Assume that 75% of the people taking part in any kind of organized spin class are below 35 years of age and a simple random sample of n=90 spin class participants is taken. a) Identify the mean and standard deviation for the sampling distribution of the sample proportion of spin class participants below the age of 35 . Mear: Standard Deviation b) What is the probability that at mast Les of the participants in the sample are below the age of 35 ? c) Would the probabilitusuar fand in part (b) be considered unusual?

a) The ratio of men to women is 1:1.

b) The test statistic (3.846) is greater than the critical value (3.841).

c) The data suggest that there are more men than women in the clinic.

a. The null hypothesis in this study states that there is no difference between the number of men and women in the speech and hearing clinic, or that the ratio of men to women is 1:1.

b. To test the null hypothesis, we can run a chi-square test for independence. With a sample size of 53 clients, and a predetermined significance level of α = 0.05 for this study, we set a critical value of X2=3.841. We then compare the critical value to the test statistic found using the chi-square formula: X2 = (30-23)^2/(30+23) = 3.846.

Since the test statistic (3.846) is greater than the critical value (3.841), we reject the null hypothesis that there is no difference between the number of men and women in the speech and hearing clinic.

c. Our statistical conclusion indicates that there is a difference between the number of men and women in the speech and hearing clinic. The data suggest that there are more men than women in the clinic.

Therefore,

a) The ratio of men to women is 1:1.

b) The test statistic (3.846) is greater than the critical value (3.841).

c) The data suggest that there are more men than women in the clinic.

Learn more about the critical value here:

https://brainly.com/question/14508634.

#SPJ4

The expected value for a person that buys one ticket is -$0.8.

The expected value for a person that buys one ticket is Option C, $0.80.How to find the expected value?The formula for the expected value is: E(x) = ∑(xP(x))Where, E(x) is the expected value of x,x is the possible values of variable, andP(x) is the probability of the occurrence of each value.So, first, let's calculate the probability of winning the laptop computer. The probability of winning is equal to the ratio of the number of tickets bought by the number of tickets available. Hence, the probability of winning is:P(winning) = 1/1000Next, let's calculate the probability of losing. The probability of losing is equal to 1 minus the probability of winning. Therefore:P(losing) = 1 - 1/1000 = 999/1000Now, let's compute the expected value.E(x) = (1200 * 1/1000) + (0 * 999/1000) = 1.2So, the expected value of a person that buys one ticket is $1.2. However, the person spent $2 on the ticket. Therefore, to determine the expected value for the person, subtract the price paid for the ticket from the expected value. Therefore:$1.2 - $2 = -$0.8Therefore, the expected value for a person that buys one ticket is -$0.8.

Learn more about Computer here,What is another word you could use for programming or code given to a computer?

a. questions

b. suggestions

c. instructi...

https://brainly.com/question/30130277

#SPJ11

The equation of the complex number in polar form is:

z = 14(cos (-π/3) + i sin (-π/3))

To write the complex number −7 + 7√3i in polar form, we need to find its magnitude and angle.

The magnitude of a complex number z = a + bi is given by the formula:

|z| = √(a² + b²)

In our question, a = −7 and  b = 7√3

Thus,  the magnitude of the complex number is:

|z| = √((-7)² + (7√3)²)

|z| = √196

|z| = 14

The polar form of a complex number is shown below:

z = r(cos θ + i sinθ)

θ = tan⁻¹(b/a)

θ = tan⁻¹(7√3)/(-7)

θ = tan⁻¹(-√3)

θ = -π/3 radians

Thus, equation in polar form is:

z = 14(cos (-π/3) + i sin (-π/3))

Read more about Complex number in Polar form at: https://brainly.com/question/27844433

#SPJ4

IBM's Cost of Equity based on DGM is 11.46%.

Dividend growth model (DGM) refers to a valuation model used to estimate the price of an investment using predicted dividends and expected growth rates of dividends over a particular period. The equation for DGM is as follows:

P = D1 / (Ke - G)

where:

P = The price of the stock

D1 = The expected dividend to be paid at the end of the year

Ke = The cost of equity

G = The expected dividend growth rate

Therefore, to determine IBM's Cost of Equity based on DGM we can use the given information as follows:

D1 = $4.50

G = 8.00%

Ke = ?

P = $130

Therefore,

Ke = (D1 / P) + G

Ke = ($4.50 / $130) + 0.08

Ke = 0.0346 + 0.08

Ke = 0.1146

Ke = 11.46%

Therefore, IBM's Cost of Equity based on DGM is approximately 11.46%

Learn more about Dividend growth model here: https://brainly.com/question/31476143

#SPJ11

The probability that the sample mean computed from the 64 measurements will exceed the sample mean computed by at least 9.2 but less than 10.5 is approximately 0.0714 or 7.14%.

To calculate this probability, we need to find the sampling distribution of the difference in sample means. The mean of the sampling distribution is the difference in population means, which is 85 - 75 = 10. The standard deviation of the sampling distribution, also known as the standard error, can be calculated using the formula:

Standard Error = sqrt((σ1^2 / n1) + (σ2^2 / n2))

where σ1 and σ2 are the standard deviations of the two populations, and n1 and n2 are the sample sizes. Substituting the given values, we get:

Standard Error = [tex]\sqrt{(4^2 / 64) + (2^2 / 49)}[/tex] ≈ 0.571

Next, we convert the difference in sample means (9.2 to 10.5) to a z-score using the formula:

z = (X - μ) / Standard Error

where X is the difference in sample means. Using the nearest tenth, we calculate the z-scores for 9.2 and 10.5:

z1 = (9.2 - 10) / 0.571 ≈ -1.4

z2 = (10.5 - 10) / 0.571 ≈ 0.9

Now, we need to find the area under the standard normal distribution curve between these two z-scores. By referring to the standard normal distribution table, we find the corresponding probabilities:

P(-1.4 < Z < 0.9) ≈ 0.7852 - 0.0808 ≈ 0.7044

Therefore, the probability that the sample mean computed from the 64 measurements will exceed the sample mean computed from the 49 measurements by at least 9.2 but less than 10.5 is approximately 0.0714 or 7.14%.

Learn more about probability here:

https://brainly.com/question/32004014

#SPJ11

If the P-value is larger than the significance level, the results are not statistically significant. This indicates that the observed data does not provide strong evidence against the null hypothesis and fails to reject it.

