Positive-sequence components consist of three phasors with _____ magnitudes and _____ phase displacement in positive sequence; negativesequence components consist of three phasors with _____ magnitudes and _____ phase displacement in negative sequence; and zero-sequence components consist of three phasors with _____ magnitudes and _____ phase displacement.

Answer:

The question is incomplete as some details are missing; Here is the complete question; If vector u = 3i and v = 4i + 4j, Find the measure of the angle θ between u and v. Express the answer in radians rounded to two decimal places, if it is not possible to express it exactly.

The measure of the angle θ between u and v = 0.785radians

Step-by-step explanation:

The detailed steps is as shown in the attachment.

Answer:

a) P=1/116280

b) P=143/38760

c) P=441/2584

Step-by-step explanation:

We have seven of these telephones are cellular, seven are cordless, and the other seven are corded phones.

a)  We calculate the number of possible combinations

{21}_C_{14}=\frac{21!}{14! · (21-14)!}=116280

The number of favorable combinations is 1.

Therefore, the probability is

P=1/116280

b) We calculate the number of possible combinations

{21}_C_{14}=\frac{21!}{14! · (21-14)!}=116280

We calculate  the number of favorable combinations  

{14}_C_{7}=\frac{14!}{7! · (14-7)!}=429

Therefore, the probability is

P=429/116280

P=143/38760

c) We calculate the number of possible combinations

{21}_C_{6}=\frac{21!}{6! · (21-6)!}=54264

We calculate  the number of favorable combinations  

{7}_C_{2} · {7}_C_{2} · {7}_C_{2} =

=\frac{7!}{2!·(7-2)!} · \frac{7!}{2!·(7-2)!}  · \frac{7!}{2!·(7-2)!}

=21 · 21 · 21=9261

Therefore, the probability is

P=9261/54264

P=441/2584

Answer:

336 different predictions.

1/336 probability of making the correct prediction by​ chance

Step-by-step explanation:

The order is important.

For example, Team A, B and C is a different outcome than team B, A, C.

So we use the permutations formula to solve this problem:

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

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

In this problem, we have that:

Permutations of 3 from a set of 8. So

[tex]P_{(8,3)} = \frac{8!}{(8-3)!} = 336[/tex]

What is the probability of making the correct prediction by​ chance?

There are 336 possible outcomes.

By chance, you predict 1.

So there is a 1/336 probability of making the correct prediction by​ chance

Final answer:

You can make 336 different predictions for the top 3 soccer teams out of 8, and the chance of making the correct prediction by chance is approximately 0.298%.

Explanation:

To calculate the number of different predictions you could make for the top 3 teams out of 8, without considering ties, you use permutations since the order matters. The formula for permutations is P(n, r) = n! / (n-r)!, where n is the total number of teams and r is the number of positions to fill.

In this case, n = 8 teams and r = 3 positions. Therefore, the calculation is P(8, 3) = 8! / (8-3)! = 8 x 7 x 6 = 336 different predictions.

To find the probability of making the correct prediction by chance, since there is only one correct prediction out of all possible predictions, the probability is 1 / 336. Thus, the probability is approximately 0.00298, or 0.298%.

Answer:

a) [tex] T(t) = \frac{<-5 sin(5t), 5cos(5t)>}{5}= <-sin(5t), cos(5t)>[/tex]

[tex] T(4) = <-sin(20), cos(20)>[/tex]

b) [tex] T(t) = \frac{<t^2, 3t^2>}{8\sqrt{37}}[/tex]

[tex] T(4) = <\frac{2\sqrt{37}}{37},\frac{6\sqrt{37}}{37} >[/tex]

c) [tex] T(t) = \frac{<5e^{5t}, -4e^{-4t}, 1>}{2425825977}[/tex]

[tex] T(4) = \frac{1}{2425825977}<5e^{50}, -4e^{-16},1 >[/tex]

Step-by-step explanation:

The tangent vector is defined as:

[tex] T(t) = \frac{r'(t)}{|r'(t)|}[/tex]

Part a

For this case we have the following function given:

[tex] r(t) = <cos(5t), sin(5t)>[/tex]

The derivate is given by:

[tex] r'(t) = <-5 sin(5t), 5cos(5t)>[/tex]

The magnitude for the derivate is given by:

[tex] |r'(t)| = \sqrt{25 sin^2(5t) +25 cos^2 (5t)}= 5\sqrt{cos^2 (5t) + sin^2 (5t)} =5[/tex]

