A computer sends a packet of information along a channel and waits for a return signal acknowledging that the packet has been received. If no acknowledgment is received within a certain time, the packet is re-sent.
Let X represent the number of times the packet is sent. Assume that the probability mass function of X is given by
p(x)={ cx for x=1,2,3,4, or 5
{ 0 otherwise
where c is a constant.
Find P(X = 2).

Final answer:

A function rule that states "The output is four more than the input" is expressed as y = x + 4, where x is the input and y is the output.

Explanation:

To write a function rule that describes "The output is four more than the input," we let x represent the input and y represent the output. According to the statement, for any given value of x, the value of y will always be 4 units larger. Therefore, the function rule can be written as y = x + 4.

This means that if you have an input value, simply add 4 to it to get the output value. For example, if the input, x, is 5, the output, y, would be 5 + 4, which equals 9.

Final answer:

The 1-day growth factor is found using the exponential growth formula. Given the sunflower's height after 2.79 days and knowing it grows by 7% daily, we calculate the initial height and then apply the rate to find a growth factor of 1.07.

Explanation:

The question is asking for the 1-day growth factor for the height of the sunflower given that it increases by 7% per day. To find the growth factor, we need to use the formula for exponential growth, which is:

Final height = Initial height x (1 + rate of growth) ^ time

We know the final height (15.7 inches) and the time (2.79 days), but we need the initial height to calculate the growth factor for one day. We can rearrange the formula to solve for the initial height first:

Initial height = Final height / (1 + rate of growth) ^ time

Once we find the initial height, we can insert the 7% growth rate as 0.07 and solve for the 1-day growth factor, which would be 1 + 0.07, or 1.07.

Answer:

Standard deviation is 25.4

Step-by-step explanation:

The standard deviation is a metric that determines the variance a set of data has both above and below the mean. A standard deviation of 25.4 means that the values in a given data set are dispersed in a range of 25.4 units both above and below the mean.

The standard error refers to the Standard Error of the Mean (SEM) which measures the precision of the mean in terms of how much a sample mean is likely to differ from the population mean. By using SEM, individuals can estimate how sure they can be that the mean of the sample reflect the true mean of the population. The standard error always is lower than the standard deviation. Since, 25.4 is the higher number, this number would be the standard deviation.

Using the normal distribution table, determine the percentile based on the z-score for each graduation rate. The organization started at the 30th percentile (2000-2009) and is now at the 79th percentile (2010-2018).

To determine percentiles, we can use a standard normal distribution table.

1. Initial Percentile (2000-2009):

  - From the normal distribution table, a graduation rate of 85.1% corresponds to a z-score.

  - Find the z-score for 85.1% and determine the percentile associated with it.

  - For instance, if the z-score is -0.5, the organization started at the 30th percentile.

2. Current Percentile (2010-2018):

  - Repeat the process for the new graduation rate of 88.3%.

  - Find the z-score for 88.3% and determine the current percentile.

  - If the z-score is 0.8, for instance, the organization is now at the 79th percentile.

The question probable may be:

A flight academy had a graduation rate of 85.1% for 27-year-old candidates from 2000-2009. With new instructors since 2010, the rate improved to 88.3%. Determine the percentile the organization initially started at (2000-2009) and its current percentile (2010-2018).

Answer:

[tex] P(62< X< 72)= P(X<72) -P(X<62)=0.975-0.5=0.475 [/tex]

Step-by-step explanation:

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

Let X the random variable who represent the courtship time (minutes).

From the problem we have the mean and the standard deviation for the random variable X. [tex]E(X)=62, Sd(X)=5[/tex]

So we can assume [tex]\mu=62 , \sigma=5[/tex]

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

• The probability of obtain values within one deviation from the mean is 0.68

• The probability of obtain values within two deviation's from the mean is 0.95

• The probability of obtain values within three deviation's from the mean is 0.997

So we need values such that

[tex]P(X<\mu -\sigma)=P(X <57)=0.16[/tex]    

[tex]P(X>\mu +\sigma)=P(X >67)=0.16[/tex]  

[tex]P(X<\mu -2*\sigma)=P(X<52)=0.025[/tex]    

[tex]P(X>\mu +2*\sigma)=P(X>72)=0.025[/tex]

[tex]P(X<\mu -3*\sigma)=P(X<47)=0.0015[/tex]

[tex]P(X>\mu +3*\sigma)=P(X>77)=0.0015[/tex]

For this case we want to find this probability:

[tex] P(62 < X< 72) [/tex]

And we can find this probability on this way:

[tex] P(62< X< 72)= P(X<72) -P(X<62) [/tex]

Since [tex] P(X>72) =0.025[/tex] by the complement rule we have that:

[tex] P(X<72) = 1-0.025 =0.975[/tex]

And [tex] P(X<62) =0.5[/tex] because for this case 62 is the mean.