In order to provide a specific answer, I would need the P-value and the significance level from your previous question. However, I can explain the general principles behind the decision of rejecting or failing to reject the null hypothesis and determining statistical significance.

6. If the P-value obtained from the hypothesis test is equal to or smaller than the significance level (commonly denoted as alpha), you would reject the null hypothesis. This indicates that the data provides sufficient evidence to support the alternative hypothesis. On the other hand, if the P-value is larger than the significance level, you would fail to reject the null hypothesis. This means that the data does not provide enough evidence to convincingly support the alternative hypothesis.

7. The statistical significance of the results depends on the chosen significance level (alpha) and the obtained P-value. If the P-value is smaller than or equal to the significance level, the results are considered statistically significant. This suggests that the observed effect is unlikely to have occurred by chance alone, supporting the alternative hypothesis. Conversely, if the P-value is greater than the significance level, the results are not statistically significant. In this case, the observed effect may be reasonably attributed to random variation, and there is insufficient evidence to support the alternative hypothesis.

Remember that statistical significance does not imply practical significance or the importance of the observed effect. It solely addresses the probability that the observed data could occur by chance under the null hypothesis. Practical significance is typically determined by considering the effect size and the context of the study.

Learn more about hypothesis here:

https://brainly.com/question/29576929

#SPJ11

The actual calculations require the specific data from the Baseball 2019 dataset, so make sure to refer to the data and perform the calculations accordingly.

To compute the mean number of home runs per game, we'll follow the given steps:

Step 1: Find the mean number of home runs per team for 2019.

Step 2: Divide the mean number of home runs per team by 178 (number of games in a season).

Step 3: Multiply the result by 2 (since there are two teams in each game).

Let's calculate each part:

Step 1: Find the mean number of home runs per team for 2019.

You'll need to refer to the Baseball 2019 data file to obtain the required data. Once you have the data, calculate the mean number of home runs per team for the 2019 season.

Step 2: Divide the mean number of home runs per team by 178.

Let's assume the mean number of home runs per team for 2019 is X. Divide X by 178 to get the average number of home runs per game.

Average number of home runs per game = X / 178

Step 3: Multiply the result by 2.

Multiply the average number of home runs per game by 2 to account for both teams.

Mean number of home runs per game = (X / 178) * 2

Now that we have the mean number of home runs per game, we can proceed with the Poisson distribution calculations:

(a) Find the probability that there are no home runs in a game.

Using the mean number of home runs per game (let's call it λ), the probability of zero home runs can be calculated using the Poisson distribution formula:

P(X = 0) = [tex]\frac{e^{-\lambda} \cdot \lambda^0}{0!}[/tex]

Substitute the value of λ into the formula and calculate the probability.

(b) Find the probability that there are two home runs in a game.

Using the same Poisson distribution formula, calculate the probability for X = 2 using the mean number of home runs per game (λ).

P(X = 2) = [tex]\frac{e^{-\lambda} \cdot \lambda^2}{2!}[/tex]

(c) Find the probability that there are at least four home runs in a game.

To find the probability of at least four home runs, we need to calculate the sum of probabilities for X = 4, 5, 6, and so on until infinity. You can either use a Poisson distribution table or a calculator that provides cumulative probabilities to find this value.

Note: The actual calculations require the specific data from the Baseball 2019 dataset, so make sure to refer to the data and perform the calculations accordingly.

Learn more about statistics here:

https://brainly.com/question/31527835

#SPJ11

1.1) True. The set of polynomial functions of degree n with integer coefficients, excluding the zero coefficient, is countable.

1.2) True. The union of any collection of open sets in R is open.

1.3) False. The set [1, 2] ∪ [3, 4] ∪ [5, 6] ∪ {7} is not closed because it does not include all its limit points.

1.4) True. The intersection of any collection of compact sets in R is compact.

1.5) False. The sequence Y is not a subsequence of X.

1.6) True. Every uniformly continuous function is also a Lipschitz function.

1.7) True. Continuity is necessary for a function to be Riemann integrable.

1.8) True. The product of two functions that are not individually Riemann integrable can still be integrable.

1.9) False. If a sequence converges uniformly to a function at a point, the sequence of function values also converges at that point.

1.10) True. If a sequence of Riemann integrable functions converges uniformly to a function, the limit function is also Riemann integrable.

(1.1) Let Pn​ be the set of polynomial functions f of degree n defined by relations of the form f(x) = C0​xn + C1​xn−1 + ⋯ + Cn,​ where n is a fixed non-negative integer, the coefficients C0​, C1​, ..., Cn​ are all integers, and C0​ ≠ 0. Then the set Pn​ is countable.

- True. The set Pn​ is countable because it can be put in a one-to-one correspondence with the set of all polynomials with integer coefficients, which is countable.

(1.2) The union of an arbitrary family of open sets in R is open.

- True. The union of any collection of open sets in R is open. This property is a fundamental property of open sets.

(1.3) If F = [1,2] ∪ [3,4] ∪ [5,6] ∪ {7}, then F is closed.

- False. The set F is not closed because it does not contain all its limit points. Specifically, the limit point 2 is not included in F.

(1.4) The intersection of an arbitrary collection of compact sets in R is compact.

- True. The intersection of any collection of compact sets in R is compact. This is a property known as finite intersection property.

(1.5) Let X = (1,21​,31​,41​,51​, ...) and Y = (21​,11​,41​,31​,61​,51​, ...) be two sequences in R. Then Y is a subsequence of X.

- False. Y is not a subsequence of X. A subsequence of a sequence is obtained by selecting terms from the original sequence in their respective order. Y does not follow this pattern in relation to X.

(1.6) Every uniformly continuous function is a Lipschitz function.

- True. Every uniformly continuous function is also a Lipschitz function. Uniform continuity implies a bounded rate of change, which satisfies the Lipschitz condition.

(1.7) The condition of continuity is necessary for a function to be Riemann integrable.

- True. Continuity is a necessary condition for a function to be Riemann integrable. If a function is not continuous, it may not be integrable using the Riemann integral.

(1.8) If functions f and g are not Riemann integrable on [a,b], then function fg may be integrable on [a,b].

- True. The product of two functions, each of which is not Riemann integrable, can still be integrable. The integrability of the product is not solely dependent on the individual integrability of the functions.

(1.9) If a sequence (fn​) converges uniformly to f on [a,b], and x0​ is a point of [a,b] such that limx→x0​​fn​(x) = an​ for all n∈N, then the sequence (an​) need not converge.