And then the tangent vector for this case would be:

[tex] T(t) = \frac{<-5 sin(5t), 5cos(5t)>}{5}= <-sin(5t), cos(5t)>[/tex]

And for the case when t=4 we got:

[tex] T(4) = <-sin(20), cos(20)>[/tex]

Part b

For this case we have the following function given:

[tex] r(t) = <t^2, t^3>[/tex]

The derivate is given by:

[tex] r'(t) = <2t, 3t^2>[/tex]

The magnitude for the derivate is given by:

[tex] |r'(t)| = \sqrt{4t^2 +9t^4}= t\sqrt{4 + 9t^2} [/tex]

[tex] |r'(4)| = \sqrt{4(4)^2 +9(4)^4}= 4\sqrt{4 + 9(4)^2} = 4\sqrt{148}= 8\sqrt{37}[/tex]

And then the tangent vector for this case would be:

[tex] T(t) = \frac{<t^2, 3t^2>}{8\sqrt{37}}[/tex]

And for the case when t=4 we got:

[tex] T(4) = <\frac{2\sqrt{37}}{37},\frac{6\sqrt{37}}{37} >[/tex]

Part c

For this case we have the following function given:

[tex] r(t) = <e^{5t}, e^{-4t} ,t>[/tex]

The derivate is given by:

[tex] r'(t) = <5e^{5t}, -4e^{-4t}, 1>[/tex]

The magnitude for the derivate is given by:

[tex] |r'(t)| = \sqrt{25e^{10t} +16e^{-8t} +1} [/tex]

[tex] |r'(t)| = \sqrt{25e^{10*4} +16e^{-8*4} +1} =2425825977 [/tex]

And then the tangent vector for this case would be:

[tex] T(t) = \frac{<5e^{5t}, -4e^{-4t}, 1>}{2425825977}[/tex]

And for the case when t=4 we got:

[tex] T(4) = \frac{1}{2425825977}<5e^{50}, -4e^{-16},1 >[/tex]

To compute the unit tangent vector T(t) for the given position vector r(t) at a given value t, take the derivative of r(t) with respect to t and divide the resulting vector by its magnitude.

To compute the unit tangent vector T(t) for the given position vector r(t) at a given value t, we need to take the derivative of r(t) with respect to t and then divide the resulting vector by its magnitude.

For part A, r(t) = (cos(5t), sin(5t)), so r'(t) = (-5sin(5t), 5cos(5t)). Plugging in t = -4, we get r'(-4) = (-5sin(-20), 5cos(-20)). To find T(-4), we divide r'(-4) by its magnitude.

T(-4) = (-5sin(-20)/sqrt((-5sin(-20))^2 + (5cos(-20))^2), 5cos(-20)/sqrt((-5sin(-20))^2 + (5cos(-20))^2)).

Answer: 126 orderings

Step-by-step explanation:

Here let's suppose that total devices is given by n= 12

and defective ones are given by m= 4

Now, to find the number of orderings in which no two defective devices are consecutive is given by following relation.

¹²⁻⁴⁺¹₄C

= ⁹₄C

= 126 orderings

Answer:

The required system of equations to the given parametric equations are:

5x1 - x2 = 6

x1 + x2 = -6

Step-by-step explanation:

Given the parametric equations:

x1 = t

x2 = -6 + 5t

Eliminating the parameter t, we obtain one of the equations of a system in two variables, x1 and x2 that has the solution set given by the parametric equations.

Doing that, we have:

5x1 - x2 = 6

Again a second equation can be a linear combination of x1 and x2

x1 + x2 = -6 + 6t

x1 + x2 = -6 (putting t=0)

And they are the required equations.

Answer:

Part a: [tex]f , \, f_y[/tex] is continuous at the initial value (0,0) so due to Picardi theorem there exists an interval such that the IVP has a unique solution.

Part b: [tex]f_y[/tex] is not continuous at the initial value (0,0) so due to Picardi theorem there does not exist an interval such that the IVP has a unique solution.

part c: [tex]f , \, f_y[/tex] is continuous at the initial value (0,1) so due to Picardi theorem there exists an interval such that the IVP has a unique solution.

