Find the equation of the plane that is parallel to the vectors left angle 3 comma 0 comma 3 right angle and left angle 0 comma 1 comma 3 right angle​, passing through the point (2 comma 0 comma negative 1 ).

Answer:

Null hypothesis: [tex]\mu_{2002}\leq 3.15[/tex]

Alternative hypothesis :[tex]\mu_{2002} > 3.15[/tex]  

C. The mean Rutgers College GPA has increased to more than 3.15

Step-by-step explanation:

Data given and notation  

[tex]\bar X_{2000}=3.15[/tex] represent the average score for Rutgers College in 2000

[tex]\bar X_{2002}=3.46[/tex] represent the average score for Rutgers College in 2002

[tex]s_{2000}[/tex] represent the sample standard deviation  in 2000

[tex]s_{2002}[/tex] represent the sample standard deviation  in 2002

[tex]n[/tex] sample size  

State the null and alternative hypotheses.  .  

What are H0 and Ha for this study?  

Null hypothesis:  [tex]\mu_{2002}\leq 3.15[/tex]

Alternative hypothesis :[tex]\mu_{2002} > 3.15[/tex]  

For this case we need to take in count that the alternative hypothesis can't have and = sign, and based on the information a good hypothesis here would be if the average GPA increased between 2000 and 2002, so based on this the best option for this case would be:

C. The mean Rutgers College GPA has increased to more than 3.15

The best alternative hypothesis based on the data provided is that the mean GPA for Rutgers College students has increased to more than 3.15.

The correct alternative hypothesis for the GPA of Rutgers College students based on the information given is: The mean Rutgers College GPA has increased to more than 3.15. This is because the average GPA in 2002 was reported to be 3.46, which is greater than the 2000's average of 3.15. If we are hypothesizing a change in the population mean, the new mean would be tested against the old mean. Since there was an increase observed, we hypothesize that the new mean GPA is greater than the one from 2000.

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After calculating the interquartile range (IQR) and identifying the thresholds for outliers, it is clear that there are no low outliers but at least one high outlier, as the maximum age exceeds the high outlier threshold. Therefore, only statement (ii) is correct.

The question asks to consider two statements regarding outliers in a five-number summary of the ages of passengers on a cruise ship, which includes a minimum age (Min), first quartile (Q1), median, third quartile (Q3), and maximum age (Max). The statements to consider are:

To determine whether these statements are correct, we use the interquartile range (IQR) method for identifying outliers. The IQR is calculated as Q3 - Q1. An age is considered a low outlier if it is less than Q1 - 1.5 * IQR, and it is a high outlier if it is greater than Q3 + 1.5 * IQR.

Using the given data:

Q1 = 20
Q3 = 38
IQR = Q3 - Q1 = 38 - 20 = 18

Therefore, the cutoff for low outliers is 20 - 1.5 * 18 = 20 - 27 = -7, and for high outliers, it is 38 + 1.5 * 18 = 38 + 27 = 65.

Since the minimum age is 1, which is well above -7, there are no low outliers. However, the maximum age is 80, which is above the high outlier threshold of 65. Hence, there is at least one high outlier.

The correct answer to the question given the provided data is:

Only statement (ii) is correct.

Answer:

Mean salary=$17818.68

Step-by-step explanation:

Salary($)          Employees(f)

5001-10,000     22

10,001-15,000    20

15,001-20,000   21

20,001-25,000  23

25,001-30,000  24

We know that company had 110 employees so ∑f should be equal to 110.

∑f=22+20+21+23+24=110

The mean salary can be computed as

[tex]xbar=\frac{sum(fx)}{sum(f)}[/tex]

The x be the midpoint can be calculated by taking the average of upper and lower class limit.

Class Interval Frequency(f)       x                fx

5001-10,000            22               7500.5     165011

10,001-15,000          20               12500.5     250010

15,001-20,000   21               17500.5     367510.5

20,001-25000   23              22500.5     517511.5

25,001-30,000 24      27500.5 660012

fx can be computed by multiplying each x value with frequency in the respective class.

∑fx=165011+250010+367510.5+517511.5+660012=1960055

[tex]xbar=\frac{1960055}{110}=17818.68[/tex]

Thus, the mean salary is $17818.68.

The mean salary is approximately $17,818.18.

