write the recursive rule for the sequence shown in each table.
position, n 1 2 3 4 5
term, f(n) 5 18 31 44 57


1 2 3 4 5 6 7
65 54 43 32 21 10 -1


1 2 3 4 5 6 7
-9 6 21 36 51 66 81


1 2 3 4 5 6 7 8
17 13 9 5 1 -3 -7 -11

In this scenario, there are two workers in a team, and each worker can either work hard or shirk. The probability of the overall project succeeding is dependent on the efforts of each worker. If both workers shirk, the probability of success is p0. If one worker shirks and the other works hard, the probability of success is p1. Finally, if both workers work hard, the probability of success is p2, where p2>p1>p0.

The cost of effort is c, and the principal cannot observe the individual efforts of each worker, but only the success or failure of the whole project. The challenge is to design an optimal contract that encourages both workers to exert effort all the time.

The optimal contract would offer a payment scheme to both workers that would incentivize them to work hard. If the workers work hard and the project succeeds, they receive a reward. If the workers shirk, they receive no reward.

The workers' efforts in this scenario are substitutes for each other. This is because if one worker shirks, the probability of success decreases, and the other worker would have to work harder to compensate for the first worker's lack of effort. Therefore, both workers must work hard to maximize the probability of success.

In conclusion, an optimal contract must be designed that encourages both workers to work hard and rewards them for the successful completion of the project. Additionally, the efforts of both workers in this scenario are substitutes for each other.

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Answer:

-1

Step-by-step explanation:

4x + 13 = 9

4x + 13 - 13 = 9 -13

4x = -4

-4/4 = -1

X = -1

Answer:

-1.

Step-by-step explanation:

4x + 13 = 9

4x =  9 -13

4x = -4

x = -4/4

X = -1

The required, equation to determine the number of people is T + (T + 5,500) + 2T + (2T + 300) = 36,700 and there were 10,300 people at the fair on Saturday.

Let's call the number of people who attended the fair on Thursday "T". Then we can use the information in the problem to set up an equation:

Friday's attendance = T + 5,500

Saturday's attendance = 2T

Sunday's attendance = 2T + 300

Total attendance = T + (T + 5,500) + 2T + (2T + 300) = 36,700

Simplifying the equation, we get:

6T + 5,800 = 36,700

6T = 30,900

T = 5,150

Therefore, the attendance on Thursday was 5,150 people. We can use this information to find the attendance on Saturday:

Saturday's attendance = 2T = 2(5,150) = 10,300

Thus, there were 10,300 people at the fair on Saturday.

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If by the baking of 81 muffins, the baker has completed 37.5% of the order. The customer placed an order for 216 muffins.

Percent refers to the ratio of numbers where the denominator is 100. The percentage of a given number is calculated as the percentage divided by 100 and multiply the given number.

Percentage of baked muffins = 37.5%

Number of the baked muffin = 81

Let the number of muffins ordered be x

37.5% of x = 81

37.5/100 * x = 81

0.375x = 81

x = 81/0.375

x = 216

Thus, 216 muffins were ordered by the customer.

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Answer:1/2

Step-by-step explanation:

If they both ate 2/8 of the pie, that means they ate 4/8 of the pie in total, which means that they ate 1/2 when simplified. If they ate one half of the pie, it also means there is 1/2 of the pie left.

The length of the two diagonals of a rhombus if one is twice as long as the other are 10 and 20.

Given that the length of two diagonals of the rhombus are 10 and 20.

Let d1 and d2 be the lengths of the diagonals of the rhombus. Since the area of the rhombus is 100, we have:

d1 * d2 / 2 = 100

We are also given that one diagonal is twice as long as the other, so:

d1 = 2d2

Substituting d1 = 2d2 into the first equation, we get:

2d2 * d2 / 2 = 100

d2^2 = 100

d2 = 10 or -10 (we ignore the negative solution)

Since d1 = 2d2, we have d1 = 20.

Therefore, the length of the two diagonals of the rhombus are 10 and 20.