Step-by-step explanation:

Part a

as [tex]y^{' }=ty^{4/3}[/tex]

Let

[tex]f(t,y)=ty^{4/3}[/tex]

Now derivative wrt y is given as

[tex]f_y=\frac{4}{3}ty^{1/3}[/tex]

Finding continuity via the initial value

[tex]f[/tex] is continuous on [tex]R^2[/tex] also [tex]f_y[/tex] is also continuous on [tex]R^2[/tex]

Also

[tex]f , \, f_y[/tex] is continuous at the initial value (0,0) so due to Picardi theorem there exists an interval such that the IVP has a unique solution.

Part b

as [tex]y^{' }=ty^{1/3}[/tex]

Let

[tex]f(t,y)=ty^{1/3}[/tex]

Now derivative wrt y is given as

[tex]f_y=\frac{1}{3}ty^{-2/3}[/tex]

Finding continuity via the initial value

[tex]f[/tex] is continuous on [tex]R^2[/tex] also [tex]f_y[/tex] is also continuous on [tex]R^2[/tex]

Also

[tex]f_y[/tex] is not continuous at the initial value (0,0) so due to Picardi theorem there does not exist an interval such that the IVP has a unique solution.

Part c

as [tex]y^{' }=ty^{1/3}[/tex]

Let

[tex]f(t,y)=ty^{1/3}[/tex]

Now derivative wrt y is given as

[tex]f_y=\frac{1}{3}ty^{-2/3}[/tex]

Finding continuity via the initial value

[tex]f[/tex] is continuous on [tex]R^2[/tex] also [tex]f_y[/tex] is also continuous on [tex]R^2[/tex] when [tex]y\neq 0[/tex]

Also

[tex]f , \, f_y[/tex] is continuous at the initial value (0,1) so due to Picardi theorem there exists an interval such that the IVP has a unique solution.

Answer and Step-by-step explanation:

a) The press release uses some weird relative risk method to arrive at this value.

P(B) = 0.847, Probability that a black is safe from being referred for cardiac carthetirization, P(B') = 1 - 0.847 = 0.153

P(W) = 0.907, P(W') = 1 - 0.907 = 0.094

The press release's relative risk = (0.847/0.153)/(0.907/0.094) = 0.574 = 57.4%

b) This is interpretation for relative risk, not the odds ratio. The actual relative risk is

(0.847/0.906) = 0.935: i.e., 60% should have been 93.5%.

Hope this helps!

Answer:

For this case the strongest linear association is given by the greatest correlation coeffcient in absolute value from the list provided. We have:

[tex] |r_3|>|r_2| > |r_4| > |r_1|[/tex]

So on this case we can conclude that the strongest linear association with number of wins is for runs allowed.

Step-by-step explanation:

Previous concepts

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

And in order to calculate the correlation coefficient we can use this formula:  

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]  

Solution to the problem

For this case we have a list of correlation coefficients given:

[tex] r_1 = 0.51[/tex] represent the correlation between number of wins and shutouts

[tex] r_2 = 0.61[/tex] represent the correlation between number of wins and hits made

[tex] r_3 = -0.7[/tex] represent the correlation between number of wins and runs allowed

[tex] r_4 = -0.56[/tex] represent the correlation between number of wins and homeruns allowed

When we analyze linear association we are interested just in the absolute value for r since if r is near to +1 we have positive linear association but on the case that r is near to -1 we have an strong linear association but inversely proportional.

For this case the strongest linear association is given by the greatest correlation coeffcient in absolute value from the list provided. We have:

[tex] |r_3|>|r_2| > |r_4| > |r_1|[/tex]

So on this case we can conclude that the strongest linear association with number of wins is for runs allowed.

In relation to the number of wins for 14 baseball teams, the variable 'runs allowed' holds the strongest linear association. This is represented by its correlation coefficient of -0.70, indicating a strong inverse relationship. As 'runs allowed' increase, the 'number of wins' decrease.

In context of these 14 baseball teams, correlations are being determined with the number of wins in the regular season and four variables: shutouts, hits made, runs allowed, and homeruns allowed. The correlation coefficient represents the strength and direction of a linear relationship between two variables. Coefficients close to +1 or -1 indicate a strong linear association, while those near 0 suggest a weak association. The sign of the correlation indicates the direction of the relationship, either positive or negative.

The variable with the strongest linear association with the number of wins is 'runs allowed', which bears a correlation of -0.70. This implies a strong inverse relationship where as 'runs allowed' increase, the 'number of wins' tends to decrease.

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The vector represented by the directed line segment with initial point A(1, −4, 1) and terminal point B(−2, 5, 4) is (-3, 9, 3).

