Find the height of the ramp and the length of the base of the ramp

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:

For each n get its posting then AND them

Step-by-step explanation:

1. Lets presume we have multiple n terms

2. We get the posting from each n term

3. We use the function AND and apply it to each n term

4. We start with the tiniest set and we continue from there

5. We have a pair of n terms

6. We get the posting from 1st n

7. We get the posting from 2nd n

8. We apply n1 AND n2

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.

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:

a) 120

b) 24

c) 18

Step-by-step explanation:

part a

The sample space is defined by the rank of the horses.

Hence, 5 ranks would permute to = 5! = 120 outcomes

part b

Fixing the first position for horse A we are left with four horses and 4 positions, the position are permutated to 4! = 24 outcomes

part c

Fixing the first position for horse A we are left with four horses and 4 positions, and horse B can not finish second hence:

A _ B _ _  the rest can permute hence, 3! = 6 outcomes

A _ _ B _ the rest can permute hence, 3! = 6 outcomes

A _ _ _ B the rest can permute hence, 3! = 6 outcomes

Total outcomes is sum of 3 cases above = 18 outcomes

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:

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: The equation of the sphere is:

(x-2)^2 + (y-2)^2 + (z-5/2)^2 = sqrt(245)/2

Step-by-step explanation:

The centre of the sphere is the midpoint of the diameter, which is

1/2

[(-4,7,6) + (8,-3,5)] =(2,2,5/2)

The length of the diameter is

=sqrt |(8,-3,5) - (-4,7,6)|^2

=sqrt (12^2 + (-10)^2 + (-1)^2)

=sqrt (144+100+1)

=sqrt(245)

so the radius of

the sphere is:

1/2(sqrt(245)) = sqrt(245)/2.

The equation of the sphere is:

(x-2)^2 + (y-2)^2 + (z-5/2)^2 = sqrt(245)/2

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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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:

a)0.08  , b)0.4  , C) i)0.84  , ii)0.56

Step-by-step explanation:

Given data

P(A) =  professor arrives on time

P(A) = 0.8

P(B) =  Student aarive on time

P(B) = 0.6

According to the question A & B are Independent  

P(A∩B) = P(A) . P(B)

Therefore  

[tex]{A}'[/tex] & [tex]{B}'[/tex] is also independent

[tex]{A}'[/tex] = 1-0.8 = 0.2

[tex]{B}'[/tex] = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B')  (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

[tex]P(\frac{B'}{A})[/tex] = [tex]\frac{P(B'\cap A)}{P(A)}[/tex] = [tex]\frac{0.4\times 0.8}{0.8}[/tex] = 0.4

Part c)

Assume the events are not independent

Given Data

P[tex](\frac{{A}'}{{B}'})[/tex] = 0.4

=[tex]\frac{P({A}'\cap {B}')}{P({B}')}[/tex] = 0.4

[tex]P[/tex][tex]({A}'\cap {B}')[/tex] = 0.4 x P[tex]({B}')[/tex]

= 0.4 x 0.4 = 0.16

[tex]P({A}'\cap {B}')[/tex] = 0.16

i)

The probability that at least one of them is on time

[tex]P(A\cup B)[/tex] = 1- [tex]P({A}'\cap {B}')[/tex]  

=  1 - 0.16 = 0.84

ii)The probability that they are both on time

P[tex](A\cap B)[/tex] = 1 - [tex]P({A}'\cup {B}')[/tex] = 1 - [tex][P({A}')+P({B}') - P({A}'\cap {B}')][/tex]

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

Answer:

east

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

a) [tex] E(T) = 0.2 E(Y) -1000= 0.2*20000 -1000=3000[/tex]

b) [tex] Sd(T) = \sqrt{0.2^2 Var(Y)}=\sqrt{0.2^2 8000^2}= 1600[/tex]

c) Assuming 20 million of families and each one with a mean of income of 20000 for each family approximately then total income would be:

[tex] E(T) = 20000000*20000= 40000 millions[/tex]

And if we replace into the formula of T we have:

[tex] T = 0.2*400000x10^6 -1000= 790000 millions[/tex]

Approximately.  

Step-by-step explanation:

For this case we knwo that Y represenet the random variable "Income" and we have the following properties:

[tex] E(Y) = 20000, Sd(Y) = 8000[/tex]

We define a new random variable T "who represent the taxes"

[tex] T = 0.2(Y-5000) = 0.2Y -1000[/tex]

Part a

For this case we need to apply properties of expected value and we have this:

[tex] E(T) = E(0.2 Y -1000)[/tex]

We can distribute the expected value like this:

[tex] E(T) = E(0.2 Y) -E(1000)[/tex]

We can take the 0.2 as a factor since is a constant and the expected value of a constant is the same constant.

