For the month of May in a certain​ city, the probability that the weather on a given day is cloudy is 0.77. Also in the month of May in the same​ city, the probability that the weather on a given day is cloudy and snowy is 0.27. What is the probability that a randomly selected day in May will be snowy if it is cloudy​?The probability is approximately nothing. ​(Round to three decimal places as​ needed.)

Answer: 2.4

Step-by-step explanation:

Given : A large disaster cleaning company estimates that 30 percent of the jobs it bids on are finished within the bid time.

i..e The proportion of  the jobs it bids on are finished within the bid time: p = 30%= 0.30

Sample size of jobs that it has contracted : n= 8

Then ,  the mean number of jobs completed within the bid time : [tex]\mu=np[/tex]

[tex]= 8\times0.30=2.4[/tex]

Hence, the mean number of jobs completed within the bid time is 2.4 .

Answer:

The only solution is r=2 or r=1

Step-by-step explanation:

Check the attachment

To determine the values of r for which the given differential equation has solutions of the form y = tr for t > 0, substitute y = tr into the given differential equation and solve for r. The values of r that satisfy the equation are r = 1.

To determine the values of r for which the given differential equation has solutions of the form y = tr for t > 0, we need to substitute y = tr into the given differential equation and solve for r.

First, we find the first and second derivatives of y = tr:

y' = r

y'' = 0

Substituting these derivatives into the differential equation, we get:

t2(0) - 2t(r) + 2(tr) = 0

Simplifying the equation, we have:

(2 - 2r)t = 0

This equation is satisfied when t = 0 or r = 1. Therefore, the values of r for which the given differential equation has solutions of the form y = tr for t > 0 are r = 1.

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Final answer:

The z-score for a jug with 128 fl. Oz. of milk and the probability of a jug containing less than 128 fl. Oz. are calculated using the mean and standard deviation provided.

Explanation:

1. Find the z-score corresponding to a jug containing 128 fl. Oz. of milk:

2. Probability that a jug contains less than 128 fl. Oz. of milk:

Answer:

A. X-intercepts

Step-by-step explanation:

The zeros of a function are the values of x for which y is 0.

For example:

y = 2x - 4

Has zeros

2x - 4 = 0

2x = 4

x = 2

Which means that when x = 2, y = 0. Looking at the graphic of this function, it crosses the x line when x = 2. So the zeros are also called x intercepts.

So the correct answer is

A. X-intercepts

Answer:

X-intercepts

Step-by-step explanation:

TRUST ME

Answer:

z(max)  =  5000

x   =  100

y   =  0

Step-by-step explanation:

Let call

x  floral arrangements  and

y  fruit baskets

Then Objective function is according to profits in each gift

z   =   50*x    +    35*y

Constraints:

1.- Hours available in Assembly department  40  in minutes   is  2400 minutes

20*x  +  15*y  ≤ 2400

2.- Hours available in packaging department 10 in minutes  is  600

6*x  + 2*y  ≤  600

3.- x    and    y must be    x ≥ 0    y  ≥ 0

Then the system is:

z      - 50*x       -   50*y                       =  0        To maximize subject to:

         20*x       +  15*y                        ≤  2400

           6*x        +   2*y                        ≤  600

x ≥ 0    y ≥ 0

Simplex Method:

z          x         y       s₁       s₂         Cte

1         -50     -35     0        0    =     0

0         20       15      1        0    =  2400

0           6         2      0        1    =  600

First iteration: 6 is a pivot   we dvede R3  by 6

z          x         y       s₁       s₂         Cte

1          0        15      0        50/6 =  5000

0         0      -50/6  -1        20/6 = -400                                                                

0         20       15      1        0    =  2400

0          1          2/6     0       1/6   =  100

We have done no negative number in the objective function we stop iteration and

z(max)  =  5000

x   =  100

y   =  0

                                         to add to R1  50*R3 [ 0  50  100/6  0  50/6  5000

                                         to add to R2 20*R3 [ 0   20  40/6   0  20/6  2000

Answer:

The controllable canonical form of the system =

Y = (5 0 0 0) (x1)

                     (x2)

                     (x3)

                     (x4)

Step-by-step explanation:

The detailed explanation is as shown in the attached file.

Answer:

(a) [tex]p(x) = 17-\frac{x}{3,500}[/tex]

(b) $8.50

Step-by-step explanation:

(a) The slope of the demand function, p(x), is determined by:

[tex]m=\frac{11-9}{21,000-28,000}=-\frac{1}{3,500 }[/tex]

Applying the point (21,000; 11) to the general linear equation formula gives us the demand function:

[tex]p(x) - 11 = -\frac{1}{3,500}*(x-21,000)\\p(x) = 17-\frac{x}{3,500}[/tex]

(b) The revenue function, r(x), is given by:

[tex]r(x) =x*p(x) = 17x-\frac{x^2}{3,500}[/tex]

The value of x for which the derivate of the revenue function is zero gives us the attendance for which revenue is maximized:

[tex]\frac{dr(x)}{dx} =0= 17-\frac{2x}{3,500}\\x=29,750[/tex]

At an attendance of 29,750, the price is:

[tex]p = 17-\frac{29,750}{3,500}\\p=\$8.50[/tex]

Tickets should be set at a price of $8.50.

