5 years will be $2,187.91.

Answer: 2188

Step-by-step explanation:

so the person deposited 1800 and expects an annual raise in the amount of 5 percent

so the equation for 1 year is

1800(1+(5/100))=the answer

but for four years u will have to power the bract by 4

180(1+(5/100))^4= 2187.9

aka 2188

Answer: it will take Sebastian's mom 45 minutes to catch up with the bus.

Step-by-step explanation:

By the time Sebastian's mom catches up with the bus, she would have covered the same distance with the bus.

Let t represent the time it will take for Sebastian's mom to catch up with the bus.

Distance = speed × time

Sebastian's school bus averages 28 miles per hour.

Distance covered by Sebastian's school bus in t hours is

28 × t = 28t

Sebastian's mom travelled at an average speed of 42mph. Since she left home 0.25 hour after Sebastian, the time it would take her to cover the same distance is

(t - 0.25) hour. Distance covered in

(t - 0.25) hour is

42(t - 0.25)

Since the distance covered is the same, then

28t = 42(t - 0.25)

28t = 42t - 10.5

42t - 28t = 10.5

14t = 10.5

t = 10.5/14

t = 0.75

Converting to minutes, it becomes

0.75 × 60 = 45 minutes

Answer:

Jacinta was paid $127.5 for babysitting.

Step-by-step explanation:

This problem can be solved by consecutive rules of three.

Jacinta babysat for 2 1/2 hours each Saturday for 6 weeks

How many hours did she work?

She worked for 6 Saturdays, 2.5 hours in each. So

1 Saturday - 2.5 hours

6 Saturdays - x hours

[tex]x = 6*2.5[/tex]

[tex]x = 15[/tex]

She worked 15 hours.

Paid $8.50 each hour

15*$8.50 = 127.5

Jacinta was paid $127.5 for babysitting.

Answer:

~ is reflexive.

~ is asymmetric.

~ is transitive.

Step-by-step explanation:

~ is reflexive:

i.e., to prove [tex]$ \forall (a, b) \in \mathbb{R}^2 $[/tex], [tex]$ (a, b) R(a, b) $[/tex].

That is, every element in the domain is related to itself.

The given relation is [tex]$\sim: (a,b) \sim (c, d) \iff a^2 + b^2 \leq c^2 + d^2$[/tex]

Reflexive:

[tex]$ (a, b) \sim (a, b) $[/tex] since [tex]$ a^2 + b^2 = a^2 + b^2 $[/tex]

This is true for any pair of numbers in [tex]$ \mathbb{R}^2 $[/tex]. So, [tex]$ \sim $[/tex] is reflexive.

Symmetry:

[tex]$ \sim $[/tex] is symmetry iff whenever [tex]$ (a, b) \sim (c, d) $[/tex] then [tex]$ (c, d) \sim (a, b) $[/tex].

Consider the following counter - example.

Let (a, b) = (2, 3) and (c, d) = (6, 3)

[tex]$ a^2 + b^2 = 2^2 + 3^2 = 4 + 9 = 13 $[/tex]

[tex]$ c^2 + d^2 = 6^2 + 3^2 = 36 + 9 = 42 $[/tex]

Hence, [tex]$ (a, b) \sim (c, d) $[/tex] since [tex]$ a^2 + b^2 \leq c^2 + d^2 $[/tex]

Note that [tex]$ c^2 + d^2 \nleq a^2 + b^2 $[/tex]

Hence, the given relation is not symmetric.

Transitive:

[tex]$ \sim $[/tex] is transitive iff whenever [tex]$ (a, b) \sim (c, d) \hspace{2mm} \& \hspace{2mm} (c, d) \sim (e, f) $[/tex] then [tex]$ (a, b) \sim (e, f) $[/tex]

To prove transitivity let us assume [tex]$ (a, b) \sim (c, d) $[/tex] and [tex]$ (c, d) \sim (e, f) $[/tex].