[tex] E(T) = 0.2 E(Y) -1000= 0.2*20000 -1000=3000[/tex]

Part b

For this case we need to first find the variance of T we need to remember that if a is a constant and X a random variable [tex] Var(aX) = a^2 Var(X)[/tex]

[tex] Var(T) = Var (0.2Y -1000)[/tex]

[tex] Var(T)= Var(0.2Y) -Var(1000) + 2 Cov(0.2Y, -1000)[/tex]

The covariance between a random variable and a constant is 0 and a constant not have variance so then we have this:

[tex] Var(T) =0.2^2 Var(Y)[/tex]

And the deviation would be:

[tex] Sd(T) = \sqrt{0.2^2 Var(Y)}=\sqrt{0.2^2 8000^2}= 1600[/tex]

Part c

Assuming 20 million of families and each one with a mean of income of 20000 for each family approximately then total income would be:

[tex] E(T) = 20000000*20000= 40000 millions[/tex]

And if we replace into the formula of T we have:

[tex] T = 0.2*400000x10^6 -1000= 790000 millions[/tex]

Approximately.  

Answer:

a)   (y+ Oz)

b) (ab + ac)(abc + cd + ce)

Step-by-step explanation:

a) xy + xOz  (First find the common factor between xy and xOz, and it is x as it appears in both xy and xOz. Factoring out x will result in

 x(y+ Oz)

There are no more common factors thus the answer is x(y + Oz)

Since the question was asking after removing redundant terms it will on be y + Oz

b) (ab0 f + a 0 cf)(a 0 bcf + c 0 f(d + e))  simplfying the expression

(ab0 f + a 0 cf)(aObcf + cOfd + cOfe) common factor Of is factored out as below

Of[ (ab + ac)(abc + cd + ce) ] therefore after removing redundant terms the answer is (ab + ac)(abc + cd + ce)

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:

a) Continuous since the weight can take decimals for example 2.3 pounds.

b) Discrete since the number of chapters in a book represent an integer, we can't say that abook have 4.3 chapters because not makes sense, we say the book have 4 or 5 chapters

c) Continuous since we can measure the width and we can get 8.3 inches so then the variable can takes decimals and for this reason is Continuous

d) Discrete since that's a binary variable and only take integers.

e) Discrete since the number of errors can't be 4.3 for example, and can't take decimals.

Step-by-step explanation:

Previous concepts

A continuous random variable by definition is a "random variable where the data can take infinitely many values" defined on a interval.

And a discrete random variable is a random variable that only can takes integers and is defined over a domain

Solution to the problem

a. Weight of the book (e.g., 2.3 pounds)

Continuous since the weight can take decimals for example 2.3 pounds.

b. Number of chapters in the book (e.g., 10 chapters)

Discrete since the number of chapters in a book represent an integer, we can't say that abook have 4.3 chapters because not makes sense, we say the book have 4 or 5 chapters

c. Width of the book (e.g., 8 inches)

Continuous since we can measure the width and we can get 8.3 inches so then the variable can takes decimals and for this reason is Continuous

d. Type of book (0=hardback or 1=paperback)

Discrete since that's a binary variable and only take integers.

e. Number of typographical errors in the book (e.g., 4 errors)

Discrete since the number of errors can't be 4.3 for example, and can't take decimals.

Answer:

In general, 50 % of the values in a data set lie at or below the median. 75 % of the values in a data set lie at or below the third quartile (Q3). If a sample consists of 1600 test scores, 0.5*1600 = 800 of them would be at or below the median. If a sample consists of 1600 test scores, 0.75*800 = 1200 of them would be at or above the first quartile (Q1)

Step-by-step explanation:

The median separates the upper half from the lower half of a set. So 50% of the values in a data set lie at or below the median, and 50% lie at or above the median.

The first quartile(Q1) separates the lower 25% from the upper 75% of a set. So 25% of the values in a data set lie at or below the first quartile, and 75% of the values in a data set lie at or above the first quartile.

The third quartile(Q3) separates the lower 75% from the upper 25% of a set. So 75% of the values in a data set lie at or below the third quartile, and 25% of the values in a data set lie at or the third quartile.

The answer is:

In general, 50 % of the values in a data set lie at or below the median. 75 % of the values in a data set lie at or below the third quartile (Q3). If a sample consists of 1600 test scores, 0.5*1600 = 800 of them would be at or below the median. If a sample consists of 1600 test scores, 0.75*800 = 1200 of them would be at or above the first quartile (Q1)

Using the principle of quartile distribution, the percentage amount of data at or below the median and third quartile are 50% and 75% respectively. Number of scores below the median and at or above the first quartile is 800 and 1200 respectively.