So then we have this:

[tex] P(62< X< 72)= P(X<72) -P(X<62)=0.975-0.5=0.475 [/tex]

Answer:

Step-by-step explanation:

To solve the differential equation

dy/dx = 5x^4(1 + y²)^(3/2)

First, separate the variables

dy/(1 + y²)^(3/2) = 5x^4 dx

Now, integrate both sides

To integrate dy/(1 + y²)^(3/2), use the substitution y = tan(u)

dy = (1/cos²u)du

So,

dy/(1 + y²)^(3/2) = [(1/cos²u)/(1 + tan²u)^(3/2)]du

= (1/cos²u)/(1 + (sin²u/cos²u))^(3/2)

Because cos²u + sin²u = 1 (Trigonometric identity),

The equation becomes

[1/(1/cos²u)^(3/2) × 1/cos²u] du

= cos³u/cos²u

= cosu

Integral of cosu = sinu

But y = tanu

Therefore u = arctany

We then have

cos(arctany) = y/√(1 + y²)

Now, the integral of the equation

dy/(1 + y²)^(3/2) = 5x^4 dx

Is

y/√(1 + y²) = x^5 + C

So

y - (x^5 + C)√(1 + y²) = 0

is the required implicit solution

Step-by-step explanation:

We have two cases for Ф,

1.  Ф=1; it implies that Pr(Ф=1)=0.5, while y~N(1,α²)

2. Ф=2; it implies that Pr(Ф=2)=0.5, while y~N(2,α²)

Now,

For 1st case of α=2,

We have marginal probability density formula

p(y)=∑p(yIФ)p(Ф)

=p(yIФ=1)p(Ф=1)+p(yIФ=2)p(Ф=2)

=N(yI1,2²)(1/2)+N(yI2,2²)(1/2)

=(1/2)[N(yI1,2²)+N(yI2,2²)]

Now.

For Pr(Ф=1Iy=1) at α=2

We have,

=p(Ф=1Iy=1)

=[p(y=1,Ф=1)]/[p(y=1)]

=[p(y=1IФ=1)p(Ф=1)]/[p(y=1)]

={(1/[tex]\sqrt{2x-2}[/tex])exp[(-1/(2*2²))(1-1)²(1/2)]}/{(1/[tex]\sqrt{2x-2}[/tex])(1/2)[exp[(-1/(2*2²))(1-1)²]+exp[(-1/(2*2²))(1-2)²]}

=0.53 Answer

Now, to describe the changes in shape of Ф when α is increased and decreased:

The formula for posterioir density is p(ФIy)=p(yIФ)p(Ф)/p(y)

=exp[(-1/(2α²)(y-Ф)²]/{exp[(-1/(2α²)(y-1)²]+exp[(-1/(2α²))(y-2)²]}

Now at Ф=1 and solving the equation, we get

p(Ф=1Iy)=1 / {1+exp[(2y-3)/2α²]}

Similarly at Ф=1 and solving the equation, we get

p(Ф=2Iy)=1 / {1+exp[(2y-3)/2α²]}

Conclusion:

α² → ∞ ⇒p(ФIy) → p(Ф) = 1/2

α² → 0 ⇒ two cases

y > 3/2, α² → 0 ⇒p(Ф=2Iy) → 1

y < 3/2, α² → 0 ⇒p(Ф=1Iy) → 1

The value of θ determines the mean of the normal distribution for y, while σ remains constant. The probabilities of θ being 1 or 2 are both 0.5.

The given information states that if θ = 1, then y has a normal distribution with a mean of 1 and standard deviation σ, and if θ = 2, then y has a normal distribution with a mean of 2 and standard deviation σ.

The probabilities of θ being 1 or 2 are both 0.5.

This means that there is a 50% chance of θ being 1, and a 50% chance of θ being 2.

This information allows us to understand how the value of θ affects the distribution of y. When θ is 1, y follows a normal distribution with mean 1 and standard deviation σ.

When θ is 2, y follows a normal distribution with mean 2 and standard deviation σ. The probabilities of these scenarios happening are equal.

Learn more about normal distribution here:

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The question discusses a binomial problem. The cumulative distribution function (CDF) for a binomial distribution is defined as the summed probability of all outcomes up to and including X. To compute the CDF, add up the probabilities of all outcomes up to X.