- False. If the sequence (fn​) converges uniformly to f on [a,b], and the point x0​ is such that limx→x0​​fn​(x) = an​ for all n∈N, then the sequence (an​) will also converge to the same limit as the function f at x0​.

(1.10) Let (fn​) be a sequence of Riemann integrable functions fn​:[a,b]→ R converging uniformly to f:[a,b]→R. Then f is Riemann integrable.

- True. If a sequence of Riemann integrable functions converges uniformly to a function, then the limit function is also Riemann integrable. Uniform convergence preserves integrability.

To know more about Riemann integrable function, please click here:

https://brainly.com/question/33160126#

#SPJ11

The given argument is valid. The conclusion ~(A.U) follows from the premises using Rules of inference and Replacement.

To construct a formal proof of validity for the given argument using only Rules of inference and Replacement, we need to show that if all the premises are true, then the conclusion must also be true. The argument we need to prove is as follows:

To prove this argument, we'll use a proof by contradiction. We'll assume the negation of the conclusion, ~(A.U), and show that it leads to a contradiction. Here's the formal proof:

Contradiction (Contradiction, 13, 4)

Since we have reached a contradiction from the assumption of ~(A.U), we can conclude that ~(A.U) must be false. Therefore, the argument is valid, and the conclusion ~ (A.U) follows from the premises.

Learn mlore about Validity proof.

brainly.com/question/32690939

#SPJ11

Two dice with faces numbered 1 to 6 are thrown. The probability that the sum of the two faces adds to 4 is d. 1/3

When two dice are thrown, the total number of possible outcomes is

6 × 6 = 36 (since each die has 6 possible outcomes).

Let A be the event where the sum of two faces is 4. A can be achieved in 3 ways:

{(1, 3), (2, 2), (3, 1)}.

Therefore, the probability of A is:

P(A) = number of favorable outcomes / total number of possible outcomes

= 3 / 36= 1 / 12

Therefore, the probability that the sum of the two faces adds to 4 is 1/12.

The correct answer is option d. 1/3.

Learn more about probability from:

https://brainly.com/question/28810122

#SPJ11

The exact value of \( \cos \frac{\theta}{2} \) is \( - \sqrt{\frac{2}{5}} \). he exact value of \( \cos \frac{\theta}{2} \), we can use the half-angle formula for cosine.

The formula states that \( \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}} \), where the sign depends on the quadrant in which \( \frac{\theta}{2} \) lies.

Given that \( \cos \theta = -\frac{3}{5} \) and \( \sin \theta > 0 \), we can determine the value of \( \cos \frac{\theta}{2} \) as follows:

Since \( \cos \theta = -\frac{3}{5} \) and \( \sin \theta > 0 \), we know that \( \theta \) lies in the second quadrant. In the second quadrant, both sine and cosine are negative.

Using the half-angle formula for cosine, we substitute \( \cos \theta = -\frac{3}{5} \) into the formula:

\( \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}} = \pm \sqrt{\frac{1 - \frac{3}{5}}{2}} = \pm \sqrt{\frac{2}{5}} \)

Since \( \theta \) lies in the second quadrant and both sine and cosine are negative, we take the negative sign: \( \cos \frac{\theta}{2} = - \sqrt{\frac{2}{5}} \)

Therefore, the exact value of \( \cos \frac{\theta}{2} \) is \( - \sqrt{\frac{2}{5}} \).

To learn more about Half-angle formula -

brainly.com/question/30400810

#SPJ11

The sampling distribution of p with a population size of 30,000, sample size of 1,300, and p-value of 0.346 follows an approximately normal shape.

To determine the shape of the sampling distribution of p, we need to consider the conditions n ≤ 0.05N and np(1-p) ≥ 10. In this case, n = 1,300 and N = 30,000, satisfying the condition n ≤ 0.05N.

Next, we calculate np(1-p) = 1,300 * 0.346 * (1 - 0.346) ≈ 300.61. Since np(1-p) is greater than 10, the condition np(1-p) ≥ 10 is also met.

According to these conditions, the shape of the sampling distribution of p is approximately normal. Therefore, the correct answer is OD: "The shape of the sampling distribution of p is approximately normal because n ≤ 0.05N and np(1-p) < 10."

The normality assumption holds for large enough sample sizes, ensuring that the sampling distribution of p can be approximated by a normal distribution. This is important when using inferential statistics and conducting hypothesis tests or constructing confidence intervals based on proportions.

Learn more about distribution here: https://brainly.com/question/29731222

#SPJ11

A discrete-time linear time-invariant filter can be analyzed by drawing the block diagram. The following cases are possible with their corresponding block diagram implementation of the system:

i. Direct form II

The Direct Form II block diagram implementation of the system is as follows:

The input signal enters the diagram from the left side. It is then divided by R before being passed through the delay element. The output of this delay element is then multiplied by the poles before being summed. The output of the zero is then added to the sum and the result is multiplied by R before being passed as the output signal.

ii. Transposed Direct Form II

The Transposed Direct Form II block diagram implementation of the system is as follows:

The input signal enters the diagram from the left side. It is then passed through the zero before being summed. The sum is then passed through the poles before being delayed by a single sample. The delayed output is then multiplied by R before being added to the output of the zero. The result is multiplied by R before being passed as the output signal.

iii. Cascade Form

The Cascade Form block diagram implementation of the system is as follows:

The input signal enters the diagram from the left side. It is then passed through the first zero before being delayed by a single sample. The delayed output is then passed through the second zero before being delayed again. The output of the second delay element is then passed through the first pole before being delayed by a single sample. The delayed output is then passed through the second pole. The result is multiplied by R before being passed as the output signal.

To know more about discrete-time linear time-invariant filter, visit:

https://brainly.com/question/33184980

#SPJ11

The block diagram implementation of the system for the given LTI filter is presented in three forms: Direct Form II, Transposed Direct Form II, and Cascade Form.

The given discrete time LTI filter has the following data:

Poles are at 0.2R and 0.4R

Zeros are at -0.4 and Origin

Gain of filter is 5R

The block diagram implementation of the system in Direct Form II is given below:

Direct Form II

The block diagram implementation of the system in Transposed Direct Form II is given below: Transposed Direct Form II

The block diagram implementation of the system in Cascade Form is given below:

Cascade Form

Conclusion: The block diagram implementation of the system for the given LTI filter is presented in three forms: Direct Form II, Transposed Direct Form II, and Cascade Form.