To find the vector represented by the directed line segment with initial point A(1, −4, 1) and terminal point B(−2, 5, 4), we subtract the coordinates of A from the coordinates of B. Subtraction of vectors is equivalent to adding a negative vector, so we have:

AB = B - A = (-2 - 1, 5 - (-4), 4 - 1) = (-3, 9, 3)

The vector AB is represented as (-3, 9, 3).

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Answer: Convenience sample.

Step-by-step explanation:

Convenience sample is also known as grab or accident or opportunity samples. It is a example of non probability sample that involves selecting of subjects because of the proximity, convenience and accessibility a researcher as to them. This type of samples are not reliable for data gathering when it involves a very large sample space, let's say a global audience.

Answer:

OUR ANSWER IS =0.0467  

Step-by-step explanation:

P(detect/defect)=0.98

P(detect/not defect)=0.01

P(defect)=1/2000=0.0005

P(not defect)=1-0.0005=0.9995

P(defect/detect)=P(detect/defect)*P(defect)/[P(detect/defect)*P(defect)+P(detect/not defect)*P(not defect)]

=0.98*0.0005/[0.98*0.0005+0.01*0.9995]

=0.00049/[0.00049+0.009995]

=0.00049/0.010485

=0.0467

Answer:

a. 0.35197 or 35.20%; b. 0.1230 or 12.30%; c. 0.48784 or 48.78%; d. $250.20 or more.

Step-by-step explanation:

In general, we can solve this question using the standard normal distribution, whose values are valid for any normally distributed data, provided that they are previously transformed to z-scores. After having these z-scores, we can consult the table to finally obtain the probability associated with that value. Likewise, for a given probability, we can find, using the same table, the z-score associated to solve the value x of the equation for the formula of z-scores.

We know that the room rates are normally distributed with a population mean and a population standard deviation of (according to the cited source in the question):

[tex] \\ \mu = \$204[/tex] (population mean)

[tex] \\ \sigma = \$55[/tex] (population standard deviation)

A z-score is the needed value to consult the standard normal table. It is a transformation of the data so that we can consult this standard normal table to obtain the probabilities associated. The standard normal table has a mean  of 0 and a standard deviation of 1.

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

After having all this information, we can proceed as follows:

1. We need to calculate the z-score associated with x = $225.

[tex] \\ z_{score}=\frac{225-204}{55}[/tex]

[tex] \\ z_{score}=0.381818[/tex]

[tex] \\ z_{score}=0.38[/tex]

We rounded the value to two decimals since the cumulative standard normal table (values for cumulative probabilities from negative infinity to the value x) to consult only have until two decimals for z values.

Then

2. For a z = 0.38, the corresponding probability is P(z<0.38) = 0.64803. But the question is asking for values greater than this value, then:

[tex] \\ P(z>038) = 1 - P(z<0.38)[/tex] (that is, the complement of the area)

[tex] \\ P(z>038) = 1 - 0.64803[/tex]

[tex] \\ P(z>038) = 0.35197[/tex]

So, the probability that a hotel room costs $225 or more per night is P(x>$225) = 0.35197 or 35.20%, approximately.

We follow a similar procedure as before, so:

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

[tex] \\ z_{score}=\frac{140-204}{55}[/tex]

[tex] \\ z_{score}=\frac{140-204}{55}[/tex]

[tex] \\ z_{score}= -1.163636 \approx -1.16[/tex]

This value is below the mean (it has a negative sign). The standard normal tables does not have these values. However, we can find them subtracting the value of the probability obtained for z = 1.16 from 1, since the symmetry for normal distribution permits it. Then, the probability associated with z = -1.16 is:

[tex] \\ P(z<1.16) = 0.87698[/tex]

[tex] \\ P(z<-1.16) = 1 - 0.87698 [/tex]

[tex] \\ P(z<-1.16) = 0.12302 \approx 0.1230[/tex]

Then, the probability that a hotel room costs less than $140 per night is P(x<$140) = 0.1230 or 12.30%.