To find the mean salary, we need to calculate the average of all the salaries. Here’s the step-by-step process:

Here's the detailed calculation:

Now, multiply each midpoint by the number of employees in that range:

Add these values together to get the total sum:

165,000 + 250,000 + 367,500 + 517,500 + 660,000 = 1,960,000

Now, divide by the total number of employees:

1,960,000 / 110 ≈ 17,818.18

Therefore, the mean salary is approximately $17,818.18.

Answer:

Manuel is correct, since adding 10 is the same as the tens digit going up by 1.

Step-by-step explanation:

It is important to know the concepts of units, tenths and cents.

For example

1 = 1 unit

10 = 1*10 + 0 = The tens digit is one the unit digit is 0

21 = 2*10 + 1 = The tens digit is two and the unit digit is 1.

120 = 1*100 + 2*10 + 0 = The cents digit is 1, the tens digit is two and the unit digit is 0.

So

Adding 1 is the same as the unit digit going up by 1.

Adding 10 is the same as the tens digit going up by 1.

Adding 100 is the same as the cents digit going up by 1.

In this problem, we have that:

460,470,480, 490,500,510

Each number is 10 added to the previous, that is, the tens digit going up by 1.

Manuel thinks the tens digit goes up by 1 in these numbers. Do you agree?

Manuel is correct, since adding 10 is the same as the tens digit going up by 1.

The statement was given by Manuel that the tens digit goes up by 1 in these numbers, is true.

The second place from the right of the number before the decimal is the tens place of a number.

If we look at the series of the number that is given to us then we will notice that in series 460,470,480, 490,500,510, the second place of the number is increasing, therefore, from 6 to 7 to 8 and further. In the last value, the tens value becomes one because after 490, when we add 10, the number will become 500.

Hence, the statement was given by Manuel that the tens digit goes up by 1 in these numbers, is true.

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Answer:

a. the probability of picking a navy sock and a black sock = P (A & B)

= (8/13 ) * (5/12) = 40/156 = 0.256

b. the probability of picking two navy or two black is

= 56/156 + 20/156 = 76/156 = 0.487

c. the probability of either 2 navy socks is picked or one black  & one navy socks.

= 40/156 + 56/156 = 96/156 = 0.615

Step-by-step explanation:

A sock drawer contains 8 navy blue socks and 5 black socks with no other socks.

If you reach in the drawer and take two socks without looking and without replacement, what is the probability that:  

Solution:

total socks = N = 8 + 5 + 0 = 13

a) you will pick a navy sock and a black sock?

Let A be the probability of picking a navy socks first.

Then P (A) = 8/13

without replacing the navy sock, will pick the black sock, total number of socks left is 12.

Let B be the probability of picking a black sock again.

 P (B) = 5/12.

Then, the probability of picking a navy sock and a black sock = P (A & B)

= (8/13 ) * (5/12) = 40/156 = 0.256

b) the colors of the two socks will match?

Let A be the probability of picking a navy socks first.

Then P (A) = 8/13

without replacing the navy sock, will pick another navy sock, total number of socks left is 12.

Let B be the probability of another navy sock again.

 P (B) = 7/12.

Then, the probability of picking 2 navy sock = P (A & B)

= (8/13 ) * (7/12) = 56/156 = 0.359

Let D be the probability of picking a black socks first.

Then P (D) = 5/13

without replacing the black sock, will pick another black sock, total number of socks left is 12.

Let E be the probability of another black sock again.

 P (E) = 4/12.

Then, the probability of picking 2 black sock = P (D & E)

= (5/13 ) * (4/12) = 5/39 = 0.128

Now, the probability of picking two navy or two black is

= 56/156 + 20/156 = 76/156 = 0.487

c) at least one navy sock will be selected?

this means, is either you pick one navy sock and one black or two navy socks.

so, if you will pick a navy sock and a black sock, the probability of picking a navy sock and a black sock = P (A & B)

= (8/13 ) * (5/12) = 40/156 = 0.256

also, if you will pick 2 navy sock, Then, the probability of picking 2 navy sock = P (A & B)

= (8/13 ) * (7/12) = 56/156 = 0.359

now either 2 navy socks is picked or one black  one navy socks.

= 40/156 + 56/156 = 96/156 = 0.615

Answer:

Google it

Step-by-step explanation:

google is very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very useful

Answer:it is false.

Step-by-step explanation:

Let us assume that the regular price of the shirt is $x.