The demand function for the baseball team can be found using the given data points. The revenue function can be derived using the demand function, and its maximum can be determined by finding the critical points. The ticket price should be set at $8.50 to maximize revenue for the baseball team.

(a) To find the demand function, we can use the point-slope form of a linear equation. Let's denote the ticket price as p and the attendance as a. We have two data points: (11, 21000) and (9, 28000). We can use the formula: a - a1 = m(p - p1) where (p1, a1) represents one of the given data points and m represents the slope. Let's use the first data point:

21000 - 28000 = m(11 - 9)

-7000 = 2m

m = -3500

So, the demand function is: a(p) = -3500p + b. To find the constant term b, we can substitute one of the data points into the equation. Let's use the first data point (11, 21000):

21000 = -3500(11) + b

21000 = -38500 + b

b = 59500

Therefore, the demand function is: a(p) = -3500p + 59500.

(b) Revenue is the product of the ticket price and the attendance. Let's denote the revenue as R and the ticket price as p. The revenue function can be written as: R(p) = p * a(p). Substituting the demand function into the revenue function, we have: R(p) = p * (-3500p + 59500). To maximize revenue, we can find the critical points of the function by taking the derivative and setting it equal to zero.

R'(p) = -7000p + 59500 = 0

-7000p = -59500

p = 8.5

The critical point is p = 8.5. To confirm that this gives a maximum, we can take the second derivative and check its sign.

R''(p) = -7000

Since the second derivative is negative, the critical point p = 8.5 corresponds to the maximum revenue. Therefore, the ticket price should be set at $8.50 to maximize revenue.

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Answer:

x  =  7

y  =  3

z (max)  =  4950/3  =  1650

Step-by-step explanation:

Let call  

x numbers of church goup and

y numbers of Union Local

Then  

First contraint

2*x + 2*y ≤ 20

Second one

1*x + 3*y ≤ 16

Objective Function

z = 150*x + 200*y

Then the system is

z = 150*x + 200*y To maximize

Subject to:

2*x + 2*y ≤ 20

1*x + 3*y ≤ 16

x ≥ 0 y ≥ 0

We will solve by using the Simplex method

z - 150 *x - 200*y = 0

2*x + 2*y + s₁ = 20  

1*x + 3*y + 0s₁ + s₂ = 16

First Table

z           x          y          s₁           s₂          Cte

1        -150    -200       0            0     =      0

0          2         2           1            0     =    20

0          1          3          0            1      =     16

First iteration:

Column pivot   ( y column ) row pivot (third row) pivot 3

Second table

z           x                y          s₁           s₂                Cte

1       -250/3           0          0         200/3    =    3200/3

0      - 4/3               0          -1           2/3       =    -20/3

0         1/3               1            0           1/3       =    -20/3

Second iteration:

Column pivot  ( x column ) row pivot  (second row)  pivot  -4/3

Third table

z           x                y          s₁           s₂                Cte

1            0               0      750/12    700/6    =   4950/3

0            1                0        3/4          -1/2     =    7

0            0                1         -1/4          1/2     =  9/3

Donna should contact four church groups and four labor unions to maximize her fundraising, which would yield $1400. The maximum money she could 'lose' (or not earn) is $0 if she does not contact any group.

We need to set up linear inequalities to represent the constraints described by Donna's situation. We will let x represent the number of church groups and y represent the number of labor unions she contacts.

The time required for letter writing can be described by the inequality: 2x + 2y <= 20, and the time for follow-ups is x + 3y <= 16.

We want to maximize the sum z = 150x + 200y which represents the total money she can raise.

To find the maximum, one method is to graph the feasible region defined by the constraints and find the vertices. Then evaluate the objective function at the vertices. In this case, the vertices are (0,0), (0,5), (8,0) and (4,4).

Upon calculation, contacting four church groups and four labor unions yields the highest amount of $1400.

As for the loss, given the nature of the task, the most money she could 'lose' in a month is just not raising any money at all, which would be $0 if she doesn't contact any group.

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Answer:

0.25 or 25%

Step-by-step explanation:

3, 5 and 7 are prime numbers.

There are two possible outcomes for which the experimenter ends up with exactly twenty-five dollars:

A) Choosing urn 5 (5 x 5 = 25).

[tex]P(A) = \frac{1}{5}[/tex]

B) Choosing urn 4 and then urn 3 ([4 x 4] + [3 x 3] = 25).

[tex]P(B)= \frac{1}{5} *\frac{1}{4}=\frac{1}{20}[/tex]

The probability that the experimenter ends up with exactly $25 is:

[tex]P(x=\$25)=P(A)+P(B)= \frac{1}{5}+\frac{1}{20}\\P(x=\$25)=0.25=25\%[/tex]

Answer:

720

Step-by-step explanation:

If every person has to choose a different character, the first person to choose a character has 6 options, the second has 5, the third has 4, the fourth has 3, and the last person has only two options. Therefore, the total number of ways you can do this if all 5 people dress up as a different character is:

[tex]n=6*5*4*3*2\\n=720[/tex]

There are 720 ways.

