Although beginning salaries vary greatly according to your field of study, the equation s = 2806.6t + 32,558 can be used to approximate and to predict average
beginning salaries for candidates with bachelor's degrees. The variable s is the starting salary and t is the number of years after 1995. Approximate when
beginning salaries for candidates will be greater than $60,000

Beginning salaries for candidates will be greater than $60,000 in the year
(Round to the nearest whole year.)
​

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.

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

Answer:

The probability that there will be a total of 7 defects on four units  is 0.14.

Step-by-step explanation:

A Poisson distribution describes the probability distribution of number of success in a specified time interval.

The probability distribution function for a Poisson distribution is:

[tex]P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}, x=0,1,2,3,...[/tex]

Let X = number of defects in a unit produced.

It is provided that there are, on average, 2 defects per unit produced.

Then in 4 units the number of defects is, [tex](2\times4)=8[/tex].

Compute the probability of exactly 7 defects in 4 units as follows:

[tex]P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}\\P(X=7)=\frac{e^{-8}8^{7}}{7!}\\=\frac{0.0003355\times2097152}{5040}\\ =0.1396\\\approx0.14[/tex]

Thus, the probability of exactly 7 defects in 4 units is 0.14.

Answer: Hello! Apparently, your question is incomplete since the rest of it is missing and we need the alternatives to work on. Fortunately, we were able to find the whole of it so we can help you! Here it goes:

Wes and Tina are a married couple and provide financial assistance to several persons during the current year. In all of the situations​ below, assume that any dependency tests not mentioned have been met.

Requirement

For each​ situation, determine whether the individuals qualify as Wes and​ Tina's dependents. ​(Select the best possible answer in each​ case.)

Requirement a. Brian is age 24 and Wes and​ Tina's son. He is a​ full-time student and lives in an apartment near campus. Wes and Tina provide over​ 50% of his support. Brian works as a waiter and earned​ $4,200.

Answer: He can't be considered a dependent because his income is higher than $4050. Also, he can't be considered a child since he is older than 23.

Requirement b. Same as Part a except that Brian is a​ part-time student.

Answer: The status of beig a student doesn't change the fact that he is over 23, so he is still not be claimed a dependent since he can't be considered a child.

Requirement c. Sherry is age 22 and Wes and​ Tina's daughter. She is a​ full-time student and lives in the college dormitory. Wes and Tina provide over​ 50% of her support. Sherry works​ part-time as a bookkeeper and earned​ $5,000.

Answer: She may be claimed as a dependent since she meets the four relationship requirements, age, abode and support. Also, being a student is not a strong influence in this case.

Requirement d. Same as Part c except that Sherry is a​ part-time student.

Answer: Under this condition, she can't be considered a dependent since she becomes an unqualified child. She's not a full-time student and over 18 years old.

Requirement e. ​Granny, age​ 82, is​ Tina's grandmother and lives with Wes and Tina. During the current​ year, Granny's only sources of income were her Social Security of​ $4,800 and interest on U.S. bonds of​ $4,500. Granny uses her income to pay for​ 40% of her total​ support, Wes and Tina provide the remainder of​ Granny's support.

Answer: Granny wouldn't be eligible, since her bonds are higher than $4050 , not considering her Social Security income.

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:

He will have to invest $20,519.84 today.

Step-by-step explanation:

We can solve this question using the simple interest formula:

This is a simple interest problem.

The simple interest formula is given by:

[tex]E = P*I*t[/tex]

In which E are the earnings, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.

After t years, the total amount of money is:

[tex]T = E + P[/tex].

In this problem, we have that:

[tex]I = 0.077, t = 6, T = 30,000[/tex]

So

[tex]T = E + P[/tex].

[tex]E + P = 30000[/tex]

[tex]E = 30000 - P[/tex]

So

[tex]E = P*I*t[/tex]

[tex]30000 - P= P*0.077*6[/tex]

[tex]30000 - P = 0.462P[/tex]

[tex]1.462P = 30000[/tex]

[tex]P = \frac{30000}{1.462}[/tex]

[tex]P = 20519.84[/tex]

He will have to invest $20,519.84 today.

Kevin Hall needs to invest approximately $19,249.38 today to have $30,000 in six years at an annual interest rate of 7.70%, compounded annually.