We have to show [tex]$ (a, b) \sim (e, f) $[/tex]

Since [tex]$ (a, b) \sim (c, d) $[/tex] we have: [tex]$ a^2 + b^2 \leq c^2 + d^2 $[/tex]

Since [tex]$ (c, d) \sim (e, f) $[/tex] we have: [tex]$ c^2 + d^2 \leq e^2 + f^2 $[/tex]

Combining both the inequalities we get:

[tex]$ a^2 + b^2 \leq c^2 + d^2 \leq e^2 + f^2 $[/tex]

Therefore, we get:  [tex]$ a^2 + b^2 \leq e^2 + f^2 $[/tex]

Therefore, [tex]$ \sim $[/tex] is transitive.

Hence, proved.

Answer:

a) there is s such that r>s and s is positive

b) For any r>0 , there exists s>0 such that s<r

Step-by-step explanation:

a) We are given a positive real number r. We need to wite that there is a positive real number that is smaller. Call that number s. Then r>s (this is equivalent to s<r, s is smaller than r) and s is positive (or s>0 if you prefer). We fill in the blanks using the bold words.

b) The last part claims that s<r, that is, s is smaller than r. We know that this must happen for all posirive real numbers r, that is, for any r>0, there is some positive s such that s<r. In other words, there exists s>0 such that s<r.

The statement with variables filled in is: (a) Given any positive real number r, there is s such that s is a positive real number and s is less than r. (b) For any positive real number r, there exists s such that s < r.

The statement can be rewritten with variables as follows:

(a) Given any positive real number r, there is s such that s is a positive real number and s is less than r.

(b) For any positive real number r, there exists s such that s < r.

The concept here is that for any positive real number, you can always find another positive real number that is less than the given number. This is fundamental in understanding the density of real numbers on the number line, where between any two distinct real numbers, there are infinitely many other real numbers.

Answer:

(a) Probability that the second one selected is defective given that the first one was defective = 0.00450

(b) Probability that both are defective = 0.0112461

(c) Probability that both are acceptable = 0.986

    2. When Three containers are selected

(a) Probability that the third one selected is defective given that the first and second one selected were defective = 0.002.

(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay = 0.00451

(c) Probability that all three are defective = 6.855 x [tex]10^{-8}[/tex] .  

Step-by-step explanation:

We are given that a batch of 445 containers for frozen orange juice contains 3 defective ones i.e.

                  Total containers = 445

                   Defective ones   = 3

           Non - Defective ones = 442 { Acceptable ones}

(a) Probability that the second one selected is defective given that the first one was defective is given by;

  Since we had selected one defective so for selecting second the available

  containers are 444 and available defective ones are 2 because once

   chosen they are not replaced.

Hence, Probability that the second one selected is defective given that the first one was defective = [tex]\frac{2}{444}[/tex] = 0.00450

(b) Probability that both are defective = P(first being defective) +

                                                                     P(Second being defective)

                 = [tex]\frac{3}{445} + \frac{2}{444}[/tex] = 0.0112461

(c) Probability that both are acceptable = P(First acceptable) +  P(Second acceptable)

Since, total number of acceptable containers are 442 and total containers are 445.

 So, Required Probability = [tex]\frac{442}{445}*\frac{441}{444}[/tex] = 0.986

(a) Probability that the third one selected is defective given that the first and second one selected were defective is given by;

Since we had selected two defective containers so now for selecting third defective one, the available total containers are 443 and available defective container is 1 .

Therefore, Probability that the third one selected is defective given that the first and second one selected were defective = [tex]\frac{1}{443}[/tex] = 0.002.

(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay is given by;

Since we had selected two containers so for selecting third container to be defective, the total containers available are 443 and available defective containers are 2 as one had been selected.

Hence, Required probability = [tex]\frac{2}{443}[/tex] = 0.00451 .

(c) Probability that all three are defective = P(First being defective) +

                              P(Second being defective) +  P(Third being defective)

        = [tex]\frac{3}{445}* \frac{2}{444} * \frac{1}{443}[/tex] = 6.855 x [tex]10^{-8}[/tex] .                

Final answer:

The probability values are found by considering the number of defective and non-defective containers remaining at each step of selection.

Explanation:

The probability that the second one selected is defective given that the first one was defective can be found by looking at the remaining pool of containers. Initially, there are 3 defective containers out of 445.

After picking one, there are 2 defective containers left out of 444. The probability is therefore calculated as 2/444 = 0.00450 (rounded to five decimal places).

The probability that both containers are defective is calculated by multiplying the probability of selecting a defective container on the first draw with the probability of selecting a defective container on the second draw: (3/445) * (2/444) = 0.00003 (rounded to seven decimal places).