The shirt is on sale for 40% off the regular price. The amount that is taken off the shirt would be

40/100 × x = 0.4 × x = 0.4x

The new price of the shirt would be x - 0.4x = $0.6x

you have an additional 20% off coupon. The value of the coupon would be

20/100 × 0.6x = 0.2 × 0.6x = 0.12x

The cost of the shirt if the coupon is applied would be

0.6x - 0.12x = 0.48x

If you assumed that the shirt will ultimately be 60% off the original price, the cost of the shirt would be

x - 60/100 × x = x - 0.6x = 0.4x

Therefore, they are not equal and si, it is false.

Answer:

The complex numbers computed are:

A) [tex]q=0.8752+j0.4838=1e^{-j0.5049}[/tex]

B) [tex]r=-8-j8\sqrt{3} =16e^{j\pi \frac{4}{3}}[/tex]

The sketches are attached to this answer

Step-by-step explanation:

To compute these complex numbers you have to remember these rules:

[tex]Z=a+jb=(a^2+b^2)^{\frac{1}{2}}e^{jtan^{-1}(b/a)}[/tex] (a)

[tex]Z=|z|e^{j\alpha}=|z|cos(\alpha)+j|z|sin(\alpha)[/tex] (b)

Also for multiplication, division, and powers, if W and U are complex numbers and k is a real number:

[tex]{W}\cdot{U}={|W|e^{j\alpha}}{|U|e^{j\beta}}={|W|}{|U|}e^{j(\alpha+\beta)}[/tex]   (1)

[tex]\frac{W}{U}=\frac{|W|e^{j\alpha}}{|U|e^{j\beta}}=\frac{|W|}{|U|}e^{j(\alpha-\beta)}[/tex]                        (2)

[tex]W^{k}=|w|^{k}e^{j(\alpha\cdot k)}[/tex]                                      (3)

With these rules we will do the followings steps:

for A:

1) We solve first the divition, writing the 2 complex numbers exponential form (equation (a)).

2) With the rule (2) we solve the division.

3) with rule (3) we solve the power.

For B:

1)We write the numbers a, b, c, d, and f in exponential form (equation (a)).

2) We use the rule (1) for the product.

Answer:

The expression of the field E as the point P becomes very far from the ring is:

[tex]\vec{E}(x)=\displaystyle\frac{q}{4\pi\epsilon_0} \frac{sgn(x)}{x^2}\vec{x} \\\left \{ {{\vec{E}(x)=\frac{q}{4\pi\epsilon_0} \frac{1}{x^2}\vec{x} \mapsto x>0} \atop {\vec{E}(x)=\frac{q}{4\pi\epsilon_0} \frac{-1}{x^2}\vec{x} \mapsto x< 0 }} \right.[/tex]

Step-by-step explanation:

The Electric field expression is:

[tex]\vec{E}(x)=\displaystyle\frac{q}{4\pi\epsilon_0} \frac{x}{(R^2+x^2)^{\frac{3}{2}}}\vec{x}[/tex]

To determine the asked expression we use limits. If we consider that x≫R, this is the same as considering the radius insignificant respect the x distance. Therefore we can considerate than from this distance X, the radius R tends to zero:

[tex]\displaystyle\lim_{R \to{}0}{\vec{E}(x)}=\lim_{R \to{}0}{\frac{q}{4\pi\epsilon_0} \frac{x}{(R^2+x^2)^{\frac{3}{2}}}\vec{x}}\rightarrow\frac{q}{4\pi\epsilon_0} \frac{x}{(0^2+x^2)^{\frac{3}{2}}}\vec{x}=\frac{q}{4\pi\epsilon_0} \frac{x}{(x^2)^{\frac{3}{2}}}\vec{x}=\frac{q}{4\pi\epsilon_0} \frac{\cancel{x}}{|x|^{\cancel{3}}}\vec{x}=\displaystyle\frac{q}{4\pi\epsilon_0} \frac{sgn(x)}{x^2}\vec{x}[/tex]

The expression for the electric field of the ring as the point P becomes very far from the ring is [tex]E_{z} = \frac{k\cdot Q}{x^{2}}[/tex].