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

The z-score and probability for x = $200:

[tex] \\ z_{score}=\frac{200-204}{55}[/tex]

[tex] \\ z_{score}= -0.072727 \approx -0.07[/tex]

[tex] \\ P(z<0.07) = 0.52790[/tex]

[tex] \\ P(z<-0.07) = 1 - 0.52790 [/tex]

[tex] \\ P(z<-0.07) = 0.47210 \approx 0.4721[/tex]

The z-score and probability for x = $300:

[tex] \\ z_{score}=\frac{300-204}{55}[/tex]

[tex] \\ z_{score}=1.745454[/tex]

[tex] \\ P(z<1.75) = 0.95994[/tex]

[tex] \\ P(z<1.75) - P(z<-0.07) = 0.95994-0.47210 [/tex]

[tex] \\ P(z<1.75) - P(z<-0.07) = 0.48784 [/tex]

Then, the probability that a hotel room costs between $200 and $300 per night is 0.48784 or 48.78%.

A way to solve this is as follows: we need to consult, using the cumulative standard normal table, the value for z such as the probability is 80%. This value is, approximately, z = 0.84. Then, solving the next equation for x:

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

[tex] \\ 0.84=\frac{x-204}{55}[/tex]

[tex] \\ 0.84*55=x-204[/tex]

[tex] \\ 0.84*55 + 204 =x[/tex]

[tex] \\ x = 250.2[/tex]

That is, the cost of the most expensive 20% of hotel rooms in New York City are of $250.20 or more.

This answer uses the normal distribution to calculate probabilities and thresholds in hotel room rates. The main steps are to convert room rates to z-scores, look up probabilities or percentiles in the standard normal distribution, and convert back to room rates if necessary.

To answer these questions, we can use the properties of the normal distribution. First, we convert the given hotel room rate to a z-score by subtracting the mean and dividing by the standard deviation. Then we can look up these z-scores in a standard normal distribution table (or use a calculator with a normal distribution function) to get probabilities.

a. To find the probability that a hotel room costs $225 or more per night, we convert $225 to a z-score: z = ($225-$204)/$55 = 0.38. The probability of getting a z-score of 0.38 or more (which means a cost of $225 or more) is 0.3520. So the probability that a room costs $225 or more per night is 0.3520 or 35.20%.

b. To find the probability that a hotel room costs less than $140 per night, we convert $140 to a z-score: z = ($140-$204)/$55 = -1.16. The probability of getting a z-score of -1.16 or less (which means a cost of $140 or less) is 0.1230. So the probability that a room costs $140 or less per night is 0.1230 or 12.30%.

c. To find the probability that a hotel room costs between $200 and $300 per night, we convert $200 to a z-score (z1 = -0.073) and $300 to a z-score (z2 = 1.745). We then subtract the probabilities: P($200 < Rate < $300) = P(z1 < Z < z2) = P(Z < z2) - P(Z < z1) = 0.9591 - 0.4713 = 0.4878. So the probability that a room costs between $200 and $300 per night is 0.4878 or 48.78%.

d. To find the cost of the most expensive 20% of hotel rooms in New York City, we look for the z-score corresponding to the 80th percentile (because everything above this point is the top 20%). This z-score is 0.84. We then convert this z-score back to a cost using the mean and standard deviation: Cost = Mean + z*SD = $204 + 0.84*$55 = $250.20. So the cost of the most expensive 20% of hotel rooms in New York City is $250.20 per night.

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Answer:

The number of different lab groups possible is 84.

Step-by-step explanation:

Given:

A class consists of 5 engineers and 4 non-engineers.

A lab groups of 3 are to be formed of these 9 students.

The problem can be solved using combinations.

Combinations is the number of ways to select k items from a group of n items without replacement. The order of the arrangement does not matter in combinations.

The combination of k items from n items is: [tex]{n\choose k}=\frac{n!}{k!(n-k)!}[/tex]

Compute the number of different lab groups possible as follows:

The number of ways of selecting 3 students from 9 is = [tex]{n\choose k}={9\choose 3}[/tex]

                                                                                         [tex]=\frac{9!}{3!(9 - 3)!}\\=\frac{9!}{3!\times 6!}\\=\frac{362880}{6\times720}\\ =84[/tex]

Thus, the number of different lab groups possible is 84.

Answer:

[tex]4 - 2.14 \frac{0.7}{\sqrt{15}}=3.61[/tex]

[tex]4 + 2.14 \frac{0.7}{\sqrt{15}}=4.39[/tex]

The 95% confidence interval is given by (3.61;4.39)

Step-by-step explanation:

Notation and definitions

n=15 represent the sample size

[tex]\bar X=4[/tex] represent the sample mean  

[tex]s=0.7[/tex] represent the sample standard deviation

m represent the margin of error

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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".  