The problem described in the question is a binomial problem. The binomial distribution model is suitable because we have 4 (N) independent trials (computers) with two possible outcomes (right or left roll), and each trial's outcome does not affect any other trial's outcome. The probability that a computer will ask for a roll to the left when a roll to the right is appropriate is 0.0005 (p). The random variable X represents the number of computers asking for an incorrect roll.

The cumulative distribution function (CDF) for a random variable X in a binomial distribution is the probability that X will take a value less than or equal to x.

The binomial distribution's CDF can be computed by calculating the probability for all values up to X and adding them together. An additional thing to note is that calculator with statistical functions or a software can be used in doing this computation.

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The number of incorrect votes follows a binomial distribution with parameters n = 4 and p = 0.0005. The CDF is determined by summing the binomial probabilities up to a given value. The CDF values for X = 0 through X = 4 are calculated step-by-step.

The question asks us to determine the cumulative distribution function (CDF) of the random variable X, which represents the number of computers voting for a left roll when a right roll is appropriate.

This scenario follows a binomial distribution, where each computer vote is an independent trial with a probability of 0.0005 of voting incorrectly.

Let X be the number of computers voting incorrectly. Since there are 4 independent computers, X can take on values 0, 1, 2, 3, and 4.

The probability mass function (PMF) for X is given by:

[tex]P(X = k) = \((4},{k) \times (0.0005)^k \times (0.9995)^{4-k}[/tex]

where C(4, k) = 4!/(k! * (4-k)!).

To find the CDF, F(x), of X, we sum the probabilities for all values up to x:

Using the binomial probability formula, calculate each PMF and its cumulative sum to get the CDF:

Hence, the CDF of X encompasses a step-by-step summation of the binomial probabilities up to the desired value.

Answer:

Θ =tan⁻¹ (4KQ²/mgr²), Q = r[mgtanΘπ∈₀][tex]\frac{1}{2\\}[/tex]

Step-by-step explanation:

initially the angle Θ=0° ,the vertical forces were equal to product of mass and gravity(m*g) and there was no horizontal or lateral force in action. But after the displacement of balls new forces are induced.

X-Axis:

Fe = TsinΘ

[KQ²/(r/2)²] = TsinΘ         where r₁=r/2, r₁ = new distance

(4KQ²/r²) = TsinΘ

Y-Axis

TcosΘ = mg

As we know that tanΘ=sinΘ/cosΘ

We have, tanΘ = 4KQ²/mgr²

By adjusting this equation and putting K=1/4π∈₀ we get,

Q = r[mgtanΘπ∈₀][tex]\frac{1}{2\\}[/tex]

Answer:

16.80% probability of 4 flaws in 100 feet.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

[tex]e = 2.71828[/tex] is the Euler number

[tex]\mu[/tex] is the mean in the given interval.

An average of 3 flaws every 100 feet.

So [tex]\mu = 3[/tex]

Find the probability of 4 flaws in 100 feet.

This is [tex]P(X = 4)[/tex]

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 4) = \frac{e^{-3}*(3)^{4}}{(4)!} = 0.1680[/tex]

16.80% probability of 4 flaws in 100 feet.

The Poisson distribution formula is used to calculate the probability of a specific number of flaws in a fixed interval based on the average rate of flaws.

Poisson Probability Distribution:

The probability of 4 flaws in 100 feet is approximately 0.8153.

Answer:

at 0.05 significance level she cannot conclude for certain there is a difference in the mean number of hours spent per week by industry because the level of significance is large and there is a possibility she might be wrong.

Step-by-step explanation:

The significance level: is the probability of rejecting the null hypothesis when it is true. For example, a significance level of 0.05 indicates a 5% risk of concluding that a difference exists when there is no actual difference

Final answer:

The probability that a component is at least 12 centimeters long, given that the lengths follow a normal distribution with mean 14 cm and variance 9, is approximately 74.86%.

Explanation:

To calculate the probability that a component is at least 12 centimeters long given that X (the length of a component) follows a normal distribution with mean 14 centimeters and variance 9, we first need to standardize the random variable X to convert it to the standard normal distribution Z.

The variance provided is 9, so the standard deviation is the square root of the variance, which is 3. We standardize using the formula

Z = (X - µ) / σ,

where µ is the mean and

σ is the standard deviation.

For X = 12 centimeters, Z = (12 - 14) / 3 = -2 / 3 ≈ -0.67.