Let suppose that the ring has an uniform linear electric density ([tex]\lambda[/tex]). A formula for the electric field at point P ([tex]E[/tex]) in rectangular coordinates is shown below:

[tex]\vec E = (E_{x}, E_{y}, E_{z})[/tex] (1)

Where:

Each component of the electric field are defined by the following integral formulae:

[tex]E_{x} = \int\limits^{2\pi}_{0} {\sin \theta \cdot \cos \phi} \, dE[/tex] (2)

[tex]E_{y} = \int\limits^{2\pi}_{0} {\sin \theta\cdot \sin\phi} \, dE[/tex] (3)

[tex]E_{z} = \int\limits^{2\pi}_{0} {\cos \theta} \, dE[/tex] (4)

Where:

By Coulomb's law and trigonometric and geometric relationships, we expand and solve each integral as following:

[tex]E_{x} = \frac{R}{\sqrt{x^{2}+R^{2}}}\int\limits^{2\pi}_{0} {\cos \phi} \, dE = \frac{k\cdot \lambda\cdot R^{2}}{(x^{2}+R^{2})^{3/2}}\int\limits^{2\pi}_{0} {\cos \phi} \, d\phi = 0[/tex]

[tex]E_{y} = \frac{R}{\sqrt{x^{2}+R^{2}}}\int\limits^{2\pi}_{0} {\sin \phi} \, dE = \frac{k\cdot \lambda\cdot R^{2}}{(x^{2}+R^{2})^{3/2}}\int\limits^{2\pi}_{0} {\sin \phi} \, d\phi = 0[/tex]

[tex]E_{z} = \frac{k\cdot \lambda\cdot x \cdot R}{(x^{2}+R^{2})^{3/2}} \int\limits^{2\pi}_{0}\, d\phi = \frac{x\cdot k \cdot (2\pi\cdot \lambda\cdot R)}{(x^{2}+R^{2})^{3/2}} = \frac{x\cdot k\cdot Q}{(x^{2}+R^{2})^{3/2}}[/tex] (5)

 Where [tex]k[/tex] is the electrostatic constant.

If [tex]x >> R[/tex], (5) is simplified into the following expression:

[tex]E_{z} = \frac{k\cdot Q}{x^{2}}[/tex] (6)

Where [tex]Q[/tex] is the electric charge of the entire ring.

Please notice that (6) tends to be zero when [tex]x \to \infty[/tex]. The expression for the electric field of the ring as the point P becomes very far from the ring is [tex]E_{z} = \frac{k\cdot Q}{x^{2}}[/tex]. [tex]\blacksquare[/tex]

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Answer:

The true mean [tex]\mu[/tex] it probably could be larger, smaller, or equal to 22

Step-by-step explanation:

False.

By definition the sample mean is defined as:

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

For this case the value for the sample size is n =190 and the calculated sample mean is [tex] \bar X=22[/tex]. This value represent the sample and for this case we can't assume that this value represent at all the population as the population mean [tex]\mu[/tex] since we probably have variability from the data of the students at City College.

So we can conclude that the true mean [tex]\mu[/tex] it probably could be larger, smaller, or equal to 22

The true mean  it probably could be larger, smaller, or equal to 22.

Given that:

Total student, n = 190.

Average age of 190 students, [tex]\bar X = 22\\[/tex].

By definition the sample mean is defined as:

[tex]\bar X =\dfrac{{\sum_{i=1}^nX_i}}{n}[/tex]

For this case, the sample size n =190 and the calculated sample mean is [tex]\bar X = 22\\[/tex]. This value represent the sample and for this case can't assume that this value represent at all the population as the population mean  since probably have variability from the data of the students at city college.

Hence, conclude that the true mean  it probably could be larger, smaller, or equal to 22

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Answer:

[tex]\sum_{n=0}^9cos(\frac{\pi n}{2})=1[/tex]

[tex] \sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0[/tex]

[tex] \sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}[/tex]

Step-by-step explanation:

[tex] \sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))[/tex]

[tex]=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})[/tex]

[tex]=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1[/tex]

2nd

[tex]\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}[/tex]

[tex]=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0[/tex]

3th

[tex] \sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))[/tex]

[tex]=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}[/tex]

What we use?

We use that

[tex] e^{i\pi n}=cos(\pi n)+i sin(\pi n)[/tex]

and

[tex]\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}[/tex]

Geometric sum formulas are used to evaluate sums of a geometric series, with the result expressed in Cartesian form (a + bi) where a is the real part and bi is the imaginary part. The sum of a geometric series is calculated with the formula: Sum = a * (1 - r^n) / (1 - r), where a is the first term and r is the ratio. Please provide the specific sums for a detailed step-by-step calculation.

The problem at hand revolves around the usage of geometric sum formulas to evaluate sums and to express the result in Cartesian form. The critical point to remember is that a geometric series is a series with a constant ratio between successive terms. The sum of the first 'n' terms of a geometric sequence can be calculated using the formula:

Sum = a * (1 - rⁿ) / (1 - r)

Assuming 'a' represents the first term in the series and 'r' is the ratio.To convert a complex number into Cartesian form, you simply map the real and imaginary parts of the number 'a + bi', where 'a' is the real part, and 'bi' is the imaginary part.Unfortunately, without the specifics of the sums you're looking to evaluate, it's impossible to give a concrete step-by-step calculation. However, understanding the formulas and how they're applied should provide you with a good start.