The probability that both containers are acceptable is the probability of not selecting a defective container twice: (442/445) * (441/444) = 0.98876.

If the first and second containers selected are defective, the probability that the third one selected is defective comes from the remaining one defective container out of 443: 1/443 = 0.00226.

If the first container is defective and the second one is not, the probability that the third one selected is defective is 2/443 = 0.00451 (rounded to five decimal places).

The probability that all three containers selected are defective is the product of the probabilities of selecting a defective container each time without replacement: (3/445) * (2/444) * (1/443) = 0.00002.

Answer:

[tex]a=285 -0.674*9.1=278.87[/tex]

So the value of height that separates the bottom 25% of data from the top 75% is 278.87.  

[tex]a=285 +0.524*9.1=289.77[/tex]

So the value of height that separates the bottom 70% of data from the top 30% is 289.77.  

The answer would be 278.87 and 289.77

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the amount of energy of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(285,9.1)[/tex]  

Where [tex]\mu=285[/tex] and [tex]\sigma=9.1[/tex]

Lowest 25%

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.75[/tex]   (a)

[tex]P(X<a)=0.25[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.25[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.25[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-0.674<\frac{a-285}{9.1}[/tex]

And if we solve for a we got

[tex]a=285 -0.674*9.1=278.87[/tex]

So the value of height that separates the bottom 25% of data from the top 75% is 278.87.  

Highest 30%

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.30[/tex]   (a)

[tex]P(X<a)=0.70[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.7 of the area on the left and 0.3 of the area on the right it's z=0.524. On this case P(Z<0.524)=0.7 and P(z>0.524)=0.3

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.7[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.7[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=0.524<\frac{a-285}{9.1}[/tex]

And if we solve for a we got

[tex]a=285 +0.524*9.1=289.77[/tex]

So the value of height that separates the bottom 70% of data from the top 30% is 289.77.  

The lower limit for the energy amounts of the remaining computers is 279.03 kwh, and the upper limit is 290.74 kwh.

To find the upper and lower limits for the energy amounts of the remaining computers, we need to find the z-scores corresponding to the lowest 25% and highest 30% of the distribution.

Using the invNorm function in a calculator, we find that the z-score for the lowest 25% is approximately -0.674 and the z-score for the highest 30% is approximately 0.524.

To find the corresponding energy amounts, we use the formula:

Lowest Energy Amount = Mean + (Z-score * Standard Deviation)

Highest Energy Amount = Mean + (Z-score * Standard Deviation)

Substituting the z-scores and the given values, we get:

Lowest Energy Amount = 285 + (-0.674 * 9.1) = 279.03 kwh

Highest Energy Amount = 285 + (0.524 * 9.1) = 290.74 kwh

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Answer:

The answer to your question is 2.8 x 10¹¹

Step-by-step explanation:

Data

1 tree produces 2.6 x 10² pounds of oxygen/year

number of trees = 3.9 x 10¹¹

pounds of oxygen per day = ?

Process

1.- Divide the pounds of oxygen by 365, to get the pounds of oxygen per day.

                      2.6 x 10² / 365   = 0.712

2.- Multiply the number of trees by the pounds of oxygen per day

                     3.9 x 10¹¹  x   0.712  = 2.8 x 10¹¹ pounds of oxygen

Answer:

The amount at the end of 4 years is $2,252.79.

Step-by-step explanation:

The amount formula for the compound interest compounded quarterly is:

[tex]A=P[1+\frac{r}{4}]^{4t}[/tex]

Here,

A = Amount after t years

P = Principal amount

t = number of years

r = interest rate

Given:

P = $1,250, r = 0.15, t = 4 years.

The amount at the end of 4 years is:

[tex]A=P[1+\frac{r}{4}]^{4t}\\=1250\times[1+\frac{0.15}{4}]^{4\times4}\\=1250\times1.80223\\=2252.7875\\\approx2252.79[/tex]

Thus, the amount at the end of 4 years is $2,252.79.

Final answer:

To calculate the amount of money you will have at the end of four years with quarterly deposits and compounded interest, use the formula for compound interest:

. Substituting the given values, the result is approximately $1,776.40.