Now, we look up the value of -0.67 on the standard normal distribution table or use a calculator with the standard normal distribution function. Let's denote this value as P(Z < -0.67).

Since we're looking for the probability that a component is at least 12 centimeters long, we need to find the complement of this probability, which is 1 - P(Z < -0.67).

Using the standard normal distribution table or a calculator, we find P(Z < -0.67) ≈ 0.2514.

Thus, the probability that a component is at least 12 centimeters long is 1 - 0.2514 ≈ 0.7486, or approximately 74.86%.

Answer:

Please read the answer below.

Step-by-step explanation:

1. Let's represent the relationship between red pepper flakes, in teaspoons, to tomato mix, in cups in a table

Red pepper (teaspoons)   1/2    1    1 1/2   2  2 1/2   3  3 1/2   4  4 1/2   5      

Tomato mix (cups)              2     4       6     8    10     12   14     16   18    20

2. Let's represent the relationship between red pepper flakes, in teaspoons, to tomato mix, in cups writing a equation:

t = amount of tomato mix cups

r = amount of red pepper flakes teaspoons

As we can see in the table,

t = 4r

Answer:

Infinitely many solutions

Step-by-step explanation:

The given system is

2y = 14 - 2x

y = -x + 7

Let us substitute the second equation into the first one to get:

2(-x+7)=14-2x

Expand to get:

-2x+14=14-2x

This means

x=x

This tells us that the system has infinitely many solutions.

The two lines coincide

Answer:

infinatly many

Answer:

$4928.00

Step-by-step explanation:

This question is solved by the compound interest formula:

[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]

In which A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.

In this problem, we have that:

Nicole deposited $4400, so [tex]P = 4400[/tex]

6% compounded monthly, which means that [tex]r = 0.06, n = 12[/tex]

How much will she have in her account in two years?

This is A when [tex]t = 2[/tex].

So

[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]

[tex]A = 4400(1 + \frac{0.06}{12})^{12*2}[/tex]

[tex]A = 4959.50[/tex]

So the correct answer is:

$4928.00

Answer:

Their total revenue the previous year was $54,000.

Step-by-step explanation:

This question can be solved by a simple rule of three.

This year revenue was $62,100. It was 15% more than last year, so 115% = 1.15 of last year. How much was the revenue last year, that is, 100% = 1?

62,100 - 1.15

x - 1

[tex]1.15x = 62100[/tex]

[tex]x = \frac{62100}{1.15}[/tex]

[tex]x = 54000[/tex]

Their total revenue the previous year was $54,000.

Answer:

False

Step-by-step explanation:

De Morgan's laws are a pair of transformation rules that are both valid rules of inference.

not (A or B) = not A and not B; and

not (A and B) = not A or not B

From the above law, the statement:

Kwame will not take a job in industry or will not go to graduate school; the or is supposed to be And. Hence the statement is False.

Answer:

Normal Good

Step-by-step explanation:

A normal good is a good in which a rise in income comes with bigger increases in its quantity demanded. In the demand function, M which is the income is positive and has the highest value.

Therefore Good X is a Normal Good.

The equation represents the demand function for good X. The coefficients of the variables indicate how demand for X is influenced by changes in the price of X itself (PX), the price of a related good (PY), income (M), and advertising (A).

The function Qxd = 100 - 2PX + 4PY + 10M + 2A represents the demand function for a particular good, X. PX represents the price of good X, PY the price of a related good (Y), M is income, and A is the amount of advertising on good X. The coefficients of these variables determine how the demand for good X responds to changes in these variables. For example, the demand for good X decreases with an increase in its own price (as indicated by the negative coefficient -2) and increases with an increase in the price of good Y, income, and the amount of advertising (as indicated by positive coefficients).

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Answer:

[tex]P(X>\mu +2*\sigma)P(X>102)=0.025[/tex]

Step-by-step explanation:

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

Let X the random variable who represent the diastolic blood pressure readings of adult males

From the problem we have the mean and the standard deviation for the random variable X. [tex]E(X)=84, Sd(X)=9[/tex]

So we can assume [tex]\mu=84 , \sigma=9[/tex]

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

• The probability of obtain values within one deviation from the mean is 0.68

• The probability of obtain values within two deviation's from the mean is 0.95

• The probability of obtain values within three deviation's from the mean is 0.997

So we need values such that

[tex]P(X<\mu -\sigma)=P(X <75)=0.16[/tex]    

[tex]P(X>\mu +\sigma)=P(X >93)=0.16[/tex]  

[tex]P(X<\mu -2*\sigma)P(X<66)=0.025[/tex]    

[tex]P(X>\mu +2*\sigma)P(X>102)=0.025[/tex]

[tex]P(X<\mu -3*\sigma)=P(X<57)=0.0015[/tex]

[tex]P(X>\mu +3*\sigma)=P(X>211)=0.0015[/tex]

So for this case the answer would be:

[tex]P(X>\mu +2*\sigma)P(X>102)=0.025[/tex]

The empirical rule indicates that about 2.5% of adult males have diastolic blood pressure readings greater than 102 mmHg, as 102 mmHg is two standard deviations above the mean diastolic blood pressure of 84 mmHg.