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Answer:

the probability that a student chosen at random plays football or cricket or both = [tex]\frac{1}{5} + \frac{2}{5} + \frac{4}{15} = \frac{13}{15}[/tex]

Step-by-step explanation:

i) 8 students play football and cricket

ii) 4 students do not play football or cricket

iii) total of 14 students play football.

iv) therefore the number of students who play only football is = 14 - 8 = 6

v) total of 20 students play cricket.

vi) therefore the number of students who play only cricket is = 20 - 8 = 12

vii) therefore the total number of students = 8 + 4 + 6 + 12 = 30

viii) the probability a student chosen at random plays football = [tex]\frac{6}{30} = \frac{1}{5}[/tex]

ix) the probability a student chosen at random plays cricket = [tex]\frac{12}{30} = \frac{2}{5}[/tex]

x) the probability a student chosen at random plays both football and cricket = [tex]\frac{8}{30} = \frac{4}{15}[/tex]

xi) therefore the probability that a student chosen at random plays football or cricket or both = [tex]\frac{1}{5} + \frac{2}{5} + \frac{4}{15} = \frac{13}{15}[/tex]

The probability that a student chosen at random plays football or cricket or both is [tex]\frac{13}{15}[/tex].

We have

Number of students play football and cricket = 8

Number of students do not play football or cricket = 4

Total Number of students play football = 14

 Therefore, the number of students who play only football

= 14 - 8

= 6

Total Number of students play cricket = 20

Therefore, the number of students who play only cricket

= 20 - 8

= 12

So, the total number of students

= 8 + 4 + 6 + 12

= 30

Now, the probability that a student chosen at random plays football

[tex]=\frac{6}{30} \\=\frac{1}{5}[/tex]

The probability that a student chosen at random plays cricket

[tex]=\frac{12}{30} \\=\frac{2}{5}[/tex]

The probability a student chosen at random plays both football and cricket  [tex]=\frac{8}{30} \\=\frac{4}{15}[/tex]

Therefore, the probability that a student chosen at random plays football or cricket or both

[tex]=\frac{1}{5} +\frac{2}{5}+\frac{4}{15}\\=\frac{3}{15} +\frac{6}{15}+\frac{4}{15}\\=\frac{13}{15}[/tex]

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Answer:B

Step-by-step explanation:

Answer:

The standard deviation for given sample is 7.97          

Step-by-step explanation:

We are given the following data set:

23, 20, 14, 35, 28

Formula:

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{120}{5} = 24[/tex]

Sum of squares of differences = 1 + 16 + 100 + 121 + 16 = 254

[tex]S.D = \sqrt{\dfrac{254}{4}} = 7.97[/tex]

The standard deviation for given sample is 7.97

To find the sample standard deviation for the data sets 23, 20, 14, 35, 28, we calculate the mean, subtract the mean from each data point and square the result, sum these squares, divide by one less than the sample size to find the variance, and finally take the square root to find the standard deviation which is approximately 7.97.

To calculate the sample standard deviation (s), follow these steps:

Find the mean (average) of the sample data.

Subtract the mean from each data point and square the result.

Sum all the squared values.

Divide this sum by the sample size minus one (n-1) to get the sample variance.

Take the square root of the sample variance to find the sample standard deviation.

Let's apply these steps to the given data: 23, 20, 14, 35, 28.

Mean = (23 + 20 + 14 + 35 + 28) / 5 = 120 / 5 = 24.

Subtract the mean and square: (23 - 24)² = 1, (20 - 24)² = 16, (14 - 24)² = 100, (35 - 24)² = 121, (28 - 24)² = 16.

Sum of squares = 1 + 16 + 100 + 121 + 16 = 254.

Variance = 254 / (5 - 1) = 254 / 4 = 63.5.

Standard Deviation = √63.5 ≈ 7.97 (rounded to the nearest hundredth).

The sample standard deviation s is approximately 7.97.

Answer:

0.9964 is the probability  that at least one of them meets specifications.

Step-by-step explanation:

We are given the following in the question:

B: Bolts meet the specification

N: Nails meet the specification

P(B) = 96% = 0.96

P(N) = 91% = 0.91

One bolt and one nail are chosen independently.