Final answer:

There are 720 different ways for 5 people to dress up as the 6 main Avengers characters; this combinatorics problem uses permutations to find the answer.

Explanation:

Combinations for Avengers Characters

The question is about calculations of combinations, which falls under the subject of Mathematics. More specifically, this is a combinatorics problem that can typically be found at a High School level. We want to find out how many different ways 5 people can dress up as any of the 6 main Avengers characters, assuming each person dresses up as a different character. To solve this, we can use the concept of permutations because the order in which we assign the characters to the 5 friends does matter.

In this case, we have 6 characters to choose from, and we want to assign these characters to 5 friends. We are therefore looking for the number of permutations of 6 characters taken 5 at a time, which is calculated using the formula:

P(n, k) = n! / (n - k)!

Here, 'n' is the total number of characters, and 'k' is the number of people to dress up. Therefore, we have:

P(6, 5) = 6! / (6 - 5)! = 6! / 1! = 720 / 1 = 720

There are 720 different ways for the 5 friends to dress up as the Avengers characters.

Answer:

[tex]n \geq 36[/tex]

Step-by-step explanation:

For this case we know the mean and the deviation:

[tex] \mu = 1.12 kg , \sigma= 0.0009[/tex]

The mean is given by this:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

If we find the expected value for the sample mean we have:

[tex] E(\bar X) = E(\frac{\sum_{i=1}^n X_i}{n}) =\frac{1}{n} \sum_{i=1}^n E(X_i)[/tex]

Since each observation in the sample [tex] X_1, X_2,...., X_{25}[/tex] have an expectation [tex] E(X_i) = \mu , i =1,2,...,25[/tex] so then we have that:

[tex] E(\bar X) = \frac{1}{n} n\mu = \mu[/tex]

Now for the variance of the sample mean we have this:

[tex]Var (\bar X) = Var(\frac{\sum_{i=1}^n X_i}{n})= \frac{1}{n^2} \sum_{i=1}^n Var(X_i)[/tex]

And again each observation have a variance [tex] \sigma^2_i = 0.0009 , i =1,2,...,25[/tex] then we have:

[tex]Var (\bar X) =\frac{1}{n^2} n(\sigma^2) =\frac{\sigma^2}{n}[/tex]

And then the standard deviation would be:

[tex] Sd(\bar X) =\sqrt{\frac{\sigma^2}{n}}= \frac{\sigma}{\sqrt{n}}[/tex]

And we want that the standard deviation for the sample no more than 0.005 so we have this condition:

[tex]\frac{\sigma}{\sqrt{n}} \leq 0.005[/tex]

And since we know the value of [tex] \sigma= \sqrt{0.0009}=0.03[/tex] we can solve for the value of n like this:

[tex] \frac{0.03}{0.005} \leq \sqrt{n}[/tex]

[tex] 6 \leq \sqrt{n}[/tex]

And if we square both sides we got:

[tex] n\geq 36[/tex]

Final answer:

To achieve an average weight standard deviation of no more than 0.005 kilograms for a certain type of brick with a variance of 0.0009 kilograms², 36 bricks need to be selected.

Explanation:

The question concerns the calculation of the number of bricks needed so that the average weight of these bricks has a standard deviation of no more than 0.005 kilograms. Given that the weight of a certain type of brick has an expectation (mean) of 1.12 kilograms and a variance of 0.0009 kilograms2, the first step is to understand that the variance of the average weight of n bricks is equal to the variance of a single brick divided by n.

To achieve a standard deviation of the average weight of no more than 0.005 kilograms, we use the formula for the standard deviation of the mean, which is √(variance/n). The variance given is 0.0009 kilograms2. Setting the equation √(0.0009/n) <= 0.005 and solving for n gives:

0.0009/n = 0.0052

0.0009/n = 0.000025

n = 0.0009 / 0.000025 = 36

Therefore, to ensure the standard deviation of the average weight is no more than 0.005 kilograms, 36 bricks would need to be selected.

Answer:

The probability that Umbrella returns any shipment is 0.0062.

Step-by-step explanation:

Let X = the impurity level in the material per drum

Then it is provided that,

[tex]X\sim N(\mu = 0.03,\ \sigma=0.004)[/tex]

Also if the impurity level is more than or equal to 4% or 0.04 in any shipment, then Umbrella returns the entire drum to the supplier.

Compute the probability that Umbrella returns any shipment as:

[tex]P(X\geq 0.04)=P(\frac{X-\mu}{\sigma}\geq \frac{0.04-0.03}{0.004}) \\=P(Z\geq 2.5)\\=1-P(Z<2.5)\\=1-0.99379\\=0.00621\\\approx0.0062[/tex]

Use the z-table fro left z-scores to determine the probability.

Thus, the probability that Umbrella returns any shipment is 0.0062.

Answer:

For the company A : [tex] P(X>33000)= P(Z> \frac{33000-27000}{3000}) =P(Z>2) = 1-P(Z<2)= 0.0228[/tex]

For the company B: [tex] P(X>33000)= P(Z> \frac{33000-27000}{7000}) =P(Z>0.857) = 1-P(Z<0.857)= 0.196[/tex]

So as we can see we have a higher probability for the company B.