The empirical rule states that for a bell-shaped distribution:

Approximately 68% of the data falls within one standard deviation of the mean.

Approximately 95% of the data falls within two standard deviations of the mean.

Approximately 99.7% of the data falls within three standard deviations of the mean.

Given that the mean diastolic blood pressure for adult males is 84 mmHg with a standard deviation of 9 mmHg, to find the percentage of adult males with diastolic blood pressure readings greater than 102 mmHg, we calculate how many standard deviations 102 is from the mean.

To calculate this, use the following formula for the z-score: z = (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation.

For 102 mmHg:

z = (102 mmHg - 84 mmHg) / 9 mmHg = 2

This means that 102 mmHg is two standard deviations above the mean. According to the empirical rule, 95% of data falls within two standard deviations of the mean, which means that 2.5% falls above this range as the data is symmetric about the mean.

Thereby, approximately 2.5% of adult males have diastolic blood pressure readings greater than 102 mmHg.

Answer:

The absolute decrease was of 33 million.

The relative decrease was of 22.45%.

Step-by-step explanation:

Absolute change

The absolute change is the number of CD's sold is the number of CD's sold in 2010 subtracted by the numbers of CD's sold in 2009.

The number is negative, which means that there was a decrease.

114 million - 147 million = -33 million

The absolute decrease was of 33 million.

Relative change

The relative change is the absolute change divided by the initial value.

So -33/147 = -0.2245

Which means that the relative decrease was of 22.45%.

Solution:

Distance, Velocity - time functions are linked easily through derivation and integration:

Distance - time function → derivation → Velocity - time function

Velocity - time function → derivation → Acceleration - time function

(and vice versa)

Let's assume we have a distance - time function:

[tex]s(t) = 4t^{2} - 2t +7[/tex]

where s is measured in feet and t in seconds.

a) To find velocity at time t, we simply derivate the distance - time function:

[tex]\frac{ds}{dt} = v(t) = 8t - 2[/tex]

b) To find velocity at t-3, we simply substitute 3 in the velocity - time function:

[tex]v (t) = 8t -2\\v(3) = 8(3) -2\\v(3) = 22 \ ft/sec[/tex]  

c) A particle will be at rest when it's velocity is zero. Thus, we substitute v = 0 in the velocity - time function:

[tex]v (t ) = 8t -2\\8t -2 = 0\\8t = 2\\\\t = \frac{2}{8}\\\\t= \frac{1}{4} seconds[/tex]

Hence, at time t = 1/4 seconds, the object will be at rest.

d) To determine the positive direction, we must understand that this is a quadratic function. Hence it has a minimum/ maximum value, after this critical point the particle must be moving either in positive or negative direction.

Hence, we find this critical point. A critical point of any function is it's derivative equalled to zero.

The derivative of distance - time function is a velocity - time function. From the previous part, we already know that a critical point exists at t = 1/4. Now, we substitute, t = 1/4, in the distance - time function to find the other co-ordinate:

[tex]s (t) = 4t^{2} - 2t +7\\s(\frac{1}{4}) = 4(\frac{1}{4})^{2} - 2(\frac{1}{4})+7\\\\s(\frac{1}{4}) = 4(\frac{1}{16}) - 2(\frac{1}{4})+7\\\\s(\frac{1}{4}) = \frac{1}{4} - \frac{2}{4}+\frac{28}{4}\\\\s(\frac{1}{4}) = \frac{27}{4} \\\\[/tex]

The function will be positive after [tex](\frac{1}{4}, \frac{27}{4})[/tex]

e) The total distance travelled in first 8 seconds can be determined by substituting t = 8 in distance - time function:

[tex]s(t) = 4t^{2} - 2t+7\\\\s(8) = 4(8)^{2} - 2(8)+7\\\\s(8) = 4 (64) - 2 (8) +7\\\\s(8) = 247 feet[/tex]

The velocity of the particle any time t is v(t) = 4t^3 - 7 ft/sec. The velocity at t=3 seconds is 98 ft/sec. The particle is at rest at t=1.323. It moves in the positive direction when t < 1.323 or t > 1.323. The total distance travelled during the first 8 seconds is approximately 4085.6 feet.

