Determine the truth values of these statements:

(a) The product of x2 and x3 is x6 for any real number x.
(b) x2 > 0 for any real number x.
(c) The number 315 − 8 is even.
(d) The sum of two odd integers is even.

Answer:

The slope is 1.

Step-by-step explanation:

A first order function has the following format

[tex]y = ax + b[/tex]

In which a is the slope

(6, -12).

This means that when [tex]x = 6, y = -12[/tex]

So

[tex]y = ax + b[/tex]

[tex]-12 = 6a + b[/tex]

(15. -3)

This means that when [tex]x = 15, y = -3[/tex]

So

[tex]y = ax + b[/tex]

[tex]-3 = 15a + b[/tex]

We have to solve the following system of equations:

[tex]-12 = 6a + b[/tex]

[tex]-3 = 15a + b[/tex]

We have to find a

In the second equation i will write as:

[tex]b = -3 - 15a[/tex]

Replacing in the first

[tex]-12 = 6a + b[/tex]

[tex]-12 = 6a - 3 - 15a[/tex]

[tex]-9 = -9a[/tex]

[tex]9a = 9[/tex]

[tex]a = \frac{9}{9}[/tex]

[tex]a = 1[/tex]

The slope is 1.

Answer:

1

Step-by-step explanation:

Slope = (y2-y1)/(x2-x1)

= (-3-(-12))/(15-6)

= (-3+12)/(9)

= 9/9

= 1

Answer:

c) Either A or B or both occur.

Step-by-step explanation:

Suppose that we have two events

Event A

Event B

We have that:

[tex]A = a + (A \cap B)[/tex]

In which a a happens and b does not and [tex]A \cap B[/tex] is the probability that aboth events happen

By the same logic, we have that:

[tex]B = b + (A \cap B)[/tex]

The union of events A and B is:

[tex](A \cup B) = a + b + (A \cap B)[/tex]

Which includes either one of them or both.

So the correct answer is:

c) Either A or B or both occur.

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:

The answer is 332.8 lb

Step-by-step explanation:

See attached picture for the solution

The force on one of the ends due to hydrostatic pressure is 332.8lb

Data;

The force due to hydrostatic pressure can be calculated as

From the attached diagram;

[tex]F = pressure * area\\density = 62.4 lb/ft^3\\depth of water = 2 - y\\pressure = (2 - y)(62.4)\\pressure = 124.8 - 62.4y\\[/tex]

We can proceed as

[tex]x^2 + (y - 2)^2 = 2^2\\x^2 = 4 - (y - 2)^2\\x = +- \sqrt{4y - y^2}\\[/tex]

this implies that

[tex]2x = 2\sqrt{4y - y^2}[/tex]

The area is given as

[tex]\delta A = (2x)*\delta y\\\delta A = 2\sqrt{4y - y^2 \delta y}[/tex]

The force would be given by

[tex]\delta F = (2-y)(62.4)(2\sqrt{4y - y^2})\delta y[/tex]

The total force is given by

[tex]F = \int\limits^2_0 {(2-y)(62.4)(2\sqrt{4y - y^2}) } \, dy\\F = 124.8\int\limits^2_0 {(2-y)(\sqrt{4y - y^2}) } \, dy\\F = 124.8[-\frac{1}{3}y(y -4)(\sqrt{4y -y^2}]_0^2\\F = 124.8[-\frac{1}{3}(2)(2-4)\sqrt{4(2)-2^2}\\ F = 332.8lb[/tex]

The force on one of the ends due to hydrostatic pressure is 332.8lb

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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.

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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.