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".

Company A

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

[tex]X \sim N(27000,3000)[/tex]  

Where [tex]\mu=27000[/tex] and [tex]\sigma=3000[/tex]

For this case if we find the probability for [tex] P(X>33000)[/tex] using the z score given by:

[tex] z= \frac{x -\mu}{\sigma}[/tex]

And if we use this formula we got:

[tex] P(X>33000)= P(Z> \frac{33000-27000}{3000}) =P(Z>2) = 1-P(Z<2)= 0.0228[/tex]

Company B

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

[tex]X \sim N(27000,7000)[/tex]  

Where [tex]\mu=27000[/tex] and [tex]\sigma=7000[/tex]

For this case if we find the probability for [tex] P(X>33000)[/tex] using the z score given by:

[tex] z= \frac{x -\mu}{\sigma}[/tex]

And if we use this formula we got:

[tex] P(X>33000)= P(Z> \frac{33000-27000}{7000}) =P(Z>0.857) = 1-P(Z<0.857)= 0.196[/tex]

The company from which we are more likely to get an offer of $33,000 or more is determined by comparing the probabilities of receiving such an offer from each of the two companies. This probability is determined using a concept called the z-score, with which we can convert any data point into a common scale, which makes comparison easier.

1. Compute the z-scores for obtaining this salary for both companies. The z-score formula is (X - μ) / σ where X is the data point we are interested in, μ stands for the mean amount of the starting salary, and σ is the standard deviation.

   For Company A:
   z = (33000 - 27000) / 3000
   We deduct the average salary for Company A (μ = 27000) from the target salary (X = 33000) and divide it by the standard deviation (σ = 3000).

   For Company B:
   Similarly, the z score is calculated by deducting the average salary for Company B (μ = 27000) the from target salary (X = 33000) and dividing by the standard deviation (σ = 7000).

2. Now we have the z-scores for both companies. The next step is to use the z-score to calculate the probability of getting an offer of $33,000 or more from both companies. We can do this by using the cumulative distribution function (CDF) which gives us the probability that a random variable is less than or equal to a certain value. But since we want the probability that the salary is more than $33,000, we subtract the result from one (1 - CDF(z)).

3. We compare these probabilities. The company with the higher probability value will be the one from which we are more likely to get an offer of $33,000 or more.

4. Based on these calculations, you are more likely to get an offer of $33,000 or more from Company B.

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Answer:

[tex]V = \frac{5\pi}{6}[/tex] or 2.62

Step-by-step explanation:

Since our region (on the left) is bounded by x = 1 and x = 0 (where [tex]y = x^2 = 0[/tex], if we take center at x = 2 then our radius will range from 1 to 2 (x=1 to x = 0). We can use the following integration to calculate the volume using shell method

[tex]V = \int\limits^2_1 {2\pi r h} \, dr[/tex]

where r = 2 - x so x = 2 - r and [tex]h = y = x^2 = (2-r)^2[/tex] for [tex]1 \leq r \leq 2[/tex]

[tex]V = \int\limits^2_1 {2\pi r(2-r)^2} \, dr[/tex]

[tex]V = \int\limits^2_1 {2\pi r(4 - 4r + r^2)} \, dr[/tex]

[tex]V = 2\pi \int\limits^2_1 {r^3 - 4r^2 +4r} \, dr[/tex]

[tex]V = 2\pi\left[\frac{r^4}{4} - \frac{4r^3}{3} + 2r^2\right]^2_1[/tex]

[tex]V = 2\pi\left[\left(\frac{2^4}{4} - \frac{4*2^3}{3} + 2*2^2\right) - \left(\frac{1^4}{4} - \frac{4*1^3}{3} + 2*1^2\right)\right][/tex]

[tex]V = 2\pi(4 - 32/3 +8 - 1/4 + 4/3 - 2)[/tex]

[tex]V = \pi(20 - 56/3 - 1/2)[/tex]

[tex]V = \pi\frac{120 - 112 - 3}{6}[/tex]

[tex]V = \frac{5\pi}{6}[/tex] or 2.62

The ball travels a distance of 52 ft during the time interval [3,3.5]. The average velocity over this interval is 104 ft/sec. When we calculate over increasingly smaller time intervals, it appears the instantaneous velocity at t=3 is 96 ft/sec.

(a) We need to find the displacement Δs over the time interval [3,3.5]. For that, we subtract the position at time 3 from the position at time 3.5:

Δs = s(3.5) - s(3) = 16(3.5^2) - 16(3^2) = 196 - 144 = 52 ft.

(b) We calculate the average velocity by dividing Δs by Δt. That is Δs/Δt = 52/0.5 = 104 ft/sec.

(c) To estimate the instantaneous velocity at t=3, we have to compute the average velocities over smaller and smaller intervals centered at t=3. Let's calculate the average velocities for [3,3.01],[2.999,3]:

For [3,3.01]: Δs/Δt = [16*(3.01)^2 - 16*3^2]/0.01 = 96.12 ft/sec. For [2.999,3]: Δs/Δt = [16*3^2 - 16*(2.999)^2]/0.001 = 96 ft/sec.