To determine how much Kevin Hall should invest today to have $30,000 in six years with a 7.70% annual interest rate, we'll use the formula for present value (PV) of a future amount, which is:

PV = FV / (1 + r)ⁿ

Where:

Plugging in the values:

PV = 30,000 / (1 + 0.077)⁶

Calculating the denominator:

(1 + 0.077)⁶ ≈ 1.5583

Thus,

PV = 30,000 / 1.5583 ≈ $19,249.38

Kevin Hall should invest approximately $19,249.38 today to have $30,000 in six years with a 7.70% annual interest rate, compounded annually.

Answer:

So to maximize profit 24 downhill and 20 cross country shouldbe produced

Step-by-step explanation:

Let X be the number of downhill skis and Y the number of cross country skis.

Time required for manufacturing and finishing each ski are: manufacturing time per ski, downhill 2.5 hours, cross country 1.5 hours

Finishing time per ski: downhill 0.5 hours, cross country 1.5 hours.

Total manufacturing time taken = (2.5) x+ (1.5+) y = 2.5x+1.5y≤90

total finishing time taken = 0.5x+1.5 y≤42

Profit function

Z = 50x+50y

Objective is to maximize Z

Solving the two equations we get intersecting point is

(x,y) = (24,20)

In the feasible region corner points are (0.28) (36,0)

Profit for these points are

i) 2200 for (24,20)

ii) 1400 for (0,28)

iii) 1800 for (36,0)

So to maximize profit 24 downhill and 20 cross country shouldbe produced.

To determine the optimal production quantity for each type of ski to achieve maximum profit, set up a system of equations using manufacturing time, finishing time, and profit. Graph the equations and find the intersection point.

To determine how many of each kind of ski should be produced to achieve a maximum profit, we can set up a system of equations.

Let x be the number of downhill skis and y be the number of cross country skis.

The manufacturing time equation is 2.5x + 1.5y ≤ 90.

The finishing time equation is 0.5x + 1.5y ≤ 42.

The profit equation is 50x + 50y ≤ P, where P is the maximum profit.

We can graph these equations and find the intersection point, which represents the optimal production quantity for each type of ski.

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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.

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:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

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

In which a is the probability that a student is proficient in reading but not mathematics and [tex]A \cap B[/tex] is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

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

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

[tex](A \cup B) + C = 1[/tex]

In which

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

65% were found to be proficient in both reading and mathematics.

This means that [tex]A \cap B = 0.65[/tex]

78% were found to be proficient in mathematics

This means that [tex]B = 0.78[/tex]

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

[tex]0.78 = b + 0.65[/tex]

[tex]b = 0.13[/tex]

85% of the students were found to be proficient in reading

This means that [tex]A = 0.85[/tex]

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

[tex]0.85 = a + 0.65[/tex]

[tex]a = 0.20[/tex]

Proficient in at least one:

[tex](A \cup B) = a + b + (A \cap B) = 0.20 + 0.13 + 0.65 = 0.98[/tex]

What is the probability that the student is proficient in neither reading nor mathematics?

[tex](A \cup B) + C = 1[/tex]

[tex]C = 1 - (A \cup B) = 1 - 0.98 = 0.02[/tex]

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Answer:

a. 0.8125

b. 0.1875

c. 0.03125

Step-by-step explanation:

from the information given, we can come up with the following data

Total No of students =128

Total Nos of students going to college = 48+56=104

Total Nos of students not going to college and on the honor roll= 52-48=4

a. To determine the probability of students going to college we have

P(going to college) = (total number of students going to college)/total number of students

P(going to college) = 104/128

P(going to college) = 0.8125

b. To determine the probability of students not going to college, we use the rule that says total probability is 1, hence

Pr(not going to college)=1-Pr(going to college)

Pr(not going to college)=1-0.8125

Pr(not going to college)=0.1875

c. To determine the probability of students NOT going to college  and on the pay roll we have

Pr = (Total Number of students not going to college and on the honor roll)/total number of students

Pr=4/128

Pr=0.03125

To find the probabilities, we use conditional probability, complement rule, and joint probability rule.

To answer this question, we need to use the concept of conditional probability. Let's break it down:

(a) To find the probability that a student is going to college, we can add the probabilities of two mutually exclusive events: being on the honor roll and going to college, and not being on the honor roll and still going to college. So, P(going to college) = P(on honor roll) * P(going to college | on honor roll) + P(not on honor roll) * P(going to college | not on honor roll).

(b) To find the probability that a student is not going to college, we can use the complement rule: P(not going to college) = 1 - P(going to college).

(c) To find the probability that a student is not going to college and on the honor roll, we can use the joint probability rule: P(not going to college and on honor roll) = P(on honor roll) * P(not going to college | on honor roll).