The first step here is to find the velocity of the particle at any given time t. Since velocity represents the rate of change in position, we'll compute this by taking the derivative of the position function s(t) = t4 - 7t + 22. This gives us the velocity function v(t) = 4t3 - 7.

Next, to find the velocity of the particle at t = 3, simply plug 3 into the velocity function: v(3) = 4(33) - 7 = 98 ft/sec.

The particle is at rest when its velocity is zero, so we set v(t) = 0, or 4t3 - 7 = 0. Solving for t reveals that the particle is at rest when t = 1.323.

The particle moves in the positive direction when the velocity is greater than zero. Looking at v(t), we see that this is the case when t < 1.323 or t > 1.323. So, using interval notation, we can say that the particle moves in the positive direction during (-inf, 1.323) and (1.323, inf).

Lastly, to find the total distance travelled during the first 8 seconds, take the absolute value of the integral of v(t) from 0 to 8. Doing the computation, we find that the particle travels approximately 4085.6 feet during this time interval.

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To find the probability of a man's score being at least 559.5 on the standardized college aptitude test, we can calculate the z-score and find the area under the normal distribution curve. The same process applies to finding the probability of the mean score of a sample of 18 men being at least 559.5.

To find the probability that a randomly selected man's score is at least 559.5, we need to calculate the z-score for this value and then find the area under the normal distribution curve to the right of that z-score.

To find the probability that the mean score of 18 randomly selected men is at least 559.5, we first need to find the mean and standard deviation of the sample mean. Then, we can calculate the z-score for the given mean score and find the area under the normal distribution curve to the right of that z-score.

P(X > 559.5) = 1 - P(X ≤ 559.5)

P(M > 559.5) = 1 - P(M ≤ 559.5)

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The probabilities of a score being above 559.5 are as follows: for a single randomly selected individual, the probability is approximately 0.3271; for a group of 18 randomly selected individuals, the probability that their mean score is above 559.5 is approximately 0.0287.

This is a problem of statistics, more specifically Normal Distribution and Standard Deviation. In a Normal Distribution, the mean (average) is the center of the distribution and standard deviation measures how spread out the scores are from the mean. The Z-Score gives us a measure of how many standard deviations an element is from the mean.

Firstly, to find the probability that a randomly selected man scores at least 559.5, we find the Z-Score using the formula Z = (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation. Thus the Z-Score is Z = (559.5 - 512) / 106 = 0.448. From the Z-table or calculator, we find that P(Z > 0.448) ≈ 0.3271. Therefore, P(X > 559.5) = 0.3271.

Secondly, for a sample of 18 men, we use the formula for the standard deviation of a sample mean, σM = σ / sqrt(n), where σ is the standard deviation, and n is the size of the sample. The new standard deviation becomes σM = 106 / sqrt(18) = 25. This gives Z = (559.5 - 512) / 25 =1.90. From the Z-table or calculator, we find that P(Z > 1.90) ≈ 0.0287. Therefore, P(M > 559.5) = 0.0287.

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Answer:

[tex]\dfrac{343}{71}[/tex]

Step-by-step explanation:

Given the equation

[tex](5x-y)^4+4y^3=2433[/tex]

Find the derivative:

[tex]((5x-y)^4+4y^3)'=(2433)'\\ \\4(5x-y)^3\cdot (5x-y)'+4\cdot 3y^2\cdot y'=0\\ \\4(5x-y)^3\cdot (5-y')+12y^2y'=0[/tex]

Substitute

[tex]x=-1\\ \\y=2,[/tex]

then

[tex]4(5\cdot (-1)-2)^3\cdot (5-y')+12\cdot 2^2\cdot y'=0\\ \\4(-5-2)^3(5-y')+48y'=0\\ \\4\cdot (-7)^3\cdot (5-y')+48y'=0\\ \\-1,372(5-y')+48y'=0\\ \\-6,860+1,372y'+48y'=0\\ \\1,420y'=6,860\\ \\y'=\dfrac{6,860}{1,420}=\dfrac{686}{142}=\dfrac{343}{71}[/tex]

Answer:

Steve should place $6,250 in the 5-year CD and $18,750 in the corporate bond

Step-by-step explanation:

System of Equations

We need to find how Steve will distribute his investments between two possible options: one of them will pay 5% per annum and the other will pay 9% per annum. We know Steve has $25,000 to invest and wants to have an overall annual rate of return of 8%.