Instantaneous velocity at t=3, v(3) can be estimated as the limit of these average velocities, which seems to be approaching about 96 ft/sec.

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(a) The ball travel Δs= 52 ft during the time interval [3,3.5].

(b) The average velocity over [3,3.5] is Δs/Δt= 104 ft/sec

(c) The object's instantaneous velocity at t=3 .V(3)= 96 ft/sec

Let's solve the problem step-by-step.

Given:
The distance traveled by the ball after t seconds is given by the function: [tex]s(t) = 16t^2[/tex]

(a) To find the distance traveled, we need to calculate the change in the distance function, which is
[tex]\Delta s = s(3.5) - s(3)[/tex]

First, compute [tex]s(3)[/tex]:
[tex]s(3) = 16(3)^2 = 16 \times 9 = 144[/tex] ft

Next, compute [tex]s(3.5)[/tex]:
[tex]s(3.5) = 16(3.5)^2 = 16 \times 12.25 = 196[/tex] ft

So,
[tex]\Delta s = 196 - 144 = 52[/tex] ft

(b) Average velocity is calculated by dividing the change in distance by the change in time, which is
[tex]\frac{\Delta s}{\Delta t} = \frac{52}{0.5} = 104[/tex] ft/sec

(c) For the time intervals, we compute each average velocity:

[3, 3.01]:
[tex]\Delta t = 0.01[/tex]
[tex]s(3.01) = 16(3.01)^2 = 16 \times 9.0601 = 144.9616[/tex] ft
[tex]\Delta s = 144.9616 - 144 = 0.9616[/tex] ft
[tex]\text{Average velocity} = \frac{0.9616}{0.01} = 96.16[/tex] ft/sec

[3, 3.001]:
[tex]\Delta t = 0.001[/tex]
[tex]s(3.001) = 16(3.001)^2 = 16 \times 9.006001 = 144.096016[/tex] ft
[tex]\Delta s = 144.096016 - 144 = 0.096016[/tex] ft
[tex]\text{Average velocity} = \frac{0.096016}{0.001} = 96.016[/tex] ft/sec

[3, 3.0001]:
[tex]\Delta t = 0.0001[/tex]
[tex]s(3.0001) = 16(3.0001)^2 = 16 \times 9.00060001 = 144.00960016[/tex] ft
[tex]\Delta s = 144.00960016 - 144 = 0.00960016[/tex] ft
[tex]\text{Average velocity} = \frac{0.00960016}{0.0001} = 96.0016[/tex] ft/sec

[2.9999, 3]:
[tex]\Delta t = 0.0001[/tex]
[tex]s(2.9999) = 16(2.9999)^2 = 16 \times 8.99940001 = 143.99040016[/tex] ft
[tex]\Delta s = 144 - 143.99040016 = 0.00959984[/tex] ft
[tex]\text{Average velocity} = \frac{0.00959984}{0.0001} = 95.9984[/tex] ft/sec

[2.999, 3]:
[tex]\Delta t = 0.001[/tex]
[tex]s(2.999) = 16(2.999)^2 = 16 \times 8.994001 = 143.904016[/tex] ft
[tex]\Delta s = 144 - 143.904016 = 0.095984[/tex] ft
[tex]\text{Average velocity} = \frac{0.095984}{0.001} = 95.984[/tex] ft/sec

[2.99, 3]:
[tex]\Delta t = 0.01[/tex]
[tex]s(2.99) = 16(2.99)^2 = 16 \times 8.9401 = 143.0416[/tex] ft
[tex]\Delta s = 144 - 143.0416 = 0.9584[/tex] ft
[tex]\text{Average velocity} = \frac{0.9584}{0.01} = 95.84[/tex] ft/sec

From these values, we can see that as the interval gets smaller, the average velocities are approaching a specific value. This helps us estimate the instantaneous velocity at [tex]t = 3[/tex].

Answer:

The information above is just a part of a long question.The main question is  found in the attached as well as the answer following a step by step methodical approach.

Step-by-step explanation:

Take a good at the full question before venturing into the statements prepared so as to have a thorough understanding of the requirements and how the answers provided have touched upon the requirements.

Answer:

150

Step-by-step explanation:

You have to divide the correct answers (111) by the total amount of questions (x) in order to find the 74% (0.74)

As per the unitary method, there are 150 questions on the test.

Let's start by breaking down the information we have been given. We know that 74% of the questions on the test were answered correctly, and the number of questions answered correctly is 111.

Step 1: Convert Percentage to Decimal

To make calculations easier, we need to convert the percentage into a decimal. To do this, we divide 74 by 100, which gives us 0.74.

Step 2: Set Up the Equation

Now we can set up an equation to solve for the total number of questions on the test. Let's denote the total number of questions as "x". Since 74% of the questions were answered correctly, we can say that 0.74x questions were answered correctly.