Explanation:

To calculate the amount of money you will have at the end of four years with quarterly deposits and compounded interest, you can use the formula for compound interest:

A = P(1 + r/n)^(nt)

Where:

A is the total amount of money after the specified time period

P is the principal amount (initial deposit)

r is the annual interest rate (15% in this case)

n is the number of times interest is compounded per year (4 in this case for quarterly compounding)

t is the specified time period in years (4 in this case)

Let's substitute the given values into the formula:

A ≈ 1250 * 1.82212

A ≈ $1,1776.4

Therefore, you will have approximately $1,776.40 at the end of four years.

Answer:

S= {(A,B,C) , (A,C,B) , (B,C,A), (B,A,C), (C,A,B), (C,B,A)}

And we have 6 possible outcomes for this case.

Step-by-step explanation:

By definition the sample space of an experiment "is the set of all possible outcomes or results of that experiment".

For the case described here: "A taste tester ranks three varieties of yogurt, A, B, and C, according to preference.".

Assuming that we have three varieties of yogurt {A,B,C}

We denote the event (A,B,C) like this: A is the first, B the second and C the third in the rank. And for example the event (C,B,A) means that C is on the 1th position of the rank, B on the 2nd position and A on the 3th position.  

The sampling space denoted by S and is given by:

S= {(A,B,C) , (A,C,B) , (B,C,A), (B,A,C), (C,A,B), (C,B,A)}

And we have 6 possible outcomes for this case.

The sample space for ranking three yogurt varieties (A, B, C) consists of 6 permutations: (A, B, C), (A, C, B), (B, A, C), (B, C, A), (C, A, B), and (C, B, A).

The problem involves finding all possible ways a taste tester can rank three varieties of yogurt: A, B, and C. In this context, a sample space represents all possible outcomes of the experiment.

Since the taste tester is ranking the yogurt varieties, we are looking for all permutations of the set {A, B, C}.

A permutation is an arrangement of all the members of a set in a specific order. For three items, the total number of permutations is given by 3! (3 factorial), which equals 6.

Therefore, the sample space S, containing all possible rankings, is:

Answer:

5 classes.

Step-by-step explanation:

You can use the [tex]2^k[/tex] rule to determine the number of classes for a frequency distribution.

The [tex]2^k[/tex] rule says that [tex]2^k\geq n[/tex] where

[tex]k[/tex] is the number of classes

[tex]n[/tex] is the number of the data points

We know that the number of data points is [tex]n[/tex] = 20.

Next, we start searching for [tex]k[/tex] so that we can get a number 2 to the [tex]k[/tex] that is larger that the number of data points.

[tex]2^4=16\\2^5=32[/tex]

This suggests that you should use 5 classes.

Final answer:

For this data, it is recommended to use 5 classes.

Explanation:

To organize the data into a frequency distribution, you need to determine the number of classes.

The recommended number of classes can vary depending on the data.

One popular rule is to use the square root of the number of data points to determine the number of classes.

In this case, there are 20 data points, so the square root is approximately 4.47.

Since you can't have a fraction of a class, you can round up to the nearest whole number, which gives you a recommendation of 5 classes.

Answer:

C. Data Mining

Step-by-step explanation:

As data mining is a technique which is used predict and to find patterns or relationships among elements of the data in a large database. It also facilitate the enterprise to predict the future trends.

Answer:

The answer is D) 0.476

Step-by-step explanation:

If 1 represents HIT, and 0 represents OUT, the probability that Sammy will get a hit = the number of HITS (1s) ÷ the output from the simulation (i.e., total number of HITs and OUTs in the simulation)

Where:

the number of HITS (1s) = 20

the output from the simulation (i.e., total number of HITs and OUTs in the simulation) = 42

therefore, the probability that Sammy will get a hit = 20 ÷ 42 = 0.476

This implies that the correct answer is D) 0.476

Answer:

0.84 square in

Step-by-step explanation:

Since the capacity of the cable is proportional to its​ cross-sectional area. If a cable that is 0.3 sq in can hold 2500 lb then per square inch it can hold

2500 / 0.3 = 8333.33 lb/in

To old 7000 lb it the cross-sectional area would need to be

7000 / 8333.33 = 0.84 square in

Answer:

0.84 square in.