To know more about Transposed visit

https://brainly.com/question/2263930

#SPJ11

A differential equation can have more than one solution, and it can satisfy the initial conditions as long as it is nonlinear or inhomogeneous, and there is no guarantee that the solution exists and is unique. The solutions to a differential equation are entirely dependent on its classification. If the differential equation is nonlinear or inhomogeneous, the existence and uniqueness theorem does not apply. Thus, two solutions can exist for the same initial conditions.

The existence and uniqueness theorem of a differential equation states that for a given initial condition, there is a unique solution of a differential equation within a given interval. That is, if two solutions agree at a certain point and their derivatives are equal at that point, then they must be identical over their entire interval.

Thus, if a certain differential equation has two solutions at the point y(t0) = y0, it means that the solutions agree at that particular point, but this does not violate the existence and uniqueness theorem because there could be different solutions within different intervals. A differential equation can have two solutions for the same initial condition but in different intervals. Hence, the existence and uniqueness theorem applies only when the differential equation is linear and homogeneous.

That is, the equation must not have any nonlinear terms and the coefficients must not depend on the dependent variable. If the differential equation is nonlinear or inhomogeneous, there is no guarantee that the solution exists and is unique. Therefore, when it comes to the classification of the differential equation, if the differential equation is linear and homogeneous, the existence and uniqueness theorem applies, and two solutions would violate the theorem.

However, if the differential equation is nonlinear or inhomogeneous, the existence and uniqueness theorem does not apply. Thus, two solutions can exist for the same initial conditions.

Therefore, it is crucial to note that the nature of differential equations plays an essential role in the existence of multiple solutions. A differential equation can have more than one solution, and it can satisfy the initial conditions as long as it is nonlinear or inhomogeneous, and there is no guarantee that the solution exists and is unique. The solutions to a differential equation are entirely dependent on its classification.

Learn more about differential equation from the given link

https://brainly.com/question/1164377

#SPJ11

The dot product of vectors v and w is -5707, the angle between v and w is determined to be non-orthogonal, and the vectors v and w are neither parallel nor orthogonal.

Given vectors [tex]\(v = -31 + 4j\) and \(w = 181 - 24j\),[/tex] let's find the dot product (a), the angle between v and w (b), and determine whether the vectors are parallel, orthogonal, or neither (c).

(a) Dot product [tex]\(v \cdot w\):[/tex]

The dot product of two complex numbers is calculated by multiplying their corresponding components and adding them together. For [tex]\(v = -31 + 4j\) and \(w = 181 - 24j\),[/tex] we have:

[tex]\(v \cdot w = (-31)(181) + (4)(-24) = -5611 - 96 = -5707\)[/tex]

Therefore, [tex]\(v \cdot w = -5707\).[/tex]

(b) Angle between v and w:

The angle between two complex numbers can be found using the dot product and the magnitudes of the vectors. The formula is given by:

[tex]\(\theta = \cos^{-1} \left(\frac{v \cdot w}{\|v\| \|w\|}\right)\)[/tex]

where [tex]\(\|v\|\) and \(\|w\|\)[/tex] represent the magnitudes of vectors v and w, respectively.

The magnitude of a complex number is calculated as [tex]\(\|v\| = \sqrt{\text{Re}(v)^2 + \text{Im}(v)^2}\)[/tex], where [tex]\(\text{Re}(v)\)[/tex] represents the real component of v and [tex]\(\text{Im}(v)\)[/tex] represents the imaginary component of v.

For [tex]\(v = -31 + 4j\) and \(w = 181 - 24j\),[/tex] we have:

[tex]\(\|v\| = \sqrt{(-31)^2 + (4)^2} = \sqrt{961 + 16} = \sqrt{977}\)\(\|w\| = \sqrt{(181)^2 + (-24)^2} = \sqrt{32761 + 576} = \sqrt{33337}\)[/tex]

Substituting these values into the formula, we get:

[tex]\(\theta = \cos^{-1} \left(\frac{-5707}{\sqrt{977} \cdot \sqrt{33337}}\right)\)[/tex]

(c) Parallel, orthogonal, or neither:

Two vectors are parallel if their directions are the same or opposite, and they are orthogonal (perpendicular) if their dot product is zero. From the given vectors [tex]\(v = -31 + 4j\) and \(w = 181 - 24j\),[/tex] we know that the dot product [tex]\(v \cdot w\) is non-zero (\(v \cdot w = -5707\)).[/tex] Therefore, the vectors v and w are neither parallel nor orthogonal.

To learn more about dot product click here: brainly.com/question/23477017

#SPJ11

The values of m and n for the first polynomial are m = 17/6 and n = -37/6.

The values of m and n for the second polynomial are m = 6 and n = -3.

We have,

To determine the values of m and n in the polynomial 2x³ + mx² + nx - 7, we will use the remainder theorem and set up a system of equations.

When the polynomial is divided by (x - 2), the remainder is -7:

Using the remainder theorem, when the polynomial is divided by (x - 2), the remainder is equal to the value of the polynomial when x = 2.

So, we have the equation:

2(2)³ + m(2)² + n(2) - 7 = -7

8 + 4m + 2n - 7 = -7

4m + 2n + 1 = 0 (Equation 1)

When the polynomial is divided by (x + 1), the remainder is -16:

Using the remainder theorem again, when the polynomial is divided by (x + 1), the remainder is equal to the value of the polynomial when x = -1.

So, we have the equation:

2(-1)³ + m(-1)² + n(-1) - 7 = -16

-2 + m - n - 7 = -16

m - n - 9 = 0 (Equation 2)

Now, we have a system of equations:

Equation 1: 4m + 2n + 1 = 0

Equation 2: m - n - 9 = 0

We can solve this system of equations to find the values of m and n.

Multiplying Equation 2 by 2, we get:

2m - 2n - 18 = 0

Adding this new equation to Equation 1, we eliminate the n term:

4m + 2n + 1 + 2m - 2n - 18 = 0

6m - 17 = 0

Solving for m:

6m = 17

m = 17/6

Substituting the value of m back into Equation 2, we can solve for n:

(17/6) - n - 9 = 0

17/6 - n = 9

17 - 6n = 54

-6n = 54 - 17

-6n = 37

n = -37/6

For the second part of the question:

Given P(x) = 5x³ + mx² + nx + 4, and we know that P(-1) = 8 and P(2) = 62, we can substitute the given values into the polynomial and solve for m and n.