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The matrix A with eigenvalues λ1 = 4, λ2 = 1, and corresponding eigenvectors v1 = (2, 1)t, v2 = (1,0)t is:

A = [v1 v2] = [2 1; 1 0]

Yes, it is possible to write a matrix A such that its eigenvalues and corresponding eigenvectors are λ1 = 4, λ2 = 1, and v1 = (2, 1)t, v2 = (1,0)t.

Let A be the matrix with columns given by the eigenvectors of A:

A = [v1 v2] = [2 1; 1 0]

Then, we can calculate the eigenvalues of A by finding the roots of its characteristic polynomial:

|A - λI| = |2-λ 1; 1 0-λ| = (2-λ)(-λ) - 1 = λ^2 - 2λ - 1

Solving for λ, we get:

λ1 = 4, λ2 = 1

which are the desired eigenvalues.

Next, we can find the corresponding eigenvectors by solving the equations (A - λI)x = 0 for each eigenvalue:

For λ1 = 4:

(A - λ1I)x = ([2 1; 1 0] - [4 0; 0 4])x = [-2 1; 1 -4]x = 0

Solving the system of equations, we get x1 = -1 and x2 = -1/2, so the eigenvector corresponding to λ1 is:

v1 = [-1; -1/2]

For λ2 = 1:

(A - λ2I)x = ([2 1; 1 0] - [1 0; 0 1])x = [1 1; 1 -1]x = 0

Solving the system of equations, we get x1 = 1 and x2 = -1, so the eigenvector corresponding to λ2 is:

v2 = [1; -1]

Therefore, the matrix A with eigenvalues λ1 = 4, λ2 = 1, and corresponding eigenvectors v1 = (2, 1)t, v2 = (1,0)t is:

A = [v1 v2] = [2 1; 1 0]

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Answer:

A. 38, 49, 60

Step-by-step explanation:

1. Find the difference between each number in the sequence

To go from -6 to 5, you add 11

To go from 5 to 16, you add 11

To go from 16 to 27, you add 11.

Therefore, the difference in all of the numbers is 11, so the pattern should continue and you should add 11 to the last number (27) making the next numbers 38, 49, 60

Answer:

  (a) Interest: $22.64; Balance: $2733.89

Step-by-step explanation:

You want a 3-month schedule of payments, interest, and the account balance for a sinking fund earning 15% APR on deposits of $900 made at the end of each month.

The interest earned by the account in any given month is the product of the monthly interest rate and the ending balance for the previous month.

The monthly interest rate is 15%/12 = 1.25%. For the second month, interest will be ...

  $900 × 1.25% = $11.25

For the third month, interest will be ...

  $1811.25 × 1.25% = $22.64

After the payment at the end of the third month, the account balance will be ...

  $1811.25 +900.00 +22.64 = $2733.89

__

Additional comment

Total interest earned is $33.89 by the end of the third month. The answer choices seem to be telling you to interpret the question as asking for the interest earned in the third month, not the total interest earned.

Answer: 17/25 miles per hour

Step-by-step explanation:

In order to find your answer, you will need to divide 3 and 2/5 by 5 hours

or (3.4/5) which equals 0.68.

As a fraction, 0.68 is 17/25 which is your answer :)

And if you want to check your work, simply multiply 0.68 x 5 and you get 3.4 (which is 3 and 2/5's)!

Good Luck!

The drive-thru that typically has more wait time, and why is C. Super Fast Food, because it has a larger median

According on the information supplied, Super Fast Food normally has a greater wait time. Although the Burger Quick box is larger, indicating a greater range of wait times, the median (15.5) is still lower than the Super Fast Food (12).

Furthermore, the Burger Quick line ends at 30, indicating that there are some extreme outliers with extremely long wait times, which could raise the mean wait time. As a result, the correct answer is Super Fast Food, which has a higher median.

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For 420 minutes of calls will the costs of the two plans be equal.

We have,

The first plan has a $21 monthly fee and charges an additional $0.09 for each minute of calls.

The second plan has no monthly fee but charges $0.14 for each minutes.