Calculate the critical value tc

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. The degrees of freedom are given by:  

[tex]df=n-1=15-1=14[/tex]  

We can find the critical values in excel using the following formulas:  

"=T.INV(0.025,14)" for [tex]t_{\alpha/2}=-2.14[/tex]  

"=T.INV(1-0.025,14)" for [tex]t_{1-\alpha/2}=2.14[/tex]  

The critical value [tex]tc=\pm 2.14[/tex]

Calculate the margin of error (m)

The margin of error for the sample mean is given by this formula:

[tex]m=t_c \frac{s}{\sqrt{n}}[/tex]

[tex]m=2.14 \frac{0.7}{\sqrt{15}}=0.387[/tex]

Calculate the confidence interval  

The interval for the mean is given by this formula:

[tex]\bar X \pm t_{c} \frac{s}{\sqrt{n}}[/tex]

And calculating the limits we got:

[tex]4 - 2.14 \frac{0.7}{\sqrt{15}}=3.61[/tex]

[tex]4 + 2.14 \frac{0.7}{\sqrt{15}}=4.39[/tex]

The 95% confidence interval is given by (3.61;4.39)

The formula for the present value (PV) of a perpetuity is \[PV = \frac{FV}{(1 + r)^n}\]. Here option C is correct.

The formula for calculating the present value (PV) of a perpetuity is given by:

\[PV = \frac{FV}{(1 + r)^n}\]

Where:

PV (Present Value) is what we want to find.

FV (Future Value) is the fixed payment that will continue indefinitely.

r (Discount Rate) represents the interest rate or required rate of return.

n represents the number of time periods (infinite in the case of a perpetuity).

This formula takes into account the infinite nature of the perpetuity and discounts future cash flows to their equivalent value in today's dollars, considering the time value of money. The discount factor \(\frac{1}{(1 + r)^n}\) ensures that the cash flows in the future are worth less in present terms. Therefore, option C is correct.

Complete question:

Which of the following equations can be used to solve for the present value (PV) of a perpetuity?

A) \(PV = PMT \cdot \left(1 - \frac{1}{{(1 + r)^{n(1+r)}}}\right)\)

B) \(PV = PV_0 \cdot (1 + r)^n\)

C) \(PV = \frac{FV}{{(1 + r)^n}}\)

D) \(PV = PMT \cdot \frac{1 - \left(\frac{1}{{(1 + r)^{n(1+r)}}}\right)}{r}\)

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Answer: $100

good luck!!

Answer:

The correct option is d i.e. scatter plot

Step-by-step explanation:

The correct option is d i.e. scatter plot

scatter plot will be the best option to display the variation of expenditure with respect to annual income.

on one axis annual income is used and on the other expenditure of the family. hence for a particular change in annual income, an impact on expenditure will easily be predicted.

A scatterplot would be a meaningful graphical display for studying the relationship between annual income and entertainment spending.

A meaningful graphical display of the relationship between the annual income of a family and the amount of money the family spends on entertainment would be a scatterplot.

A scatterplot shows the relationship between two variables by plotting each data point as a dot on a graph. In this case, the x-axis would represent the annual income and the y-axis would represent the amount spent on entertainment. Each dot on the scatterplot would represent a family and its corresponding values for income and entertainment spending.

By examining the scatterplot, it can be determined whether there is a correlation between income and entertainment spending. For example, if most dots are clustered around a certain line or pattern, it suggests a relationship between the two variables.

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The exact solution to given differential equation is y(x) = Ce^(x^2/2). Using Euler's method with step size of 0.2, the solution approximates to 1.696 at x = 1.4; while the exact solution gives y(1.4) ~= 7.63.

The differential equation given is of the form dy/dx = x*y. The exact solution to this equation can be found using the method of separation of variables. Integrating both sides, we get y(x) = Ce^(x^2/2) where C is the integration constant. Given that y(0.2) = 7, we can find C = 7*exp(-0.02).

Moving on to Euler's method, the next stage can be found via the algorithm x(n+1) = x(n) + h and y(n+1) = y(n) + h*f(x(n), y(n)). The function f(x,y) = x / y is provided. Completing the table for x = 1, y can be calculated as y(1) = y(0.8) + h*f(0.8, y(0.8)) = 1.22814 + 0.2(0.8/1.22814) = 1.3924. Following the same steps until x = 1.4 gives y(1.4) ~= 1.696.