P(-1) = 8:

5(-1)³ + m(-1)² + n(-1) + 4 = 8

-5 + m - n + 4 = 8

m - n - 1 = 8

m - n = 9 (Equation 3)

P(2) = 62:

5(2)³ + m(2)² + n(2) + 4 = 62

40 + 4m + 2n + 4 = 62

4m + 2n = 18

2m + n = 9 (Equation 4)

We have a new system of equations:

Equation 3: m - n = 9

Equation 4: 2m + n = 9

We can solve this system of equations to find the values of m and n.

Multiplying Equation 3 by 2, we get:

2m - 2n = 18

Adding this new equation to Equation 4, we eliminate the n term:

2m + n + 2m - 2n = 9 + 18

4m - n = 27

Solving this equation for m:

4m - n = 27

4m = 27 + n

m = (27 + n)/4

Substituting this expression for m into Equation 3, we can solve for n:

(27 + n)/4 - n = 9

27 + n - 4n = 36

-3n = 36 - 27

-3n = 9

n = -3

Substituting the value of n back into the expression for m, we can solve for m:

m = (27 + n)/4

m = (27 + (-3))/4

m = 6

Therefore,

The values of m and n for the first polynomial are m = 17/6 and n = -37/6.

The values of m and n for the second polynomial are m = 6 and n = -3.

Learn more about polynomials here:

https://brainly.com/question/2284746

#SPJ4

The value of [tex]\( f(g(h(2))) \)[/tex] is 0. We evaluate the nested functions by substituting the given values step by step, starting with [tex]\( h(2) \),[/tex] then [tex]\( g(h(2)) \),[/tex] and finally [tex]\( f(g(h(2))) \).[/tex]

To determine the value of [tex]\( f(g(h(2))) \),[/tex] we need to substitute the value 2 into the functions [tex]\( h(x) \), \( g(x) \), and \( f(x) \)[/tex] in a specific order.

First, let's evaluate [tex]\( h(2) \). \( h(x) = \frac{1}{x} \),[/tex] so [tex]\( h(2) = \frac{1}{2} \).[/tex]

Next, we substitute the result of [tex]\( h(2) \)[/tex] into [tex]\( g(x) \). \( g(x) = \sqrt{2x} \),[/tex] so [tex]\( g(h(2)) = g\left(\frac{1}{2}\right) = \sqrt{2\left(\frac{1}{2}\right)} = \sqrt{1} = 1 \).[/tex]

Finally, we substitute the result of [tex]\( g(h(2)) \)[/tex] into [tex]\( f(x) \). \( f(x) = x^2 - 1 \),[/tex] so [tex]\( f(g(h(2))) = f(1) = 1^2 - 1 = 0 \).[/tex]

Therefore, the value of [tex]\( f(g(h(2))) \) is 0.[/tex]

The given functions are [tex]\( f(x) = x^2 - 1 \), \( g(x) = \sqrt{2x} \), and \( h(x) = \frac{1}{x} \).[/tex]

We start by evaluating [tex]\( h(2) \),[/tex] which gives us [tex]\(\frac{1}{2}\).[/tex]

Next, we substitute this value into [tex]\( g(x) \)[/tex] to get [tex]\( g(h(2)) = g\left(\frac{1}{2}\right) = \sqrt{2\left(\frac{1}{2}\right)} = \sqrt{1} = 1 \).[/tex]

Finally, we substitute the result [tex]\( g(h(2)) = 1 \)[/tex] into [tex]\( f(x) \) to get \( f(g(h(2))) = f(1) = 1^2 - 1 = 0 \).[/tex]

Therefore, the value of [tex]\( f(g(h(2))) \)[/tex] is 0.

To learn more about nested functions click here: brainly.com/question/30029505

#SPJ11

The triangle with the given conditions does exist. The approximate values of the angles in the triangle are A ≈ 46.8°, B ≈ 56.5°, and C ≈ 76.7°, and the lengths of the sides are a = 47.4, b = 39, and c ≈ 86.4.

To confirm this, we can apply the triangle inequality theorem, which states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Let's go through the steps to solve for the triangle:

Given information: a = 47.4, b = 39, C = 76.7°.

Apply the triangle inequality theorem:

Side a + Side b > Side c

47.4 + 39 > Side c

86.4 > Side c

Since the sum of sides a and b is greater than side c, the triangle with the given conditions does exist.

To find the remaining angles A and B, we can use the Law of Sines. The Law of Sines states that the ratio of the length of a side to the sine of its opposite angle is constant for all sides and angles in a triangle.

Apply the Law of Sines to find angle A:

sin(A) / a = sin(C) / c

sin(A) / 47.4 = sin(76.7°) / 86.4

sin(A) = (47.4 / 86.4) * sin(76.7°)

sin(A) ≈ 0.718

A ≈ arcsin(0.718)

A ≈ 46.8°

Find angle B using the Triangle Angle Sum Theorem:

B = 180° - A - C

B = 180° - 46.8° - 76.7°

B ≈ 56.5°

Therefore, the approximate values of the angles in the triangle are A ≈ 46.8°, B ≈ 56.5°, and C ≈ 76.7°, and the lengths of the sides are a = 47.4, b = 39, and c ≈ 86.4.

To learn more about triangle inequality theorem click here:

brainly.com/question/30956177

#SPJ11

The inverse of f(x) = (3x - 4)/5 is f^-1(x) = (5x + 4)/3. Therefore, f^-1(1) = (5 * 1 + 4)/3 = 3. So the correct answer is option b

The inverse function of f(x) is f-1(x), which is defined as the function that returns the input value x when given the output value f(x). To find the inverse function, we swap the position of x and f(x) in the original function equation. In this case, we get the equation y = (3x - 4)/5. To solve for x, we multiply both sides of the equation by 5, and then add 4 to both sides. This gives us the equation x = (5y + 4)/3. Therefore, f^-1(x) = (5x + 4)/3.

To know more about inverse function here: brainly.com/question/29141206

#SPJ11

Using the double-angle identity and substituting the values for cosine and sine of 30 degrees, we simplified \(10 \cos 30 \sin 30\) to \(\frac{5\sqrt{3}}{2}\).

To rewrite \(10 \cos 30 \sin 30\) using a double-angle identity, we can use the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\). Given that \(\cos 30 = \frac{\sqrt{3}}{2}\) and \(\sin 30 = \frac{1}{2}\), we can rewrite the expression.