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 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:

23040 bats

Step-by-step explanation:

Let N(t) be the number of bats at time t

We know that exponential function

[tex]y=ab^t[/tex]

According to question

[tex]N(t)=ab^t[/tex]

Where t (in years)

Substitute t=2 and N(2)=180

[tex]180=ab^2[/tex]...(1)

Substitute t=5 and N(5)=1440

[tex]1440=ab^5[/tex]...(2)

Equation (1) divided by equation (2)

[tex]\frac{180}{1440}=\frac{ab^2}{ab^5}=\frac{1}{b^{5-2}}[/tex]

By using the property [tex]a^x\div a^y=a^{x-y}[/tex]

[tex]\frac{1}{8}=\frac{1}{b^3}[/tex]

[tex]b^3=8=2\times 2\times 2=2^3[/tex]

[tex]b=2[/tex]

Substitute the values of b in equation (1)

[tex]180=a(2)^2=4a[/tex]

[tex]a=\frac{180}{4}=45[/tex]

Substitute t=9

[tex]N(9)=45(2)^9=23040 bats[/tex]

Hence, after 9 years the expected bats in the colony=23040 bats

To find the number of bats expected to be in the colony after 9 years, we can use the equation for exponential growth. By plugging in the given population values and solving for the growth rate, we can then calculate the population after 9 years.

To find the number of bats expected to be in the colony after 9 years, we need to determine the growth rate. Let's use the equation for exponential growth: N = P * e^(kt), where N is the final population, P is the initial population, e is the base of the natural logarithm, k is the growth rate, and t is the time.

We are given the population after 2 years (P = 180) and after 5 years (P = 1440). Plugging these values into the equation, we can solve for k:

180 = P * e^(2k) and 1440 = P * e^(5k).

Dividing the second equation by the first equation, we can eliminate P and solve for e^(3k): 8 = e^(3k).

Taking the natural logarithm of both sides, we get: ln(8) = 3k.

Finally, solving for k, we have: k = ln(8) / 3.

Now, we can use the calculated value of k to find the population after 9 years:

N = P * e^(9k).

Plugging in the value of P and k, we get: N = 180 * e^(9 * ln(8) / 3). Calculating this expression gives us the expected number of bats in the colony after 9 years.

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Answer:

(A) The expected loss is $0.045.

(B) The variance and standard deviation of physical damage loss are $2,025 and $45 respectively.

Step-by-step explanation:

The annual probability of Bespin Car Rental's cars being destroyed is 1 in a million, i.e 0.000001.

It is assumed that the car is either destroyed or there was no loss suffered.

The loss amount in case the car is destroyed is, $45,000.

(A)

The distribution for physical damage loss is displayed in the table below.

The Expected value of physical damage loss is:

[tex]E(X)=\sum xP(X)=(45000\times0.000001)+(0\times0.999999)=0.045[/tex]

Thus, the expected loss is $0.045.

(B)

The variance of a random variable X is: Var (X) = E (X²) - [E (X)]².

The variance of physical damage loss is:

Compute the variance as follows:

[tex]Var(X)=E(X^{2})-[E(X)]^{2}\\=\sum x^{2}P(X)-[\sum xP(X)]^{2}\\=[(45000^{2}\times0.000001)+(0^{2}\times0.999999)]-(0.045)^{2}\\=2025-0.002025\\=2024.997975\\\approx2025[/tex]

The standard deviation of physical damage loss is:

[tex]SD=\sqrt{Var(X)}=\sqrt{2025}=45[/tex]

Thus, the variance and standard deviation of physical damage loss are $2,025 and $45 respectively.

Answer:

6 quarts

Step-by-step explanation:

60% 3 gallons = 1.8 gallons of broth

water = 0% broth

1.8=1.2+0.4x

0.6=0.4x

x=1.5

1.5 gallons = 6 quarts

The amount of water must be added to 3 gallons of soup which is 60% chicken broth to make the soup 40% chicken broth is 6 quartz.

In other terms, it is a mathematical statement stating that "this is equivalent to that." It appears to be a mathematical expression on the left, an equal sign in the center, and a mathematical expression on the right.