Let's call x to the amount Steve will invest in the CD paying 5% per annum and y to the amount he will invest in a corporate bond paying 9% per annum.

The total investment is $25,000 which leads to the first equation

[tex]x+y=25,000[/tex]

If x dollars are invested at 5%, then the interest return is 0.05x. Similarly, y dollars at 9% return 0.09y. The overall return is 8% on the total investment, thus

[tex]0.05x+0.09y=0.08(x+y)[/tex]

Rearranging:

[tex]0.05x+0.09y=0.08x+0.08y[/tex]

Simplifying

[tex]0.01y=0.03x[/tex]

Multiplying by 100

[tex]y=3x[/tex]

Substituting in the first equation

[tex]x+3x=25,000\\4x=25,000\\x=6,250[/tex]

And therefore

[tex]y=25,000-6,250=18,750[/tex]

Steve should place $6,250 in the 5-year CD and $18,750 in the corporate bond

Answer:

1 / √17

Step-by-step explanation:

to solve cos(tan^-1(4))

we break it into simpler terms

tan^-1(4) ------ these will be taken as an angle when dealing with cos

tan Ф = opposite / adjacent = 4 / 1 = 4

Using Pythagoras Theorem

Hypothenus ² = opposite² + adjacent ²

h² = 4² + 1²

h²  = 16 + 1

h² = 17  

a = √17

cos Ф = adjacent / hypothenus = 1 / √17

cos(tan^-1(4)) =  1 / √17

The probability of randomly selecting atleast 2 defective speakers from 7 trials is 0.18

The probability of randomly selecting a defective speaker can be calculated thus :

Using the binomial probability relation :

P(x ≥ 2 ) = P(x = 2)+P(x = 3)+P(x = 4)+P(x = 5)+P(x =6)+P(x = 7)

Using a binomial probability calculator to save time :

P(x ≥ 2 ) = 0.17785

P(x ≥ 2 ) = 0.18 ( 2 decimal places)

Therefore, the probability of selecting atleast 2 defective speakers from 7 is 0.18

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Answer:

1) Randomization: We assume that we have a random sample of students

2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size

3) np = 500*0.66= 330 >10

n(1-p) = 500*(1-0.66) =170>10

So then we can use the normal approximation for the distribution of p, since the conditions are satisfied

The population proportion have the following distribution :

[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]  

And we have :

[tex] \mu_p = 0.66[/tex]

[tex] \sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212[/tex]

Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).

Step-by-step explanation:

For this case we know that we have a sample of n = 500 students and we have a percentage of expected return for their sophomore years given 66% and on fraction would be 0.66 and we are interested on the distribution for the population proportion p.

We want to know if we can apply the normal approximation, so we need to check 3 conditions:

1) Randomization: We assume that we have a random sample of students

2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size

3) np = 500*0.66= 330 >10

n(1-p) = 500*(1-0.66) =170>10

So then we can use the normal approximation for the distribution of p, since the conditions are satisfied

The population proportion have the following distribution :

[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]  

And we have :

[tex] \mu_p = 0.66[/tex]

[tex] \sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212[/tex]

And we can use the empirical rule to describe the distribution of percentages.

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

• The probability of obtain values within one deviation from the mean is 0.68

• The probability of obtain values within two deviation's from the mean is 0.95

• The probability of obtain values within three deviation's from the mean is 0.997

Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).

The 68-95-99.7 rule can be used to describe the sampling distribution model for the percentage of college freshmen expected to return for their sophomore years. The appropriate conditions for using this rule are met.

The question asks to describe the sampling distribution model for the percentage of college freshmen expected to return for their sophomore years using the 68-95-99.7 rule. The 68-95-99.7 rule is a statistical rule that states that for a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations.

In this case, the company reported that the national college freshman-to-sophomore retention rate was 66%. Assuming that the retention rate follows a normal distribution, approximately 68% of colleges would have a retention rate within one standard deviation of 66%, approximately 95% would have a retention rate within two standard deviations, and approximately 99.7% would have a retention rate within three standard deviations.

Based on these assumptions, the appropriate conditions for using the 68-95-99.7 rule in the sampling distribution model are met.

Answer:

a ₙ = 7n

Step-by-step explanation:

This is an arithmetic sequence, the common difference between each term is 14-7 = 21-14 = 28-21 = 7

to the previous term in the sequence addition of 7 gives the next term.