Thus, we can write

[tex]P(B\cap N) = P(B) \times P(N) = 0.96\times 0.91 = 0.8736[/tex]

We have to find the probability that at least one of them meets specifications.

[tex]P(B\cup N) = P(B) + P(N) -P(B\cap N)\\P(B\cup N) =0.96 + 0.91-0.8736\\P(B\cup N) =0.9964[/tex]

0.9964 is the probability  that at least one of them meets specifications.

The cost of printing posters from 2 to 100, given the marginal cost function dc/dx = 1/2√x, is calculated by finding the anti-derivative (or integral) of the function from 1 to 100, resulting in a cost of $18.

The given function dc/dx = 1/2√x represents the marginal cost, which is the derivative of the cost function c(x). To find the cost of printing from the 2nd to the 100th poster, we need to find the integral of the marginal cost from 1 to 100, not including the cost of the first poster. This is represented mathematically as ∫ from 1 to 100 of (1/2√x) dx. Here's how to solve this:

This results in 2*(√100 - √1) = 20 - 2 = $18. This is the cost for printing posters 2 to 100.

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Option:

A) The risk factor in the population of interest

B) The participants

C) The outcome in the population of interest

D) A & C

Answer:

D) A & C

Step-by-step explanation:

A Simple Random Sample or a Stratified Random Sample of students across the CSU campuses would allow for reliable data collection on the average distance between student hometowns and their campus, taking into account that self-reporting may introduce errors.

When considering methods to sample the average distance between the hometowns of students and their California State University (CSU) campuses, there are several sampling techniques that can be considered:

Option (b) is incorrect as there are potential problems with self-reporting of distances, and option (d) is impractical for such a large population and not necessary for making inferences. Therefore, options (a), (c), and (e) are relevant to the question.

Answer:

Web searching, specialists consultations and comparisons.

Step-by-step explanation:

a. Does the Honda CR-V iclude traction control as a standard feature?

Research about the Honda CR-V on the internet, or reading an article about it.

b. How much money has our company´s philanthropic foundation donated to colleges and universities in each of the last three year?

Look over the company´s administrative records.

c.How does a 3D printer work?

Search on the web about the 3D printer function.

d. Could our Building 3 support a rooftop green space?

Consultation with an architect.

e. How can we determine whether we would save more money by switching to LED lighting in our corporate offices?

Search on the web about the LED lighting use of electricty and the use of electricty of the type of lighting that the company is already using and compare for the best one.

Answer:

east

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

D. The outlier will appear as a bar far from all of the other bars with a height that corresponds to a frequency of 1.

Step-by-step explanation:

An histogram measures how many times each value appears in the set we are studying. That is, it is a frequency measure.

Suppose we have the following set:

S = {1,1,1,1,1,1, 2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,5,5,5,5,5,5,100}

1 appears 6 times. That means that when the X axis is 1, the y axis is 6.

2 appears 8 times. The means that when the X axis is 2, the y axis is 8.

...

100 appears 1 time. This means that when the X axis is 100, the y axis is 1. The X is the outlier, and it is quite far from the other values.

So the correct answer is:

D. The outlier will appear as a bar far from all of the other bars with a height that corresponds to a frequency of 1.

Answer:  24

Step-by-step explanation:

Given : Choices for album covers = 3

Choices for television commercials = 4

Choices for album posters = 2

Now , the number of different combinations he needs in order to test each album cover, television commercial, and album poster = ( Choices for album covers ) x (Choices for television commercials) x (Choices for album posters)

= 3 x 4 x 2 = 24

Hence, the number of different combinations he needs in order to test each album cover, television commercial, and album poster is 24.

Answer:

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

Step-by-step explanation:

we know that

The equation of a linear function has no exponents higher than 1, and the graph of a linear function is a straight line.

Verify each case

case 1) we have

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

Is the equation of a line in slope intercept form

so

Is a straight line and has no exponents higher than 1

therefore

Is a linear equation

case 2) we have

[tex]y=2x^{2}+2[/tex]

Is a quadratic equation

Is a curved line and has at least one exponent higher than 1,

therefore

Is a non-linear equation

case 3) we have

[tex]y=2x^{3}+2[/tex]

Is a cubic equation

Is a curved line and has at least one exponent higher than 1,

therefore

Is a non-linear equation

case 4) we have

[tex]y=2x^{4}+2[/tex]

Is a quartic equation

Is a curved line and has at least one exponent higher than 1,

therefore

Is a non-linear equation

Answer:

For this case since since they not want to differentiate between the campuses, so is good to use a simple random sampling or SRS, that is a procedure in order to select a sample of size n from a population of size N known, and each element of the population have the sample probability of being selected [tex]p=\frac{1}{N}[/tex]

So then the administrator can select a random sample of n students from all the campuses from the CSU University and obtain the distance from hometwon to campuses for the selected sample and then calculate the average and use this to inference.