Step 3: Solve for Total Questions

We are given that 111 questions were answered correctly. So, we can set up the equation:

0.74x = 111

Step 4: Solve for "x"

To solve for "x", we need to isolate it on one side of the equation. We can do this by dividing both sides of the equation by 0.74:

x = 111 / 0.74

x = 150

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Answer = 321.7

Explanation:

radius = y

height = [tex]4y^{2} - y^{3}[/tex]

area of cylinder = 2π*r*h

Integrate the area to calculate the volume:

[tex]= \int\limits^0_4 {2\pi(4y^{3} -y^{4}) } \, dy[/tex]

[tex]= 2\pi (\int\limits^0_4 {4y^{3}dy -\int\limits^0_4 y^{4} dy )} \,[/tex]

[tex]= 2\pi (256-\frac{1024}{5} )[/tex]

[tex]=\frac{512\pi }{5}[/tex]

[tex]= 321.7[/tex]

Final answer:

To compute the volume of the solid formed by rotating the region bounded by the curves x = 4y² − y³ and x = 0 around the x-axis, utilize the cylindrical shells method with the relevant volume formula for cylindrical shells and integrate over the intersecting interval of y-values.

Explanation:

Finding the Volume of a Solid of Revolution

To find the volume of the solid obtained by rotating the region bounded by the curves x = 4y2 − y3 and x = 0 about the x-axis using the method of cylindrical shells, we follow these steps:

The relevant formulas for the volume and surface area of a sphere are (4/3)πr3 and 4πr2, respectively. It's important to note that the volume depends on the cube of the radius R3, while the surface area is a function of the square of the radius R2.

Answer:

[tex]P(k) = \frac{(n C k) [(N-n) C (m-k)]}{(NCm)}[/tex]

Step-by-step explanation:

Step 1: Number of possible combination of selecting ‘m’ deer in second sample

Total number of deer are N and therefore the combinations can be calculated as (N С m).  

Step 2: Number of possible combination of marked deer ‘k’ in second sample

Total number of marked deer in total population is ‘n’. Therefore, the possible number of selecting marked deer is (n C k).

Step 3: Number of possible combination unmarked deer in second sample

Since we have already calculated the total combinations of selecting marked deer in the second sample. Hence, we have to calculate the total unmarked deer in total population which is N-n and number of unmarked deer in the second sample which is m-k.

Therefore, total possible combination of unmarked deer in second sample is [(N-n) C (m-k)].

Step 4: Probability of selecting unmarked deer in the second sample is

Let the probability of selecting unmarked deer in the second sample be P(k)

Therefore,

                                 [tex]P(k) = \frac{(n C k) [(N-n) C (m-k)]}{(NCm)}[/tex]

Answer:  

120 ways

Step-by-step explanation: If there are three different categories of components, that is, component A, comprising four types; component B comprising, five types and component C comprising, six types. Three types are to be selected, that is, one type from each category. The number of ways one component of each type be selected is:

Component A * Component B * Component C = 4 X 5 X 6 = 120

The volume of a box is the amount of space in it.

The expression that represents volume is: [tex]\mathbf{V = (12 -2x) (9 - 2x)x}[/tex]

The dimension of the cardboard is given as:

[tex]\mathbf{Length = 12}[/tex]

[tex]\mathbf{Width = 9}[/tex]

Assume the cut-out is x.

So, the dimension of the box is:

[tex]\mathbf{Length =12-2x}[/tex]

[tex]\mathbf{Width =9 - 2x}[/tex]

[tex]\mathbf{Height = x}[/tex]

The volume of the box is:

[tex]\mathbf{V = (12 -2x) (9 - 2x)x}[/tex]

Hence, the expression that represents volume is:

[tex]\mathbf{V = (12 -2x) (9 - 2x)x}[/tex]

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The volume of a box formed by cutting a square of side x from a 9" by 12" sheet and folding the sides is given by the formula V = x(12 - 2x)(9 - 2x).

The volume of a box is formed by multiplying its length, width, and height. In this case, if you cut a square of side x from each corner of the sheet, the new dimensions of your box would be:

Therefore, the formula to express the volume V in terms of x is:

V = x(12 - 2x)(9 - 2x)

.

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Answer:

60%

Step-by-step explanation:

The percentage does not change.

Probability at least one of next 3 vehicles fail inspection: approximately 0.784 (to 3 decimal places).

To find the probability that at least one of the next three vehicles fail, we can calculate the probability of the complementary event, i.e., the probability that all three vehicles pass, and then subtract it from 1.

Given that 60% of vehicles pass the inspection, the probability that one vehicle fails the inspection is 1 - 0.60 = 0.40.

Since the vehicles pass or fail independently, the probability that all three vehicles pass is:

[tex]\[0.60 \times 0.60 \times 0.60 = 0.216\][/tex]

Now, the probability that at least one vehicle fails is:

[tex]\[1 - 0.216 = 0.784\][/tex]

So, the probability that at least one of the next three vehicles fail is approximately [tex]\(0.784\)[/tex] (to 3 decimal places).

Answer:

66.98% probability that the mean gain for the sample was between 250 and 500.

Step-by-step explanation:

To solve this problem, it is important to know the Normal probability distribution and the Central limit theorem.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

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 = 432, \sigma = 722, n = 40, s = \frac{722}{\sqrt{40}} = 114.16[/tex]

What is the probability that the mean gain for the sample was between 250 and 500?