The median of 1600 can be calculated thus :

Above the first quartile :

Therefore, the score above the first quartile is 1200.

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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:

a) There is a 59.87% probability that none of the LED light bulbs are defective.

b) There is a 31.51% probability that exactly one of the light bulbs is defective.

c) There is a 98.84% probability that two or fewer of the LED light bulbs are defective.

d) There is a 100% probability that three or more of the LED light bulbs are not defective.

Step-by-step explanation:

For each light bulb, there are only two possible outcomes. Either it fails, or it does not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

In this problem we have that:

[tex]n = 10, p = 0.05[/tex]

a) None of the LED light bulbs are defective?

This is P(X = 0).

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

[tex]P(X = 0) = C_{10,0}*(0.05)^{0}*(0.95)^{10} = 0.5987[/tex]

There is a 59.87% probability that none of the LED light bulbs are defective.

b) Exactly one of the LED light bulbs is defective?

This is P(X = 1).

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

[tex]P(X = 1) = C_{10,1}*(0.05)^{1}*(0.95)^{9} = 0.3151[/tex]

There is a 31.51% probability that exactly one of the light bulbs is defective.

c) Two or fewer of the LED light bulbs are defective?

This is

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

[tex]P(X = 2) = C_{10,2}*(0.05)^{2}*(0.95)^{8} = 0.0746[/tex]

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5987 + 0.3151 + 0.0746 0.9884[/tex]

There is a 98.84% probability that two or fewer of the LED light bulbs are defective.

d) Three or more of the LED light bulbs are not defective?

Now we use p = 0.95.

Either two or fewer are not defective, or three or more are not defective. The sum of these probabilities is decimal 1.

So

[tex]P(X \leq 2) + P(X \geq 3) = 1[/tex]

[tex]P(X \geq 3) = 1 - P(X \leq 2)[/tex]

In which

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

[tex]P(X = 0) = C_{10,0}*(0.95)^{0}*(0.05)^{10}\cong 0[/tex]

[tex]P(X = 1) = C_{10,1}*(0.95)^{1}*(0.05)^{9} \cong 0[/tex]

[tex]P(X = 2) = C_{10,1}*(0.95)^{2}*(0.05)^{8} \cong 0[/tex]

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0[/tex]

[tex]P(X \geq 3) = 1 - P(X \leq 2) = 1[/tex]

There is a 100% probability that three or more of the LED light bulbs are not defective.

The question relates to binomial distribution in probability theory. The probabilities calculated include those of none, one, two or less, and three or more LED bulbs being defective out of a random sample of 10.

This question relates to the binomial probability distribution. A binomial distribution is applicable because there are exactly two outcomes in each trial (either the LED bulb is defective or it's not) and the probability of a success remains consistent.

a) In this scenario, 'none of the bulbs being defective' means 10 successes. The formula for probability in a binomial distribution is p(x) = C(n, x) * [p^x] * [(1-p)^(n-x)]. Plugging in the values, we find p(10) = C(10, 10) * [0.95^10] * [0.05^0] = 0.5987 or 59.87%.

b) 'Exactly one of the bulbs being defective' implies 9 successes and 1 failure. Following the same formula, we get p(9) = C(10, 9) * [0.95^9] * [0.05^1] = 0.3151 or 31.51%.

c) 'Two or less bulbs being defective' means 8, 9 or 10 successes. We add the probabilities calculated in (a) and (b) with that of 8 successes to get this probability. Therefore, p(8 or 9 or 10) = p(8) + p(9) + p(10) = 0.95.

d) 'Three or more bulbs are not defective' means anywhere from 3 to 10 successes. As the failure rate is low, it's easier to calculate the case for 0, 1 and 2 successes and subtract it from 1 to find this probability. This gives us p(>=3) = 1 - p(2) - p(1) - p(0) = 0.98.

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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:

[tex] s= \sqrt{80.568}=8.97 [/tex]

[tex] Median =\frac{1+3}{2}=2[/tex]

Step-by-step explanation:

For this case we have the following data:

1.0 -5 3.0 -8 14 -11 12 0 16 -2 12 7

And ordering this data we have:

-11 -8 -5 -2 0 1 3 7 12 12 14 16  

And we are interested in find the standard deviation for the sample data. In order to do this the first step is find the mean given by this formula:

[tex] \bar X =\frac{\sum_{i=1}^n X_i}{n}=\frac{1-5+3-8+14-11+12+0+16-2+12+7}{12}=3.25[/tex]

Now with the sample mean we can calculate the sample variance with the following formula:

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

And if we replace we got:

[tex] s^2 =80.568[/tex]

And for the sample deviation we just need to take the square root of the sample variance and we got:

[tex] s= \sqrt{80.568}=8.97 [/tex]

The median on this case would be given by:

[tex] Median =\frac{1+3}{2}=2[/tex]

Using the positions 5 and 6 for the average since the sample size is an even number.