In part 2), with x = 1.4 in the exact solution we found earlier, y(1.4) = 7*exp(-0.02) * exp(1.4^2/2) ~= 7.63.

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Answer:

[tex]Z = -1.63[/tex]

Step-by-step explanation:

The z-score measures how many standard deviation a score X is above or below the mean.

it is given by the following formula:

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

In which

[tex]\mu[/tex] is the mean, [tex]\sigma[/tex] is the standard deviation

In this problem, we have that:

Mean weight of all students in a class is 165 pounds, so [tex]\mu = 165[/tex]

Variance of 234.09 square pounds. The standard deviation is the square root of the variance. So [tex]\sigma = \sqrt{234.09} = 15.3[/tex]

What is the z-value associated with a student whose weight is 140 pounds?

Z when X = 140. So

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

[tex]Z = \frac{140 - 165}{15.3}[/tex]

[tex]Z = -1.63[/tex]

To find the z-value associated with a student whose weight is 140 pounds, use the formula for calculating z-score. Substitute the values of mean, standard deviation, and the student's weight into the formula.

To find the z-value associated with a student whose weight is 140 pounds, we need to use the formula for calculating z-score:

z = (x - μ) / σ

The z-value associated with a student whose weight is 140 pounds can be calculated using the given formula and the information provided.

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Answer:

(a) 0.56

(b) 0.06

(c) 0.94

Step-by-step explanation:

P(E1) = 0.7 and P(E2) = 0.8

(a) The probability that both bids are successful is given by the product of the probability of success of each bid:

[tex]P(E1\ and\ E2) = 0.7*0.8=0.56[/tex]

(b) The probability that neither bid is successful is given by the product of the probability of failure of each bid:

[tex]P(not\ E1\ and\ not\ E2)= (1-P(E1))*(1-P(E2))\\P(not\ E1\ and\ not\ E2)=0.3*0.2=0.06[/tex]

(c) The probability that the firm is successful in at least one of the two bids is given by the sum of the probability of success of each bid subtracted by the probability that both bids are successful:

[tex]P(E1\ or\ E2)=P(E1)+P(E2) - P(E1\ and\ E2)\\P(E1\ or\ E2)=0.7+0.8-0.56\\P(E1\ or\ E2)=0.94[/tex]

Using probability concepts, it is found that there is a:

a) 0.56 = 56% probability that both bids are successful.

b) 0.06 = 6% probability that neither bid is successful.

c) 0.94 = 94% probability that the firm is successful in at least one of the two bids.

Item a:

These two events are independent, hence, the probability of both is the multiplication of the probabilities of each, thus:

[tex]p = 0.7(0.8) = 0.56[/tex]

0.56 = 56% probability that both bids are successful.

Item b:

E1 has a 1 - 0.7 = 0.3 probability of being unsuccessful, while E2 has a 0.2 probability, hence:

[tex]p = 0.3(0.2) = 0.06[/tex]

0.06 = 6% probability that neither bid is successful.

Item c:

1 - 0.06 = 0.94

0.94 = 94% probability that the firm is successful in at least one of the two bids.

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Answer:

8 < x < 40

Step-by-step explanation:

x − 8 must be more than 0, but it can't be greater than 32.

0 < x − 8 < 32

8 < x < 40

A more precise answer would require law of cosines and calculus.

Answer:

b

Step-by-step explanation:

a) Census. It would be easy enough to count all of them.

b) Sample. It would be too costly to track each can.

c) Census. You can count them all quickly and cheaply.

The sample space for a given set of events is the set of all possible values the events may assume.

A)  The number of cans of Campbell’s soup on your local supermarket's shelf today at 6:00 p.m.

Census. It would be easy enough to count all of them.

B) The proportion of soup sales last week in Boston that was sold under the Campbell's brand.

Sample. It would be too costly to track each can.

C)  The proportion of Campbell’s brand soup cans in your family's pantry.

Census. You can count them all quickly and cheaply.