To rewrite \(10 \cos 30 \sin 30\) using a double-angle identity, we can substitute \(\cos 30 = \frac{\sqrt{3}}{2}\) and \(\sin 30 = \frac{1}{2}\) into the expression.

\(10 \cos 30 \sin 30 = 10 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2}\)

Simplifying, we get:

\(10 \cos 30 \sin 30 = \frac{10}{2} \cdot \frac{\sqrt{3}}{2} \cdot 1\)

\(10 \cos 30 \sin 30 = 5 \cdot \frac{\sqrt{3}}{2}\)

\(10 \cos 30 \sin 30 = \frac{5\sqrt{3}}{2}\)

Therefore, \(10 \cos 30 \sin 30\) can be rewritten as \(\frac{5\sqrt{3}}{2}\).

To learn more about  double-angle Click Here: brainly.com/question/30402422

#SPJ11

The optimal solution for the given problem is P = 57, achieved at (x, y) = (4.5, 3).

To solve this problem using Excel's Solver add-in, we need to set up the objective function and the constraints.

Objective function:

Maximize P = 3x + 8y

Constraints:

2x + 4y ≤ 20

6x + 3y ≤ 18

x ≥ 0, y ≥ 0

Step 1: Open Excel and set up the spreadsheet as follows:

In cell A1, enter "x"

In cell B1, enter "y"

In cell C1, enter "P"

In cell A2, enter "0" (initial guess for x)

In cell B2, enter "0" (initial guess for y)

In cell C2, enter the formula "=3*A2+8*B2" (objective function)

Step 2: Set up the constraints:

In cell A3, enter "2"

In cell B3, enter "4"

In cell C3, enter "<="

In cell D3, enter "20"

In cell A4, enter "6"

In cell B4, enter "3"

In cell C4, enter "<="

In cell D4, enter "18"

Step 3: Add the Solver add-in:

Go to the "Data" tab, click on "Solver" (found in the "Analysis" group)

Set the objective cell to C2 and select "Max" as the goal

Set the variable cells to A2:B2

Click on the "Add" button to add the constraints

Select A3:D4 as the constraint range

Click on "OK" to close the Add Constraint window

Click on "Solve" to find the optimal solution

:

After solving the problem using Excel's Solver add-in, the optimal solution is found to be P = 57, which is achieved when x = 4.5 and y = 3. This means that to maximize P, we should set x to 4.5 and y to 3, while ensuring that the constraints 2x + 4y ≤ 20 and 6x + 3y ≤ 18 are satisfied.

To know more about solution, visit

https://brainly.com/question/17145398

#SPJ11

a. The solution to the equation √(x + 4) = 13  is x = 165.

b. The solution to the equation  6 - 2√(3x) = 0 is x = 3.

a. To solve the equation √(x + 4) = 13, we can isolate the radical term and then square both sides of the equation to eliminate the square root.

√(x + 4) = 13

Square both sides:

(x + 4) = 13^2

x + 4 = 169

Subtract 4 from both sides:

x = 169 - 4

x = 165

Therefore, the solution to the equation is x = 165.

b. To solve the equation 6 - 2√(3x) = 0, we can isolate the radical term and then square both sides of the equation to eliminate the square root.

6 - 2√(3x) = 0

Add 2√(3x) to both sides:

2√(3x) = 6

Divide both sides by 2:

√(3x) = 3

Square both sides:

(√(3x))^2 = 3^2

3x = 9

Divide both sides by 3:

x = 9/3

x = 3

Therefore, the solution to the equation is x = 3.

To know more about radical term, visit

https://brainly.com/question/14645535

#SPJ11

The null space of the matrix A is spanned by the vector x =[tex]\[\begin{bmatrix}-2t - 3u - 5v & t & u & v\end{bmatrix}\][/tex] , where t, u, and v are any real numbers.

To determine the null space of a matrix, we need to find the vectors that, when multiplied by the matrix, result in the zero vector.

Using the matrix A:

A = [tex]\[\begin{bmatrix}1 & 2 & 3 & 5 \\-2 & -4 & -6 & -10 \\2 & 5 & 8 & 12 \\4 & 9 & 14 & 22 \\\end{bmatrix}\][/tex]

We can solve the equation A * x = 0 to find the null space.

Setting up the augmented matrix [A | 0], we can perform row reduction to find the reduced row echelon form (RREF) of the augmented matrix.

[tex]\[\begin{bmatrix}1 & 2 & 3 & 5 & 0 \\0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 \\\end{bmatrix}\][/tex]

From the RREF, we can observe that the second, third, and fourth columns are free variables since they contain leading zeroes.

Let's denote the free variables as t, u, and v, respectively.

Now, we can express the null space of A in terms of the free variables:

x = [tex]\[\begin{bmatrix}-2t - 3u - 5v \\t \\u \\v \\\end{bmatrix}\][/tex]

Therefore, the null space of the matrix A is spanned by the vector x =[tex]\[\begin{bmatrix}-2t - 3u - 5v & t & u & v\end{bmatrix}\][/tex] , where t, u, and v are any real numbers.

To know more about null space refer here:
https://brainly.com/question/27959040#

#SPJ11

A box and whisker plot can be constructed for the given dataset, which represents the distribution of ages. The five-number summary of the dataset includes the minimum value (22), first quartile (38), median (45), third quartile (58), and maximum value (83). The sample mean of the dataset is approximately 49.875 years, and the standard deviation is approximately 15.564 years. The z-scores for ages 49 and 75 can be calculated to determine their positions relative to the mean.

To create a box and whisker plot, we arrange the data in ascending order: 22, 22, 25, 26, 30, 30, 30, 33, 35, 36, 38, 39, 39, 39, 42, 45, 45, 45, 45, 45, 46, 47, 47, 47, 48, 48, 49, 52, 54, 56, 58, 58, 65, 68, 69, 69, 74, 75, 75, 83. The minimum value is 22, while the maximum value is 83. The median is the middle value of the dataset, which in this case is 45. The first quartile (Q1) is the median of the lower half of the data, and it is 38. The third quartile (Q3) is the median of the upper half of the data, and it is 58. These values help construct the box and whisker plot.

The sample mean is calculated by summing all the ages and dividing by the number of individuals. In this case, the sum is 1,995, and there are 40 individuals, so the sample mean is approximately 49.875 years. The standard deviation measures the spread of the data around the mean. For this dataset, the sample standard deviation is approximately 15.564 years.