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Answer:

The question is incomplete, here is the complete question ; Quantity A of an ideal gas is at absolute temperature T, and a second quantity B of the same gas is at absolute temperature 2T. Heat is added to each gas, and both gases are allowed to expand isothermally. If both gases undergo the same entropy change, is more heat added to gas A or gas B?

a. More heat is added to gas A

b. More heat is added to gas B

c.The same amount of heat is added to each gas

Option B is the correct answer = more heat is added to gas B

Step-by-step explanation:

Considering dQ = dS/T

dQ(A) = dS/T

dQ(B) = dS/2T

From this, it implies that dQ(B) = dQ(A)/2

and as such, more heat is added to gas B or gas B will undergo the greater entropy change

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:

Step-by-step explanation:

3/(4x-15)/5

3÷(4x-15)/5

3 x 5/(4x-15) = 15/(4x-15)

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:

The answer to your question is [tex]\frac{(x+1)^{2}(x - 23)}{(x-7)^{3}}[/tex], because the directions says to give the answer in factor form.

Step-by-step explanation:

                               [tex]\frac{3(x+1)^{2}(x-7)^{2}- (x+1)^{3}(2)(x - 7)}{(x - 7)^{4}}[/tex]

Factor like terms   (x - 7)(x + 1)²

                               [tex]\frac{(x - 7)(x + 1)^{2}[3(x - 7) - 2(x + 1)}{(x - 7)^{4}}[/tex]

Simplify

                              [tex]\frac{(x+1)^{2}[3(x - 7) - 2(x+1)]}{(x-7)^{3}}[/tex]

Expand

                              [tex]\frac{(x+1)^{2}[3x - 21 -2x - 2]}{(x - 7)^{3}}[/tex]

Simplify

                             [tex]\frac{(x + 1)^{2}(x - 23)}{(x - 7)^{3}}[/tex]    

Answer:

The mean is 52.

Step-by-step explanation:

The mean is the sum of all elements divided by the number of elements.

In this problem, we have that:

Elements

56 47 48 52 62 59 49 56 43 48

Sum

[tex]56+47+48+52+62+59+49+56+43+48 = 520[/tex]

Number of Elements

10 elements

The mean is

[tex]M = \frac{520}{10} = 52[/tex]

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

Learn more on hydrostatic pressure here;

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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:

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]

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:

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:

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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To estimate the number of high-speed Internet plans that cost between $51.48 and $76.72, we can use the standard normal distribution and the z-score formula. The estimated number of plans is 587.

To estimate the number of plans that cost between $51.48 and $76.72, we can use the standard normal distribution and the z-score formula. First, we calculate the z-scores for both costs:

z1 = (51.48 - 64.1) / 12.62 = -1.003

z2 = (76.72 - 64.1) / 12.62 = 1.003

Next, we find the area under the standard normal curve between these two z-scores using a z-table or a calculator. Let's assume the area is approximately 0.6827.

Finally, we multiply this area by the total number of homeowners surveyed (859) to estimate the number of plans that fall within this cost range:

Number of plans = 0.6827 * 859 = 586.92

Rounding to the nearest whole number, the estimated number of plans is 587.

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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:

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:

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

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:

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:

Multiple Answers

Step-by-step explanation:

You forgot to put all the question, I attached it to the answer.

The questions we need to response are:

a)What type of decision is Ken facing?

There are three types of decision in probability. This are:

Risky that cover when the event is known and you know the chances of success.

Of uncertainty that cover for a known event but you dont know the  possibilities of success.

Of ignorance that cover a unknown event, with unknown possibilities of succes.

So the decision Kenneth is facing is of uncertainty.

(b)What decision criterion should he use?

The criterion decision he should take would be Maximax.

This states that you should select the option that have the maximum gain.

(c)What alternative is best? The best option should be sub 100 because of the decision criterion we decided to use. Sub 100 has the maximum gain.

Final answer:

Kenneth Brown faces a decision on purchasing equipment for his company, Brown Oil, Inc. Business management principles and financial implications play a key role in this decision-making process.

Explanation:

Kenneth Brown, the principal owner of Brown Oil, Inc., faces the decision to purchase equipment for the company due to competition. His alternatives are presented in the form of a table detailing the costs under favorable and unfavorable market conditions.

In this scenario, business management principles come into play as Ken evaluates the potential costs and benefits of investing in new equipment for Brown Oil, Inc. This decision-making process is crucial for the company's competitiveness and growth in the market.

Considering the financial implications and potential outcomes, Ken will need to assess the risks and rewards associated with each equipment option to make an informed decision that aligns with the company's goals and future success.