This is the pvalue of Z when X = 500 subtracted by the pvalue of Z when X = 250.

So

X = 500

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

By the Central Limit Theorem

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

[tex]Z = \frac{500 - 432}{114.16}[/tex]

[tex]Z = 0.6[/tex]

[tex]Z = 0.6[/tex] has a pvalue of 0.7257.

X = 250

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

By the Central Limit Theorem

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

[tex]Z = \frac{250 - 432}{114.16}[/tex]

[tex]Z = -1.59[/tex]

[tex]Z = -1.59[/tex] has a pvalue of 0.0559.

So there is a 0.7257 - 0.0559 = 0.6698 = 66.98% probability that the mean gain for the sample was between 250 and 500.

The probability that the mean gain for a sample of 40 years is between 250 and 500 is approximately 0.6698, or 66.98%.

To determine the probability that the mean gain for a random sample of 40 years of the Dow Jones Industrial Average is between 250 and 500, we will use the concept of the sampling distribution of the sample mean. The steps involved are:

State the given information:

  - Population mean [tex](\(\mu\))[/tex] = 432

  - Population standard deviation [tex](\(\sigma\))[/tex] = 722

  - Sample size n = 40

Find the standard error of the mean (SEM):

  The standard error of the mean is calculated as:

  [tex]\[ \text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{722}{\sqrt{40}} \approx 114.2 \][/tex]

Convert the sample means to z-scores:

  To find the probability that the sample mean [tex](\(\bar{x}\))[/tex] is between 250 and 500, we convert these values to z-scores using the formula:

  [tex]\[ z = \frac{\bar{x} - \mu}{\text{SEM}} \] For \(\bar{x} = 250\): \[ z = \frac{250 - 432}{114.2} \approx \frac{-182}{114.2} \approx -1.59 \] For \(\bar{x} = 500\):[/tex]

  [tex]\[ z = \frac{500 - 432}{114.2} \approx \frac{68}{114.2} \approx 0.60 \][/tex]

Find the probability corresponding to the z-scores:

  Using the standard normal distribution table or a calculator, we find the probabilities corresponding to these z-scores.

   [tex]\(z = -1.59\), \(P(Z \leq -1.59) \approx 0.0559\) \\\(z = 0.60\), \(P(Z \leq 0.60) \approx 0.7257\)[/tex]

Calculate the probability that the mean gain is between 250 and 500:

  \[tex][ P(250 \leq \bar{x} \leq 500) = P(Z \leq 0.60) - P(Z \leq -1.59) \] \[[/tex]

 = 0.7257 - 0.0559 = 0.6698

Conclusion:

The probability that the mean gain for a sample of 40 years is between 250 and 500 is approximately 0.6698, or 66.98%.

Answer:

e. 0.08

Step-by-step explanation:

In the question above, a certain quantity of goods was supplied while a specific quantity of goods was sold per week. In a given week, if the number of proportion sold is X, therefore:

f(x) = {Γ(4+2)/Γ(4)Γ(2) x^3 (1-x), 0≤x≤1 ; 0, elsewhere

and

P(X greater than 0.9) =  [tex]\int\limits^1_ {0.9} \, 20(x^{3} - x^{4}) dx[/tex] = 20*{(y^4/4)[1,0.9] - (y^5/5)[1,0.9]} = 20*{(0.25 - 0.164) - (0.20 - 0.118)}  = 20*{0.086 - 0.0819} = 20*0.0041 = 0.082

Therefore the probability of the proportion sold is approximately 0.082

The probability that at least 90% of the stock will be sold in a given week is approximately 0.05. The correct answer option is b. 0.05

Here's how to calculate it:

1. Given the beta distribution with parameters alpha = 4 and beta = 2, we want to find [tex]\( P(X > 0.9) \)[/tex], where X represents the proportion of stock sold in a week.

2. Since [tex]\( P(X > 0.9) = 1 - P(X \leq 0.9) \)[/tex], we need to find the cumulative distribution function (CDF) of the beta distribution and then subtract it from 1.

3. Using the incomplete beta function formula [tex]\( I_{x}(\alpha, \beta) \)[/tex], we have:

[tex]\[ P(X > 0.9) = 1 - I_{0.9}(4, 2) \][/tex]

4. Calculating [tex]\( I_{0.9}(4, 2) \)[/tex] using software or a calculator gives approximately 0.95.

5. Subtracting this from 1:

[tex]\[ P(X > 0.9) = 1 - 0.95 = 0.05 \][/tex]

So, the correct answer is b. 0.05.

The mean (average) of the data set is approximately 110.40, the median (the middle value when the data set is arranged in order from least to greatest) is 65, and there is no mode (the most frequently occurring number) in the data set.

The subject of your question is Mathematics, specifically in the field of Statistics. To compute the mean, median, and mode, you would do the following:

For the given data set your mean is approximately 110.40 , the median is 65 and there is no mode as no numbers in the data set are repeated.