Step-by-step explanation:

Since the capacity of the cable is proportional to its​ cross-sectional area.

Mathematically,

C = k * A

A1 = 0.3 sq

C1 = 2500 lb

C2 = 7000 lb

C1/A1 = C2/A2

2500/0.3 = 7000/C2

= 7000 / 8333.33

= 0.84 square in.

Answer:

answers are shown in the file attached

Step-by-step explanation:

The detailed step are as shown in the attached file.

The solution of the all the equations are:

a) x = {10 + 7k}, where k is an integer.

b) x = {2 + 23k}, where k is an integer.

c) x = {18 + 26k}, where k is an integer.

d) x = {2 + 5k}, where k is an integer.

e) x = {5 + 6k}, where k is an integer.

f) x = {1 + 6k}, where k is an integer.

To find all integer solutions (x ∈ Z) for each equation, we need to solve them using modular arithmetic.

For each equation, we will use different approaches depending on the specific form of the equation.

Let's solve them one by one:

(a) 3x ≡ 2 (mod 7):

To find x, we'll multiply both sides of the equation by the modular inverse of 3 modulo 7, which is 5 (since 3 × 5 ≡ 1 (mod 7)):

3x × 5 ≡ 2 × 5 (mod 7)

15x ≡ 10 (mod 7)

Now, we find the smallest non-negative integer solution by dividing both sides by the greatest common divisor (GCD) of 15 and 7 (which is 1 since 15 and 7 are coprime):

x ≡ 10 (mod 7)

The solution set is x = {10, 17, 24, 31, ...}.

Since we are looking for integer solutions, we can simplify it to x = {10 + 7k}, where k is an integer.

(b) 5x + 1 ≡ 13 (mod 23):

Subtract 1 from both sides:

5x ≡ 12 (mod 23)

Next, we'll find the modular inverse of 5 modulo 23, which is 14 (since 5 * 14 ≡ 1 (mod 23)):

5x × 14 ≡ 12 × 14 (mod 23)

70x ≡ 168 (mod 23)

Now, reduce the coefficients to the smallest positive residue modulo 23:

x ≡ 2 (mod 23)

The solution set is x = {2, 25, 48, ...}, which can be simplified to x = {2 + 23k}, where k is an integer.

(c) 5x + 1 ≡ 13 (mod 26):

This equation is the same as the previous one. Follow the same steps:

5x ≡ 12 (mod 26)

Find the modular inverse of 5 modulo 26, which is 21 (since 5 × 21 ≡ 1 (mod 26)):

5x × 21 ≡ 12 × 21 (mod 26)

105x ≡ 252 (mod 26)

Reduce the coefficients to the smallest positive residue modulo 26:

x ≡ 18 (mod 26)

The solution set is x = {18, 44, 70, ...}, which can be simplified to x = {18 + 26k}, where k is an integer.

(d) 9x ≡ 3 (mod 5):

To find x, we'll multiply both sides by the modular inverse of 9 modulo 5, which is 4 (since 9 × 4 ≡ 1 (mod 5)):

9x × 4 ≡ 3 × 4 (mod 5)

36x ≡ 12 (mod 5)

Reduce the coefficients to the smallest positive residue modulo 5:

x ≡ 2 (mod 5)

The solution set is x = {2, 7, 12, ...}, which can be simplified to x = {2 + 5k}, where k is an integer.

(e) 5x ≡ 1 (mod 6):

To find x, we'll multiply both sides by the modular inverse of 5 modulo 6, which is 5 (since 5 × 5 ≡ 1 (mod 6)):

5x × 5 ≡ 1 × 5 (mod 6)

25x ≡ 5 (mod 6)

Reduce the coefficients to the smallest positive residue modulo 6:

x ≡ 5 (mod 6)

The solution set is x = {5, 11, 17, ...}, which can be simplified to x = {5 + 6k}, where k is an integer.

(f) 3x ≡ 1 (mod 6):

To find x, we'll multiply both sides by the modular inverse of 3 modulo 6, which is 3 (since 3 × 3 ≡ 1 (mod 6)):

3x × 3 ≡ 1 × 3 (mod 6)

9x ≡ 3 (mod 6)

Reduce the coefficients to the smallest positive residue modulo 6:

x ≡ 1 (mod 6)

The solution set is x = {1, 7, 13, ...}, which can be simplified to x = {1 + 6k}, where k is an integer.