Learn more about sample space here:

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Answer:

Obtuse triangle

Step-by-step explanation:

Given are the vertices of a triangle

Let A (-4, 0, 0),B (0, 0, 0),C(7, 2, 6)

Let us find angles between AB, BC and CA

AB = (4, 0,0): BC = (7,2,6) : CA = (11, 2,6)

Cos B = [tex]\frac{AB.BC}{|AB||BC|} \\[/tex]

B = arc cos [tex]\frac{AB.BC}{|AB||BC|} \\[/tex]=137 deg 54 min 7 sec

Similarly

A=29 deg 53 min 53 seconds

C = 12 deg 12 min 3 sec

Obtuse triangle since one angle > 90 degrees

To determine the type of triangle, find the lengths of the sides using the distance formula. Then compare the sum of squares of the two shortest sides with the square of the longest side.

To determine whether a triangle is acute, obtuse, or right, we need to find the lengths of its three sides and then use the Pythagorean theorem. The distance formula can be used to find the lengths of the sides by finding the distances between the given vertices. After finding the lengths, we can compare the sum of the squares of the two shortest sides with the square of the longest side to determine the type of triangle.

Using the distance formula, we find that the lengths of the sides are 4,7, and 9. The shortest side is 4, so we calculate the sum of the squares of 4 and 7, which equals 65. The square of the longest side (9) is 81. Since 65 < 81, the triangle is an acute triangle.

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Answer:

Option 4) Type I error is convicting an innocent person.

Step-by-step explanation:

We are given the following in the question:

[tex]H_{0}: \text{ Innocent}\\H_A: \text{ Guilty}[/tex]

Type II error:

Thus, with the given scenario type II error will be freeing a guilty person.

That is acquitting a guilty person.

Type I error:

Thus, with respect to given scenario type I error is convicting an innocent person.

Thus, the correct answer is:

Option 4) Type I error is convicting an innocent person.

Answer:

0.19

Step-by-step explanation:

The are three candidate running for president and we know that probability of winning for first candidate and the probability of winning for second candidate and we have to find the probability of winning for third candidate

P(C1)=0.37

P(C2)=0.44

P(C3)=?

We know that sum of probabilities is always 1. So,

P(C1)+P(C2)+P(C3)=1

0.37+0.44+P(C3)=1

P(C3)=1-0.37-0.44

P(C3)=0.19

Thus, the probability of winning for third candidate is 0.19.

Answer:

a. Exercise is a suppressor.

Step-by-step explanation:

A variable that diminishes the effect of another variable on the outcome is known as a suppressor. In this particular experiment, when exercise is introduced, it was observed that when exercise was introduced, there was a decrease on the effect of sugar intake on weight. Therefore, exercise is a suppressor.

Answer:

D

Step-by-step explanation:

The population consists of all characteristics of interest and sample is portion or a subset of population. Here, 1502 adults 18 years or older are select ted from a population of all adults 18 years or older, so, 1502 adults are the sample. The measurement taken from sample is termed as statistic. The given value 69% is computed from a sample and thus it is a sample statistic.

Answer:

z-score=0.385

(See attached picture)

Step-by-step explanation:

The procedure to find the z-score will depend on the resources we have available. I have a table with the area between the mean and the value we wish to normalize, so the very first thing we need to do is precisely find this area we need to analyze.

Everything to the left of thte mean will represent 50% of the data, so we start by subtracting:

50%-35%=15%

so we need to look in the table for the value 0.15.

In my table I can see that for an area of 0.15, the z-score will be between 0.38 (z-score of 0.1480) and 0.39 (z-score of 0.1517).

By doing some interpolation, you can determine a more accurate value of the z-score to be 0.385.

The z-score corresponding to an area of 0.3500 in a standard normal distribution is approximately -0.39. This z-value indicates that the data point is 0.39 standard deviations below the mean.

In a standard Normal distribution, the z-score is equivalent to the number of standard deviations a given data point is from the mean. If you know the area to the left of the z-score (which in this case is 0.3500), you can use a z-score table (also known as a standard normal table) to find the corresponding z-score.

Normally, the z-score table gives the area to the left of the score. However, in this case, the value (0.3500) does not appear in the body of the z-score table because it corresponds to a negative z-score (since 0.3500 < 0.5). Thus, we will first find the equivalent positive area (1- 0.3500 = 0.6500) and look up that value in the z-score table. The value 0.6500 corresponds approximately to a z-score of 0.39. Since our original question gives an area less than 0.5 (indicating a z-score below the mean) the z-score is -0.39.

Please note that for illustrating the Normal curve, any standard statistics textbook or online resource will have a diagram illustrating the curve, with a vertical line indicating the z-score (in this case -0.39) and shading demonstrating the area to the left of the z-score. These images typically aren't included in text-based tutoring platforms.

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