Other possibility is use a systematic random sampling for example they can define a random number k and they can select 1 student each k individuals and then create a random sample of size n and calculate the average for this in order to do inference.

The convenience sampling is not too useful since is not a probabilistic method.

Step-by-step explanation:

For this case we can clasify this study as an enumerative study and inferential since they want to identify the average distance between the hometowns of students and their campuses.

For this case the sampling frame represent all the 23 campuses, and is known for the researcher,

For this case since since they not want to differentiate between the campuses, so is good to use a simple random sampling or SRS, that is a procedure in order to select a sample of size n from a population of size N known, and each element of the population have the sample probability of being selected [tex]p=\frac{1}{N}[/tex]

So then the administrator can select a random sample of n students from all the campuses from the CSU University and obtain the distance from hometwon to campuses for the selected sample and then calculate the average and use this to inference.

Other possibility is use a systematic random sampling for example they can define a random number k and they can select 1 student each k individuals and then create a random sample of size n and calculate the average for this in order to do inference.

The convenience sampling is not too useful since is not a probabilistic method.

Different sampling methods that can be employed to estimate the average distance between CSU student hometowns and their campuses include Simple Random Sampling, Stratified Random Sampling, Cluster Sampling, and Systematic Sampling. Each has its own advantages and potential drawbacks.

There are several sampling methods that can be employed to gauge the average distance between student hometowns and their respective California State University (CSU) campuses.

A Simple Random Sampling method could be applied, where every student from every campuse has the exact same chance of being selected. However, this method may not give a representative sample if certain campuses have more or less students than others.

Another method is Stratified Random Sampling, where students are first divided into groups or 'strata' based on their campus, and then random samples are taken from each stratum. This ensures a balanced sample from all campuses.

Then comes the Cluster Sampling, it divides the student population into 'clusters' based on their hometown, and then randomly selected clusters are surveyed. It would be useful for large-scale surveys.

Last is Systematic Sampling. In this method, every nth student on a list would be selected. This approach ensures evenly distributed selection of students across the whole population but requires a complete listing of all students at the CSU campuses.

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Answer:

a) ordinal

b) ratio

c) nominal

d) interval

e) ratio

f) nominal

Step-by-step explanation:

a) Salesperson performance is ordinal because it is ordered from below average to above average

b) Price of company stock is ratio because it has a defined zero point.

c) Names of new product is nominal because it can be classified into different categories yet cannot be ordered.

d) Temperature in CEO's office is interval because it don't have a defined zero point.  In temperature zero temperature can exists.

e) Again gross income is ratio because it has a defined zero point. Zero gross income will means that there is no gross income.

f) Color of product packaging is again nominal because it can be classified into different categories yet cannot be ordered.

The measurements can be categorized as "ordinal, ratio, nominal, and interval."

(a) Salesperson's performance: below average, average, above average. - Ordinal

(b) Price of company's stock - Ratio

(c) Names of new products - Nominal

(d) Temperature (°F) in CEO's private office - Interval

(e) Gross income for each of the past 5 years - Ratio

(f) Color of product packaging - Nominal

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Answer:

Part A:

[tex]p=0.0095[/tex]

Part B:

[tex]p=0.0043[/tex]

Step-by-step explanation:

Part A:

The number of rivets=22 rivets

Probability that no rivet is defective= (1-p)^22

The probability that at least one rivet is defective=1-(1-p)^22

For 19% of all seams need reworking, probability that a rivet is defective is given by:

[tex]1-(1-p)^{22}=0.19[/tex]

[tex](1-p)^{22}=0.81\\p=1-\sqrt[22]{0.81} \\p=0.0095[/tex]

Part B:

For 9% of all seams need reworking, probability of a defective rivet is:

[tex]1-(1-p)^{22}=0.09\\p=1-\sqrt[22]{0.91} \\p=0.0043[/tex]

To find the probability of a defective rivet in a seam and the smallest probability of a defective rivet to ensure a certain reworking percentage, we use the concept of independent events and probability calculations.

(a) To find the probability that a rivet is defective:

Let p be the probability of a defective rivet.