Learn more about modular arithmetic click;

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a. The average velocity for the time period beginning at t = 2 and lasting 0.5 seconds is -36 ft/s.

b. The average velocity for the time period beginning at t = 2 and lasting 0.1 seconds is -29.6 ft/s.

c. The average velocity for the time period beginning at t = 2 and lasting 0.05 seconds is -40.8 ft/s.

d. The average velocity for the time period beginning at t = 2 and lasting 0.01 seconds is 3.84 ft/s.

e. The estimated instantaneous velocity at t = 2 is -32 ft/s.

(a) To find the average velocity over a time period, we use the formula for average velocity:

[tex]\[ \text{Average Velocity} = \frac{\text{Change in height}}{\text{Change in time}} \][/tex]

Given [tex]\( y = 36t - 16t^2 \)[/tex], we'll find the heights at [tex]\( t = 2 \)[/tex] and [tex]\( t = 2.5 \)[/tex], then calculate the difference.

At [tex]\( t = 2 \), \( y = 36(2) - 16(2)^2 = 72 - 64 = 8 \)[/tex] ft.

At [tex]\( t = 2.5 \), \( y = 36(2.5) - 16(2.5)^2 = 90 - 100 = -10 \)[/tex] ft.

Change in height = [tex]\( -10 - 8 = -18 \)[/tex] ft.

Change in time = [tex]\( 2.5 - 2 = 0.5 \)[/tex] sec.

Average velocity = [tex]\( \frac{-18}{0.5} = -36 \)[/tex] ft/s.

(b) Following the same procedure as (a), at [tex]\( t = 2.1 \), \( y = 36(2.1) - 16(2.1)^2 = 75.6 - 70.56 = 5.04 \)[/tex] ft.

Change in height = [tex]\( 5.04 - 8 = -2.96 \)[/tex] ft.

Change in time = [tex]\( 2.1 - 2 = 0.1 \)[/tex] sec.

Average velocity = [tex]\( \frac{-2.96}{0.1} = -29.6 \)[/tex] ft/s.

(c) Continuing the process, at [tex]\( t = 2.05 \), \( y = 36(2.05) - 16(2.05)^2 = 73.8 - 67.84 = 5.96 \)[/tex] ft.

Change in height = [tex]\( 5.96 - 8 = -2.04 \)[/tex] ft.

Change in time = [tex]\( 2.05 - 2 = 0.05 \)[/tex] sec.

Average velocity = [tex]\( \frac{-2.04}{0.05} = -40.8 \)[/tex] ft/s.

(d) Applying the method to [tex]\( t = 2.01 \), \( y = 36(2.01) - 16(2.01)^2 = 72.36 - 64.3216 = 8.0384 \)[/tex] ft.

Change in height = [tex]\( 8.0384 - 8 = 0.0384 \)[/tex] ft.

Change in time = [tex]\( 2.01 - 2 = 0.01 \)[/tex] sec.

Average velocity = [tex]\( \frac{0.0384}{0.01} = 3.84 \)[/tex] ft/s.

(e) To estimate the instantaneous velocity at t = 2, we find the derivative of [tex]\( y = 36t - 16t^2 \)[/tex] with respect to [tex]\( t \)[/tex], which is [tex]\( v(t) = 36 - 32t \)[/tex]. Plugging in [tex]\( t = 2 \)[/tex],

we get [tex]\( v(2) = 36 - 32(2) = -32 \)[/tex] ft/s.

Answer:

for sample = xbar

population = μ

Step-by-step explanation:

The arithmetic mean for sample can be represented by xbar and it can be calculated as

xbar=∑xi/n

Where xi represents data values and n represents number of data values in a sample.

The arithmetic mean for population can be represented by μ and it can be calculated as

μ=∑xi/N

Where xi represents data values and N represents number of data values in a population.

The symbol for the arithmetic mean when it's a sample statistic is 'x', and 'μ' when it's a population parameter. Sample mean (x) refers to the mean of a sample group, while population mean (μ) refers to the mean of an entire population.