Final answer:

The standard deviation of the weight changes for the 12 women is approximately 8.1 pounds, after computing the variance and taking the square root.

Explanation:

To calculate the sample standard deviation of the given weight changes, we will follow these steps:

Now let's apply these steps to the given data:

The data points are: 1.0, -5, 3.0, -8, 14, -11, 12, 0, 16, -2, 12, 7

The mean (step 1) is calculated as:

(1.0 - 5 + 3.0 - 8 + 14 - 11 + 12 + 0 + 16 - 2 + 12 + 7) / 12 = (49) / 12 ≈ 4.0833 pounds

Continuing with steps 2 to 5, the squared differences from the mean, their sum, and the variance are computed. Finally, the sample standard deviation is found to be approximately 8.1 pounds.

Therefore, the correct answer is option b, which is 8.1 pounds.

Answer:

The percentage of inter-rater reliability is 80%.

Step-by-step explanation:

The rate of inter-rater reliability is the division of the number of intervals in which they agreed number of total intervals.

In this problem, we have that:

There are 10 intervals.

The observers agreed on 8 of them.

So the rate of agreement is 8/10 = 0.8.

As a percentage, we multiply the rate by 100, so 0.8*100 = 80%.

The percentage of inter-rater reliability is 80%.

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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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:

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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Answer:

Step-by-step explanation:

To prove sin(a+b)*sin(a-b)=cos^2b-cos^2a

we simplify the left side  sin(a+b)*sin(a-b) first

sin(a+b) = sin a cos b + cos a sin b

sin(a-b) = sin a cos b - cos a sin b

sin(a+b)*sin(a-b) = (sin a cos b + cos a sin b) x (sin a cos b -cos a sin b)

sin a cos b((sin a cos b + cos a sin b) - cos a sin b (sin a cos b + cos a sin b)

open the bracket

sin a cos b(sin a cos b) + sin a cos b(cos a sin b) -cos a sin b (sin a cos b)+ cos a sin b ( cos a sin b)

sin²a cos²b + sin a cos b cos a sin b - cos a sin b sin a cos b + cos²a sin²b

sin²a cos²b  + 0 + cos²a sin²b

sin²a cos²b + cos²a sin²b

(1-cos² a)cos² b - cos² a(1-cos² b)

= cos² b - cos² a cos² b - cos² a +cos² a cos² b

cos² b - cos² a - cos² a cos² b + cos² a cos² b = cos² b - cos² a  + 0

cos² b - cos² a

left hand side equals right hand side

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:

Each packet of orange can make 2 cups of juice and each packets of Apple can make 3 cups of juice.

Step-by-step explanation:

Let number of cups of juice in each packet of orange be 'x'.

Let number of cups of juice in each pack of apples be 'y'.

Given:

On Tuesday;

Number of packets of oranges = 8

Number of packets of apples = 6 packets.

Amount of fruit juice = 34 cups

So we can say that;

Amount of fruit juice is equal to sum of Number of packets of oranges multiplied by number of cups of juice in each packet of orange and Number of packets of apples multiplied by number of cups of juice in each pack of apples.

framing in equation form we get;

[tex]8x+6y=34 \ \ \ \ equation \ 1[/tex]

On Wednesday;

Number of packets of oranges = 14

Number of packets of apples = 6

Amount of fruit juice = 46 cups

So we can say that;

Amount of fruit juice is equal to sum of Number of packets of oranges multiplied by number of cups of juice in each packet of orange and Number of packets of apples multiplied by number of cups of juice in each pack of apples.

framing in equation form we get;

[tex]14x+6y=46 \ \ \ \ equation \ 2[/tex]

Now Subtracting equation 1 from equation we get;

[tex]14x+6y-(8x+6y)=46-34\\\\14x+6y-8x-6y=12\\\\6x=12[/tex]

Now dividing both side by 6 we get;

[tex]\frac{6x}{6}=\frac{12}{6}\\\\x=2\ cups[/tex]

Now we will substitute the value of 'x' in equation 1 we get;

[tex]8x+6y=34\\\\8\times2+6y=34\\\\16+6y=34[/tex]

Subtracting both side by 16 we get;

[tex]16+6y-16=34-16\\\\6y= 18[/tex]

Now dividing both side by 6 we get;

[tex]\frac{6y}{6}=\frac{18}{6}\\\\y=3[/tex]

Hence Each packet of orange can make 2 cups of juice and each packets of Apple can make 3 cups of juice.

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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