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Answer:

∞

Step-by-step explanation:

Answer:

13/4260 tons

Step-by-step explanation:

We have the rate at which they remove tons of dirt per hour. We also know that total that needs to be removed. We can determine the time by dividing  the amount of tons that need to be removed by the rate:

[tex]=(13/6)/\cdot{710}=13/4260[/tex]

will take 13/4260 hours to remove the dirt

Answer:

[tex] P(X\leq 10) = 1- 0.961525= 0.0385[/tex]  

The nearest answer for this case would be 3.1%

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=20, p=0.71)[/tex]

The probability mass function for the Binomial distribution is given as:

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

Where (nCx) means combinatory and it's given by this formula:

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

For this case we can begin finding the probability P(X>10). If we find the individual probabilities we got:

[tex]P(X=11)=(20C11)(0.71)^{11} (1-0.71)^{20-11}=0.0563[/tex]

[tex]P(X=12)=(20C12)(0.71)^{12} (1-0.71)^{20-12}=0.1034[/tex]

[tex]P(X=13)=(20C13)(0.71)^{13} (1-0.71)^{20-13}=0.1558[/tex]

[tex]P(X=14)=(20C14)(0.71)^{14} (1-0.71)^{20-14}=0.1907[/tex]

[tex]P(X=15)=(20C15)(0.71)^{15} (1-0.71)^{20-15}=0.1867[/tex]

[tex]P(X=16)=(20C16)(0.71)^{16} (1-0.71)^{20-16}=0.1429[/tex]

[tex]P(X=17)=(20C17)(0.71)^{17} (1-0.71)^{20-17}=0.082[/tex]

[tex]P(X=18)=(20C18)(0.71)^{18} (1-0.71)^{20-18}=0.036[/tex]

[tex]P(X=19)=(20C19)(0.71)^{19} (1-0.71)^{20-19}=0.0087[/tex]

[tex]P(X=20)=(20C20)(0.71)^{20} (1-0.71)^{20-20}=0.00106[/tex]

And if we add the values we got:

[tex] P(X>10)= P(X=11) +.... +P(X=20) = 0.961525[/tex]

And if we use the complement rule the probability that "no more than 10 wear glasses for driving" we can do this:

[tex] P(X\leq 10) = 1- 0.961525= 0.0385[/tex]  

The nearest answer for this case would be 3.1%

Answer:

Step-by-step explanation:

The sum of the angles in a triangle is 180 degrees. This means that in triangle ABC,

Angle A + angle B + angle C = 180

Therefore,

6x - 1 + 20 + x + 14 = 180

6x + x + 20 + 14 - 1 = 180

7x + 33 = 180

Subtracting 33 from the left hand side and the right hand side of the equation, it becomes

7x + 33 - 33 = 180 - 33

7x = 147

Dividing the left hand side and the right hand side of the equation by 7, it becomes

7x/7 = 147/7

x = 21

Therefore

Angle A = 6x - 1 = 6 × 21 - 1

Angle A = 125 degrees

Angle C = x + 14 = 21 + 14

Angle C = 35 degrees.

Answer:

36% probability that she will find an acceptable house in the first two weeks that she looks.

Step-by-step explanation:

For each house, there is a 20% probability of it being acceptable. And a 100-20 = 80% of not being acceptable

What is the probability that she will find an acceptable house in the first two weeks that she looks (round off to second decimal place)?

She only looks one house a week.

So this is the same as the probability of taking two or less weeks to find a house.

The are two outcomes

Finding an acceptable house in the first week, with 20% probability

Not finding an acceptable house in the first week, with 80% probability, and then finding an acceptable house in the second week, with 20% probability.

Probability:

[tex]P = 0.2 + 0.8*0.2 = 0.36[/tex]

36% probability that she will find an acceptable house in the first two weeks that she looks.

The correct answer is 0.36 or 36%.

To solve this problem, we need to calculate the probability that the house hunter does not find an acceptable house in the first two weeks and then subtract this probability from 1 to find the probability that she does find an acceptable house within that time frame.

 Let's denote the probability of finding an acceptable house in a given week as[tex]\( p \)[/tex] and the probability of not finding an acceptable house in a given week as [tex]\( q \)[/tex]. Since 20% of the houses are in acceptable condition, [tex]\( p = 0.20 \) and \( q = 1 - p = 1 - 0.20 = 0.80 \).[/tex]

The probability that the house hunter does not find an acceptable house in the first week is[tex]\( q \).[/tex] The probability that she does not find an acceptable house in the first two weeks is the product of the probabilities that she does not find one in each of the two weeks separately, which is [tex]\( q \times q \) or \( q^2 \).[/tex]

Now, we calculate [tex]\( q^2 \):[/tex]

[tex]\[ q^2 = (0.80)^2 = 0.64 \][/tex]

This is the probability that she will not find an acceptable house in the first two weeks. To find the probability that she will find an acceptable house in the first two weeks, we subtract this value from 1:

[tex]\[ \text{Probability of finding an acceptable house in two weeks} = 1 - q^2 \][/tex]

[tex]\[ \text{Probability of finding an acceptable house in two weeks} = 1 - 0.64 \][/tex]

[tex]\[ \text{Probability of finding an acceptable house in two weeks} = 0.36 \][/tex]