So, the equation can be set as

0.14x = 0.09x +21

where x be the number of minutes.

0.14x - 0.09x = 21

0.05x = 21

x = 420 minutes

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The index form is simplified to give 3m/2n

It is important to note that the index forms are models that are used for the representation of variables or numbers that too small or large in more convenient forms.

Some of the rules of index forms are;

From the information given, we have that;

3m³/ n -²• 2m -² n³

Add the like exponents of the denominator

3m³/2m² n³⁻²

add the values

3m³/2m⁻²n¹

Now, subtract the exponents

3/2 n⁻¹ m¹

Then, we have;

3m/2n

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The mean of all six sample means is equal to the population mean (18,750), the sample mean is an unbiased estimator of the population mean.

First, let's calculate the mean for each of the given samples:

1. R, B: (15,000 + 18,000) / 2 = 16,500 (already given)

2. R, G: (15,000 + 22,000) / 2 = 18,500 (already given)

3. R, W: (15,000 + 20,000) / 2 = 17,500 (already given)

4. B, G: (18,000 + 22,000) / 2 = 20,000 (already given)

5. B, W: (18,000 + 20,000) / 2 = 19,000 (already given)

6. G, W: (22,000 + 20,000) / 2 = 21,000 (already given)

Now, let's calculate the mean of all six sample means:

(16,500 + 18,500 + 17,500 + 20,000 + 19,000 + 21,000) / 6 = 112,500 / 6 = 18,750

The mean of all six sample means is 18,750.

Next, let's calculate the population mean:

(15,000 + 18,000 + 22,000 + 20,000) / 4 = 75,000 / 4 = 18,750

The population mean is 18,750.

Since the mean of all six sample means is equal to the population mean (18,750), the sample mean is an unbiased estimator of the population mean.

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Every natural number n > 2 is a product of prime numbers.

To prove that every natural number n > 2 is a product of prime numbers, we will use the principle of complete induction (CVI).

First, let's establish the base case. For n = 3, we know that 3 is a prime number, and therefore it is a product of prime numbers. This satisfies the base case.

Now, let's assume that for some natural number k > 2, every natural number between 3 and k (inclusive) is a product of prime numbers. We want to prove that this implies that the next natural number, k+1, is also a product of prime numbers.

There are two possibilities for k+1: either it is a prime number itself, or it is composite (i.e. not prime). Let's consider each case separately.

Case 1: k+1 is a prime number.
If k+1 is a prime number, then it is obviously a product of prime numbers (since it is itself prime). Therefore, our assumption that every natural number between 3 and k is a product of prime numbers implies that k+1 is also a product of prime numbers.

Case 2: k+1 is composite.
If k+1 is composite, then it can be written as the product of two natural numbers a and b, where a and b are both greater than 1. Since a and b are both less than k+1, we know that they are both products of prime numbers (by our assumption). Therefore, k+1 can be written as the product of prime numbers (namely, the prime factors of a and b).

Since we have established the base case and shown that our assumption implies the next natural number is a product of prime numbers, we can conclude by CVI that every natural number n > 2 is a product of prime numbers.

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If the arrow lands on yellow you win $75, blue gives $25, green gives $10, and red gives $1. The expected value of the game is $27.75, written as a decimal rounded to two places.

To find the expected value of the game, we need to multiply the value of each piece by the probability of landing on that piece, and then add up all the products.

First, let's determine the probability of landing on each colored piece of the wheel. Since each piece is equally likely, we can find the probability by dividing 1 by the number of pieces.
There are 4 colors on the wheel (yellow, blue, green, and red), so the probability of landing on any color is 1/4 or 0.25.
Now, let's calculate the expected value of the game:

Expected Value = (Probability of Yellow) × (Value of Yellow) + (Probability of Blue) × (Value of Blue) + (Probability of Green) × (Value of Green) + (Probability of Red) × (Value of Red)

The probability of landing on yellow is 1/4, so the value of yellow is $75.

The probability of landing on blue is also 1/4, so the value of blue is $25.

The probability of landing on the green is 1/4, so the value of green is $10.