Since 19% of seams need reworking, 19% of the seams have at least one defective rivet.

Therefore, 19% of all seams equals the probability that at least one rivet is defective:

P(at least one defective rivet) = 1 - P(no defective rivets) = 0.19

P(no defective rivets) = 1 - P(at least one defective rivet) = 1 - 0.19

P(no defective rivets) = 0.81

Since each rivet is defective independently of one another, the probability that a rivet is defective is:

p = 1 - P(no defective rivet)

p = 1 - 0.81

p = 0.19

Therefore, the probability that a rivet is defective is 0.19 or 19%.

(b) To find the smallest probability of a defective rivet:

Let p be the probability of a defective rivet that ensures only 9% of seams need reworking.

We need to find the value of p such that P(at least one defective rivet) = 0.09.

From part (a), we know that P(at least one defective rivet) = 1 - P(no defective rivets) = 0.19.

Therefore, we can set up the equation:

0.19 = 1 - (1 - p)22

Solving this equation will give us the smallest value of p that satisfies the condition.

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Answer:

- 2⁻⁶

Step-by-step explanation:

To simplify this we have to know the following rules.

(i) (xᵃ)ᵇ = xᵃᵇ

(ii) 1/xᵃ = x⁻ᵃ

Given: [tex]$ \frac{-1}{64} $[/tex]

= [tex]$ \frac{-1}{4^3} $[/tex]

= [tex]$ \frac{-1}{(2^2)^3} $[/tex]

= [tex]$ \frac{-1}{2^6} \hspace{10mm} $[/tex]               [using (i)]

= [tex]$ 2^{-6} $[/tex]              [using (ii)]

Hence, the simplified form would be: [tex]$ 2^{-6} $[/tex]

Answer:

D. $140.

Step-by-step explanation:

Given:

Cost of clothes = $76.17

Cost of books = $44.98

Cost of meal = $19.15

We need to find the best estimate of total cost of her shopping trip.

Solution:

First we will find the total cost of her shopping trip.

total cost of her shopping trip is equal to sum of Cost of clothes, Cost of books and Cost of meal.

framing in equation form we get;

total cost of her shopping trip = [tex]76.17+44.98+19.15 = \$140.3[/tex]

Now we can say that;

140.3 is close to 140

Hence Best estimate of total cost of shopping trip is $140.

Answer: The expected value of the water depth is 4.5 m.

Step-by-step explanation:

Let x be a random variable which is uniformly distributed in interval [a,b] .

Then the mean of the distribution is ghiven by :-

[tex]E(x)=\dfrac{a+b}{2}[/tex]

Given : While conducting experiments, a marine biologist selects water depths from a uniformly distributed collection that vary between 2.00 m and 7.00 m.

Then, the expected value of the water depth = [tex]\dfrac{2+7}{2}=\dfrac{9}{2}=4.5[/tex]

Hence, the expected value of the water depth is 4.5 m.

Answer:option A is the correct answer.

Step-by-step explanation:

Ronald can read at a constant rate of p pages per minute.

If Roland can read a total number of N pages in minutes, then the equation representing the relationship between the number of pages, N and the time, m minutes would be

p pages = 1 minute

N pages = m minutes

Crossmultiplying, it becomes

p × m = N × 1

N = pm

Answer: 5.85 years

Therefore, an average associate is employed for 5.85 years.

Step-by-step explanation:

Given:

Rate of employment yearly = 20 associates per year

Total number of associates = 117 associates

Since the total number of associates remain constant the rate at which they employ new associates is equal to the rate at which associates leave = 20 per year

If 20 new associates are employed in a particular year it would take aan average of :

Average employment year A = total number of associates divided by the rate at which associates leave

A = 117/20

A = 5.85 years

Therefore, an average associate is employed for 5.85 years.

The average length of employment for an associate at the consulting company is calculated by dividing the total number of associates (117) by the annual turnover (20), yielding an average of approximately 5.85 years.

To calculate the average length of employment for associates at the consulting company, we can use the concept of employee turnover rate, which is the rate at which employees leave an organization and are replaced. Since the company must hire 20 new associates each year to replace those who have departed and the total number of associates is 117, we can use the formula for the average employment duration: Total Number of Associates / Annual Turnover = Average Length of Employment.

Using the numbers provided: 117 associates / 20 associates per year = 5.85 years.

This result signifies that the average associate is employed at the consulting company for approximately 5.85 years before they leave the company, although this is a simplification assuming a constant replacement and turnover rate.