The symbol that is used for the arithmetic mean when it is a sample statistic is 'x' (pronounced as x-bar). This is used to denote the mean of a sample. On the other side, the symbol that is used when the arithmetic mean is a population parameter is 'μ' (pronounced as mu). This symbol is representative of the mean of an entire population.

Take for example a group of 50 students in a class who took a test. If you wanted to find the sample mean (x) of their scores, you'd add all their scores and divide by the total number of students (which is 50 in this case). However, if you were to calculate the population mean (μ) of the scores of all students in all high school classes in the nation, you'd add their scores and divide by the total number.

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The median of a set of bids can be found by arranging them in numerical order and selecting the middle value.

The median is the middle value of a set of data arranged in numerical order. In this case, we have a set of six bids: 2.2, 1.3, 1.9, 1.2, 2.4, and x (blurred number). To find the median, we first need to arrange the bids in numerical order:

Since there are six bids, the middle value will be the fourth number in the ordered list. Therefore, the median is 2.2.

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Answer:

Explanation:

Question 1. Between what two values will approximately 68% of the numbers of days be?.

You can answer based on the 68-95-99.7% rule. As per this rule, about 68% of the data of a normal distribution (bell shaped) are within one standard deviation of the mean.

Here the mean is 32 day and the standard deviation is 7 day. Then 68% are in the interval 32 days ± 7 days.

That is:

Consequently, approximately 68% of the numbers of days will be between 25 and 39 days.

Question 2. Estimate the percentage of customer accounts for which the number of days is between 18 and 46.

First must determine the Z-scores both both values X = 18 and X = 46

The formula is:

         [tex]Z-score = (X-mean)/(standard\text{ }deviation)[/tex]

        [tex]Z-score=(18-32)/7=-2[/tex]

       [tex]Z-score=(46-32)/7=2[/tex]

Hence, you want to estimate the percentage of customers accounts for which the the number of days is within 2 standard deviations of the mean.

As per the 68-95-99.7 rule about 95% of the data are within 2 standard deviations of the mean. You can calculate it also from a standard normal distribution table, finding the area to the left of Z-score = - 2 and subtracting the area to the right of Z-score equal to 2: That is: 0.9772 - 0.0228 = 0.9484 = 95.44% ≈ 95%.

Question 3. Estimate the percentage of customer accounts for which the number of days is between 11 and 53.

Again, determine the Z-scores for the two values, X = 11 and X = 53.

         [tex]Z-score=(11-32)/7=-3[/tex]

        [tex]Z-score=(53-32)/7=3[/tex]

Hence, you want to estimate the probability of the number of days s between - 3 and 3 standard deviations.

Such probability is about 99.7%, according to the 68-95-99.7 rule.

If you use a standard distritution table you will find that the area to the right of the Z-score of -3 is 0.99865, thus the probability of the Z-score be to the right of 3 is 1 - 0.99865 = 0.00135.

And the probability in between -3 and 3 standard deviations is 0.99865 - 0.00135 = 0.9973 = 99.73% ≈ 99.7%.

Estimate the percentage of bills for number 39 when a bill was sent and when payment was made Answer:

Step-by-step explanation: yes

Answer:

There is a 5.5% probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

Step-by-step explanation:

To solve this problem, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].

In this problem, we have that:

[tex]\mu = 185, \sigma = 26.7, n = 48, s = \frac{26.7}{\sqrt{48}} = 3.85[/tex]

The weights of a random sample of 48 commercial boat passengers were recorded. The sample mean was determined to be 177.6 pounds. Find the probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

The probability of an extreme value below the mean.

This is the pvalue of Z when X = 177.6.

So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{177.6 - 185}{3.85}[/tex]

[tex]Z = -1.92[/tex]

[tex]Z = -1.92[/tex] has a pvalue of 0.0274.

So there is a 2.74% of having a sample mean as extreme than that and lower than the mean.

The probability of an extrema value above the mean.

Measures above the mean have a positive z score.

So this probability is 1 subtracted by the pvalue of [tex]Z = 1.92[/tex]

[tex]Z = 1.92[/tex] has a pvalue of 0.9726.

So there is a 1-0.9726 = 0.0274 = 2.74% of having a sample mean as extreme than that and above than the mean.

Total:

2*0.0274 = 0.0548 = 0.055

There is a 5.5% probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

Answer:

Answer: 40 sq. in.