Given:

There are 3 gallons of soup that is 60% chicken broth to make the soup 40% chicken broth,

Assume the number of gallons is x then write the equation as shown below,

60% 3 gallons = 1.8 gallons of broth

water = 0% broth

1.8 = 1.2 + 0.4x

0.6 = 0.4x

x  = 1.5

As we know that 1 gallon = 4 quartz,

1.5 gallons = 6 quarts

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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:

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:

a) X1+X2 have distribution Bi(n1+n2, p)

b)

P(X1+X2 = 1 | X2 = 0) =  np(1-p)ⁿ¹⁻¹

P(X1+X2 = 1| X2 = 1) = (1-p)ⁿ¹

P(X1 + X2 = 1) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹

Step-by-step explanation:

Since both variables are independent but they have the same probability parameter, you can interpret that like if the experiment that models each try in both variables is the same. When you sum both random variables toguether, what you obtain as a result is the total amount of success in n1+n2 tries of the same experiment, thus X1+X2 have distribution Bi(n1+n2, p).

b)

Note that, if X2 = k, then X1+X2 = 1 is equivalent to X1 = 1-k. Since X1 and X2 are independent, then P(X1+X2 = 1| X2 = K) = P(X1=1-k|X2=k) = P(X1 = 1-k).

If k = 0, then this probability is equal to P(X1 = 1) = np(1-p)ⁿ¹⁻¹

If k = 1, then it is equal to P(X1 = 0) = (1-p)ⁿ¹

Thus,

P(X1+X2 = 1) = P(X1+X2 = 1| X2 = 1) * P(X2=1) + P(X1+X2 = 1| X2 = 0) * P(X2 = 0) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹

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.

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:

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. [tex]P = 1000 - Ce^{-\frac{kt^2}{2} }[/tex]

b. [tex]C = 950[/tex]

c. [tex]P = 1000 - 950e^{-\frac{kt^2}{2} }[/tex]

Step-by-step explanation:

a. Let the number of penguins who have the disease t days after the outbreak be P

Initial number of penguins = 1000

Therefore, current number of penguins = 1000 - P

And the rate of spread of disease according to the statement is

[tex]\frac{dP}{dt}\alpha t(1000-P)\\\frac{dP}{dt}=kt(1000-P)[/tex]

where k is the constant of proportionality

[tex]\frac{dP}{1000-P}=kt.dt[/tex]

Integrating both sides

[tex]-ln(1000-P) = \frac{kt^2}{2}+c\\\frac{1}{(1000-P)} = Ce^{\frac{kt^2}{2} }\\ (1000-P) = Ce^{-\frac{kt^2}{2} }\\P = 1000 - Ce^{-\frac{kt^2}{2} }[/tex]

b. Seeing as 50 penguins had the disease initially,

t = 0

P = 50

The general solution of the differential solution becomes

50 = 1000 - C (anything raised to the power of 0 is 1, hence e is equal to 1)

[tex]C = 1000 - 50 = 950[/tex]

c. Therefore, the solution that satisfies the initial condition is

[tex]P = 1000 - 950e^{-\frac{kt^2}{2} }[/tex]

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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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:

Your commission for the month is $1,553.35.

Step-by-step explanation:

You made 2 sales.

In each you got a commission of 3%. Your total commission is the sum of both commisions. So

Sale of $45,050:

You got 3% of the sale. So

0.03*45050 = $1,351.5

Sale of $6,728.25:

You got 3% of this sale. So

0.03*6728.25 = $201.85

Total commision for the month:

$1,351.5 + $201.85 = $1,553.35.

Your commission for the month is $1,553.35.

Answer:

1,553.35

The commission for the month is a total of 1,553.35

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

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: 9/64

Step-by-step explanation:

Probability is a chance of prediction. It's a measure of how an event is likely to happen.

P(A) = Number of favorable outcome/Total Number of favorable outcome

Let's make W the correct answer and C the right answer.

The probability of choosing the correct answer from multiple choice question:

P(C) = 1/4

The probability of choosing the wrong answer from multiple choice question:

P(C) =1/4

P(W)= 1 - 1/4

P(W) = 3/4

Therefore, to find P(WWC)

P(WWC) = P(W) × P(W) × P(C)

P(WWC) = 3/4 ×3/4 × 1/4

P(WWC) = 9/64

The probability is 9/64.

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

Learn more : https://brainly.com/question/12474772

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:

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

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:

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.