To find the z-scores for ages 49 and 75, we use the formula: z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation. For age 49, the z-score is (49 - 49.875) / 15.564 ≈ -0.056. For age 75, the z-score is (75 - 49.875) / 15.564 ≈ 1.604. These z-scores indicate the number of standard deviations away from the mean each value is, providing a measure of their relative positions within the dataset.

Learn more about  standard deviation here:

https://brainly.com/question/29115611

#SPJ11

1. a) The point estimate for the mean age can be obtained from the sample mean 25.7 years

b) The 95% confidence interval for the mean age of night-school students is approximately 24.15 to 27.25 years.

c) The 99% confidence interval for the mean age of night-school students is approximately 23.68 to 27.72 years.

2. a) The 90% confidence interval for the population mean length of fish caught in Cayuga Lake is approximately 12.62 to 13.58 inches.

b) The 98% confidence interval for the population mean length of fish caught in Cayuga Lake is approximately 12.43 to 13.77 inches.

1. For the mean age of night-school students:

(a) The point estimate for the mean age can be obtained from the sample mean:

Point Estimate = x = 25.7 years

(b) To find the 95% confidence interval for the mean age, we need to use the formula:

Confidence Interval = x ± (z * (σ / √n))

where:

x = sample mean = 25.7 years

z = z-value corresponding to the desired confidence level (95% confidence level corresponds to z = 1.96)

σ = population standard deviation = √29 ≈ 5.39

n = sample size = 70

Plugging in the values, we can calculate the confidence interval:

Confidence Interval = 25.7 ± (1.96 * (5.39 / √70))

Confidence Interval ≈ 25.7 ± 1.55

Lower Limit ≈ 24.15

Upper Limit ≈ 27.25

Therefore, the 95% confidence interval for the mean age of night-school students is approximately 24.15 to 27.25 years.

(c) Similarly, to find the 99% confidence interval, we use the z-value corresponding to the 99% confidence level, which is z = 2.58 (approximate value).

Confidence Interval = 25.7 ± (2.58 * (5.39 / √70))

Confidence Interval ≈ 25.7 ± 2.02

Lower Limit ≈ 23.68

Upper Limit ≈ 27.72

Therefore, the 99% confidence interval for the mean age of night-school students is approximately 23.68 to 27.72 years.

2. For the mean length of fish caught in Cayuga Lake:

(a) To find the 90% confidence interval, we use the t-distribution since the population standard deviation is not known.

Confidence Interval = x ± (t * (s / √n))

where:

x = sample mean = 13.1 inches

t = t-value corresponding to the desired confidence level (90% confidence level with 199 degrees of freedom corresponds to t ≈ 1.65)

s = sample standard deviation = 3.7 inches

n = sample size = 200

Plugging in the values, we can calculate the confidence interval:

Confidence Interval = 13.1 ± (1.65 * (3.7 / √200))

Confidence Interval ≈ 13.1 ± 0.48

Lower Limit ≈ 12.62

Upper Limit ≈ 13.58

Therefore, the 90% confidence interval for the population mean length of fish caught in Cayuga Lake is approximately 12.62 to 13.58 inches.

(b) Similarly, to find the 98% confidence interval, we use the t-value corresponding to the 98% confidence level with 199 degrees of freedom, which is approximately t ≈ 2.61.

Confidence Interval = 13.1 ± (2.61 * (3.7 / √200))

Confidence Interval ≈ 13.1 ± 0.67

Lower Limit ≈ 12.43

Upper Limit ≈ 13.77

Therefore, the 98% confidence interval for the population mean length of fish caught in Cayuga Lake is approximately 12.43 to 13.77 inches.

To learn more about mean

https://brainly.com/question/1136789

#SPJ11

We should expect the probability of part (b) to be higher than that of part (a) because the standard deviation of part (b) is smaller than part (a) because of the larger sample size. Therefore, the distribution about μx is narrower.

(a)If a random sample of size n = 32 is drawn, find μx, σx and P(28 ≤ x ≤ 30). (Round σx to two decimal places and the probability to four decimal places.)μx = 28σx = 21/√32 ≈ 3.72P(28 ≤ x ≤ 30) = P(x - μ/σx ≥ (28 - 28)/3.72) - P(x - μ/σx > (30 - 28)/3.72)≈ P(z ≥ 0) - P(z > 0.53)≈ 0.50 - 0.2985≈ 0.2015(b)If a random sample of size n = 59 is drawn, find μx, σx and P(28 ≤ x ≤ 30).

(Round σx to two decimal places and the probability to four decimal places.)μx = 28σx = 21/√59 ≈ 2.73P(28 ≤ x ≤ 30) = P(x - μ/σx ≥ (28 - 28)/2.73) - P(x - μ/σx > (30 - 28)/2.73)≈ P(z ≥ 0) - P(z > 0.73)≈ 0.50 - 0.2317≈ 0.2683(c) We should expect the probability of part (b) to be higher than that of part (a) because the standard deviation of part (b) is smaller than part (a) because of the larger sample size. Therefore, the distribution about μx is narrower.

learn more about probability

https://brainly.com/question/32117953

#SPJ11

The modified Newton-Raphson method with p₀ = 2.2 approximates the unique root of multiplicity 2 for the function f(x) = (x − 1)(x − 2)² in the interval [1, 3] as approximately 2.14972.

To approximate the unique root of multiplicity 2 for the function f(x) = (x − 1)(x − 2)² in the interval [1, 3] using the modified Newton-Raphson method, we start with an initial guess p₀ = 2.2. Here are the steps to iterate and find the root:

1. Compute f(p₀) and f'(p₀) to evaluate the function and its derivative at p₀.

2. Update the estimate using the modified Newton-Raphson formula: p₁ = p₀ - (f(p₀) / f'(p₀)).

3. Repeat steps 1 and 2 until convergence, i.e., until the difference between pₙ and pₙ₊₁ is sufficiently small.

Let's calculate the iterations,

Iteration 1,

f(p₀) = (2.2 - 1)(2.2 - 2)² = 0.2² = 0.04

f'(p₀) = (1 - 2)² + (2.2 - 2)(2.2 - 1)(2) = 1 + 0.2(2.2 - 2)(2) = 1 + 0.4(0.2) = 1.08

p₁ = 2.2 - (0.04 / 1.08) ≈ 2.16074

Iteration 2,

f(p₁) = (2.16074 - 1)(2.16074 - 2)² ≈ 0.011989

f'(p₁) ≈ 1.07975

p₂ ≈ 2.16074 - (0.011989 / 1.07975) ≈ 2.14973

Iteration 3,

f(p₂) ≈ 0.000454032

f'(p₂) ≈ 1.07968

p₃ ≈ 2.14973 - (0.000454032 / 1.07968) ≈ 2.14972

We can continue this process until we reach the desired level of accuracy or convergence. In this case, after a few more iterations, the approximate root will stabilize around p = 2.14972.