Step-by-step explanation:

First we gotta find the area of triangle part of the shape, we can see the left side of the shape is 5 in. , so we subtract the 3 from 5 which gives 2 the height of the triangle.

Now, we find the length for the base of the triangle, the top part of the shape is 7 in. , so we subtract 7 from 12 which gives us 5 in. as the base

Now, we find the area of the triangle:

A = [tex]\frac{1}{2}[/tex]b × h

A = ([tex]\frac{1}{2}[/tex] × 5 in,)  × 2 in

A = 5 sq. in.

Now we find the area for the rectangle:

A = b × h

A = 5 x 7

A = 35 sq. in

Finally, we add the areas together

35 sq. in. + 5 sq. in. = 40 sq. in.

We get our answer 40 sq. in,

Answer:

Jenna will have to work for 210 hours

Step-by-step explanation:

i) vacation is of six days

ii) $279 for airfare

iii) $300 for food and entertainment

iv) $65 per day for her share of the hotel, for 6 days = $65 [tex]\times[/tex] 6 = $390

v) Total expenses = $279 + $300 + $390 = $969

vi) Let x be the number of hours she will have to work to save for the vacation, where she earns $25 per hours

vii) She has $550 saved for the vacation

viii) Total savings + total earnings  = Total expenses

     therefore 550 + 25x  = 969

     therefore x = (969 -550) / 25  = 209.5 hours

   therefore Jenna will have to work for 209.5 hours, or 210 hours (rounded up to the nearest whole number)

Answer:

She must work 15 more hours to pay for her vacation

Step-by-step explanation:

279+300+5(65)

579+325=904

She has already made 550 dollars so for the next part of the problem;

904-550=354

354/25=14.16

Normally we would round down but since she needs to make the minimum amount for her trip we round up to 15 hours.

Hope that helps, this is the correct answer.

Answer:

-5; 4

Step-by-step explanation:

The given linear system is:

[tex]x_1+4x_2=11\\2x_1+7x_2=18[/tex]

Multiplying the first equation by -2 and adding to the second gives:

[tex]x_1+4x_2=11\\2x_1-2x_1_+7x_2-8x_2=18 -22\\\\x_1+4x_2=11\\-x_2=-4[/tex]

Multiply the second equation by 4 and add to the first to find x1:

[tex]x_1+4x_2-4x_2=11-16\\-x_2=-4\\\\x_1=-5\\x_2=4[/tex]

The order pair for the solution of the system is -5; 4.

Answer:

a) the length of the wire for the circle = [tex](\frac{60\pi }{\pi+4}) in[/tex]

b)the length of the wire for the square = [tex](\frac{240}{\pi+4}) in[/tex]

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

         b=[tex]\frac{y}{4}[/tex]

Area of the square which is L² can now be said to be;

[tex]A_S=(\frac{y}{4})^2 = \frac{y^2}{16}[/tex]

On the otherhand; let the radius (r) of the  circle be;

2πr = 60-y

[tex]r = \frac{60-y}{2\pi }[/tex]

Area of the circle which is πr² can now be;

[tex]A_C= \pi (\frac{60-y}{2\pi } )^2[/tex]

     [tex]=( \frac{60-y}{4\pi } )^2[/tex]

Total Area (A);

A = [tex]A_S+A_C[/tex]

   = [tex]\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2[/tex]

For the smallest possible area; [tex]\frac{dA}{dy}=0[/tex]

∴ [tex]\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0[/tex]

If we divide through with (2) and each entity move to the opposite side; we have:

[tex]\frac{y}{18}=\frac{(60-y)}{2\pi}[/tex]

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

[tex]y= \frac{240}{\pi+4}[/tex]

∴ since [tex]y= \frac{240}{\pi+4}[/tex], we can determine for the length of the circle ;

60-y can now be;

= [tex]60-\frac{240}{\pi+4}[/tex]

= [tex]\frac{(\pi+4)*60-240}{\pi+40}[/tex]

= [tex]\frac{60\pi+240-240}{\pi+4}[/tex]

= [tex](\frac{60\pi}{\pi+4})in[/tex]

also, the length of wire for the square  (y) ; [tex]y= (\frac{240}{\pi+4})in[/tex]

The smallest possible area (A) = [tex]\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})[/tex]

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)