Arithmetic Sequence:  

d  = 7

This is the formula of an arithmetic sequence.

a ₙ = a₁ + d(n − 1)  

Substitute in the values of  

a₁ = 7  and d = 7

a ₙ = 7 + 7(n − 1)  

a ₙ = 7 + 7n -7

a ₙ = 7 - 7 +7n = 7n

a ₙ = 7n

Answer:

Step-by-step explanation:

In an arithmetic sequence, consecutive terms differ by a common difference and it is always constant. Looking at the set of numbers,

14 - 7 = 21 - 14 = 28 - 21 = 7

Therefore, it is an arithmetic sequence with a common difference of 7.

The formula for determining the nth term of an arithmetic sequence is expressed as

Tn = a + (n - 1)d

Where

a represents the first term of the sequence.

d represents the common difference.

n represents the number of terms in the sequence.

From the information given,

a = 7

d = 7

The function that represents the sequence would be

Tn = 7 + (n - 1)7

Tn = 7 + 7n - 7

Tn = 7n

the mean speed is [tex]\( 12.75 \)[/tex] km/s, the standard deviation is approximately [tex]\( 0.433 \)[/tex] km/s, and the standard deviation of the mean is approximately [tex]\( 0.217 \)[/tex] km/s.

To compute the mean, standard deviation, and standard deviation of the mean, we'll follow these steps:

1. Calculate the mean [tex](\( \mu \))[/tex]:

[tex]\[ \mu = \frac{\text{sum of all measurements}}{\text{number of measurements}} \][/tex]

2. Calculate the standard deviation [tex](\( \sigma \))[/tex]:

[tex]\[ \sigma = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \mu)^2}{n}} \][/tex]

3. Calculate the standard deviation of the mean [tex](\( \sigma_\bar{x} \))[/tex]:

[tex]\[ \sigma_\bar{x} = \frac{\sigma}{\sqrt{n}} \][/tex]

Let's plug in the given measurements:

[tex]\[ x_1 = 12.7 \, \text{km/s} \][/tex]

[tex]\[ x_2 = 13.4 \, \text{km/s} \][/tex]

[tex]\[ x_3 = 12.6 \, \text{km/s} \][/tex]

[tex]\[ x_4 = 13.3 \, \text{km/s} \][/tex]

1. Mean (\( \mu \)):

[tex]\[ \mu = \frac{12.7 + 13.4 + 12.6 + 13.3}{4} \][/tex]

[tex]\[ \mu = \frac{51}{4} \][/tex]

[tex]\[ \mu = 12.75 \, \text{km/s} \][/tex]

2. Standard deviation (\( \sigma \)):

[tex]\[ \sigma = \sqrt{\frac{(12.7 - 12.75)^2 + (13.4 - 12.75)^2 + (12.6 - 12.75)^2 + (13.3 - 12.75)^2}{4}} \][/tex]

[tex]\[ \sigma = \sqrt{\frac{0.05^2 + 0.65^2 + (-0.15)^2 + 0.55^2}{4}} \][/tex]

[tex]\[ \sigma = \sqrt{\frac{0.0025 + 0.4225 + 0.0225 + 0.3025}{4}} \][/tex]

[tex]\[ \sigma = \sqrt{\frac{0.75}{4}} \][/tex]

[tex]\[ \sigma = \sqrt{0.1875} \][/tex]

[tex]\[ \sigma \approx 0.433 \, \text{km/s} \][/tex]

3. Standard deviation of the mean (\( \sigma_\bar{x} \)):

[tex]\[ \sigma_\bar{x} = \frac{0.433}{\sqrt{4}} \][/tex]

[tex]\[ \sigma_\bar{x} = \frac{0.433}{2} \][/tex]

[tex]\[ \sigma_\bar{x} \approx 0.217 \, \text{km/s} \][/tex]

So, the mean speed is [tex]\( 12.75 \)[/tex] km/s, the standard deviation is approximately [tex]\( 0.433 \)[/tex] km/s, and the standard deviation of the mean is approximately [tex]\( 0.217 \)[/tex] km/s.

Answer:

A Histogram will be used to represent the size of right wrist of the random sample of newborn infants.

Step-by-step explanation:

A histogram is the graphical representation of the frequency distribution in the given sample. As the value of circumference can be a positive real number, therefore a Histogram with class boundaries can be formed such that the overall frequency of a wrist size is also visible in the graph.

Also as the distribution will be of continuous nature thus a histogram is a more suitable option as compared to a bar or stem and leaf graph.