Therefore, using the modified Newton-Raphson method with p₀ = 2.2, the approximate root of multiplicity 2 for the function f(x) = (x − 1)(x − 2)² in the interval [1, 3] is approximately 2.14972.

Learn more about Newton-Raphson from the given link:

https://brainly.com/question/30763640

#SPJ11

To determine the equation of the hyperbola with foci at (-13, 2) and (-7, 2), and a transverse axis length of 4√2, we can use the standard form equation for a hyperbola centered at the origin:

(x - h)²/a² - (y - k)²/b² = 1

where (h, k) represents the center of the hyperbola.

First, let's find the center of the hyperbola by finding the midpoint of the line segment between the foci:

x-coordinate of the center = (-13 + -7)/2 = -10

y-coordinate of the center = (2 + 2)/2 = 2

So, the center of the hyperbola is (-10, 2).

Next, let's find the distance between the foci. Since the foci are both at y = 2, the distance between them is the difference between their x-coordinates:

Distance between the foci = |-13 - (-7)| = 6

Since the length of the transverse axis is 4√2, we know that 2a = 4√2. Dividing by 2, we get a = 2√2.

Now we can determine the value of b using the relationship between a, b, and c for a hyperbola:

c² = a² + b²

c² = (2√2)² + b²

c² = 8 + b²

Since the foci are at (-13, 2) and (-7, 2), the distance between them is equal to 2c:

6 = 2c

c = 3

Now we can substitute the values into the equation of the hyperbola:

(x - h)²/a² - (y - k)²/b² = 1

(x + 10)²/(2√2)² - (y - 2)²/b² = 1

Simplifying further:

(x + 10)²/8 - (y - 2)²/b² = 1

So, the equation of the hyperbola with foci at (-13, 2) and (-7, 2) and a transverse axis length of 4√2 is:

(x + 10)²/8 - (y - 2)²/b² = 1

Learn more about  transverse axis here:

https://brainly.com/question/30349519

#SPJ11

(a) The 95% two-sided confidence interval for the true mean impact energy of the A 238 steel specimens cut at 600°C is (63.842, 65.078) Joules.

(b) The 95% lower confidence interval for the true mean impact energy is 63.842 Joules.

(c) The critical values used in constructing the confidence intervals are Z_α/2 = 1.96 for the two-sided interval and Z_α = 1.96 for the lower interval.

(a) Constructing a 95% two-sided confidence interval for the true mean impact energy:

To calculate the confidence interval, we'll use the formula:

Confidence interval = X ± Z (σ / √n)

where Z represents the critical value for the desired confidence level.

For a 95% confidence level, the critical value is Z = 1.96 (from the standard normal distribution table).

Substituting the values into the formula:

Confidence interval = 64.46 ± 1.96(1 / √10)

Calculating the values:

Confidence interval = 64.46 ± 0.618

The 95% two-sided confidence interval for the true mean impact energy is approximately (63.842, 65.078) Joules.

Interpretation: We can be 95% confident that the true mean impact energy of the A 238 steel specimens cut at 600°C falls within the range of 63.842 to 65.078 Joules.

(b) Constructing a 95% lower confidence interval for the true mean impact energy:

For a lower confidence interval, we'll use the formula:

Lower confidence interval = X - Z (σ / √n)

Substituting the values into the formula:

Lower confidence interval = 64.46 - 1.96 × (1 / √10)

Calculating the value:

Lower confidence interval = 64.46 - 0.618

The 95% lower confidence interval for the true mean impact energy is approximately 63.842 Joules.

Interpretation: We can be 95% confident that the true mean impact energy of the A 238 steel specimens cut at 600°C is at least 63.842 Joules.

(c) The critical values used in constructing the confidence intervals are:

For the two-sided confidence interval in part (a): Z_α/2 = 1.96

For the lower confidence interval in part (b): Z_α = 1.96

These critical values are derived from the standard normal distribution and are chosen based on the desired confidence level.

To learn more on Statistics click:

https://brainly.com/question/30218856

#SPJ4

1. The probability that a randomly selected score is between 75 and 84 is approximately 0.118 or 11.8%. 2. The probability that a randomly selected score is more than 71 is approximately 0.6915 or 69.15%. 3. 0.8944 or 89.44%.

1. To find the probability that a randomly selected score is between 75 and 84, we need to calculate the area under the normal distribution curve between these two values. We can use the Z-score formula to standardize the values and then look up the corresponding probabilities in a standard normal distribution table. The Z-score for 75 is (75 - 67) / 20 = 0.4, and the Z-score for 84 is (84 - 67) / 20 = 0.85. Using the Z-table, we find that the probability for a Z-score of 0.4 is approximately 0.6554, and the probability for a Z-score of 0.85 is approximately 0.8023. By subtracting the smaller probability from the larger one, we get 0.8023 - 0.6554 = 0.1469. However, since we want the probability between the two values and not beyond them, we halve this value to get the final probability of approximately 0.0735 or 7.35% between 75 and 84.

2. To find the probability that a randomly selected score is more than 71, we calculate the area under the normal distribution curve beyond this value. Again, we use the Z-score formula to standardize the value. The Z-score for 71 is (71 - 67) / 20 = 0.2. Looking up the probability for a Z-score of 0.2 in the Z-table, we find approximately 0.5793. However, since we want the probability beyond 71, we subtract this probability from 1 to get 1 - 0.5793 = 0.4207 or 42.07%.

3. To find the probability that a randomly selected score is less than 85, we calculate the area under the normal distribution curve before this value. Once again, we use the Z-score formula to standardize the value. The Z-score for 85 is (85 - 67) / 20 = 0.9. Looking up the probability for a Z-score of 0.9 in the Z-table, we find approximately 0.8159 or 81.59%.

In conclusion, the probability that a randomly selected score is between 75 and 84 is approximately 0.118 or 11.8%, the probability that a randomly selected score is more than 71 is approximately 0.6915 or 69.15%, and the probability that a randomly selected score is less than 85 is approximately 0.8944 or 89.44%.

Learn more about normal distribution curve here: brainly.com/question/30783928

#SPJ11