The probability of landing on red is also 1/4, so the value of red is $1.

Now we can calculate the expected value:

Expected value = (1/4) x $75 + (1/4) x $25 + (1/4) x $10 + (1/4) x $1
Expected value = $18.75 + $6.25 + $2.50 + $0.25
Expected value = $27.75

So the expected value of the game is $27.75. Written as a decimal rounded to two places, the answer is $27.75.

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Since there is no way to list all the infinite sequences of 0s and 1s, we have proved that S is uncountable.

To show that S is uncountable, we need to show that there is no one-to-one correspondence between the set of infinite sequences of 0s and 1s and the set of natural numbers (which are used to count elements in a set).
We can use the diagonal argument, which is similar to the one used to show that the real numbers are uncountable.
Assume that S is countable, meaning that there is a way to list all the infinite sequences of 0s and 1s as a sequence: s1, s2, s3, ...
We can construct a new sequence t by choosing the opposite of each digit on the diagonal of the previous sequences: if the diagonal digit of s1 is 0, then the first digit of t is 1, and vice versa. Similarly, if the diagonal digit of s2 is 1, then the second digit of t is 0, and vice versa. And so on for all the diagonal digits.
The sequence t is guaranteed to be different from all the sequences in the original list, because it differs from each sequence at least at one position (the diagonal position). Therefore, t cannot be in the original list, which means that the original list did not include all the infinite sequences of 0s and 1s.
Since we have shown that there is no way to list all the infinite sequences of 0s and 1s, we have proved that S is uncountable.

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The translations of the functions are as follows

Reflection

Shift Right

Shift Up

Vertical Stretch: the coefficient of  the factor x when it is greater than one results to vertical stretch

Shift Left: this is translation of 3 units left

Vertical Shrink: the coefficient of  the factor x when it is less than one results to vertical shrink

Shift Left: this is translation of 2 units left

Vertical Stretch: the coefficient of  the factor x when it is greater than one results to vertical stretch. in this case 3/2 = 1.5 which is greater than one.

Shift Left: this is translation of 5 units left

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Demand refers to the quantity of a product or service that buyers are willing and able to purchase at a certain price level. A demand curve is a graphical representation of the relationship between the quantity of a good or service that buyers are willing to purchase at different price levels.

In the market for VCRs, the demand and supply equations are Qd = 1,000 - 6P and Qs = 4P respectively. The government decides to impose a tax of $30 per unit sold on VCRs. This will lead to a shift in the supply curve upwards by $30, resulting in a new equilibrium point.

1. To find the new equilibrium quantity with the tax, we set Qd = Qs + tax. Thus, 1,000 - 6P = 4P + 30. Solving for P, we get P = $105. The equilibrium quantity is therefore Q = 1,000 - 6($105) = 370.

2. The price paid by buyers is the same as the equilibrium price, which is $105.

3. The price paid by sellers is the equilibrium price minus the tax, which is $75.

4. The diagram below illustrates the impact of the tax on the market for VCRs. The original equilibrium point is E0, with a price of $70 and a quantity of 400. With the tax, the new equilibrium point is E1, with a higher price of $105 and a lower quantity of 370. The tax creates a vertical distance between the two equilibrium points, which represents the tax revenue collected by the government.

[Insert Diagram Here]

5. The deadweight loss due to the tax is the reduction in total surplus (consumer surplus + producer surplus) that results from the tax. Using the original equilibrium as a benchmark, the total surplus is (1/2)($70)(400) = $14,000. With the tax, the total surplus is (1/2)($75)(340) = $6,375. Thus, the deadweight loss is $7,625.

6. To calculate the consumer surplus due to the tax, we need to compare the original consumer surplus with the new consumer surplus. Using the original equilibrium as a benchmark, the consumer surplus is (1/2)($70)(400 - 70) = $5,950. With the tax, the consumer surplus is (1/2)($105)(370 - 105) = $12,145. Thus, the consumer surplus due to the tax is $6,195.

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The number of people expected to go to the Youth Wing, would be 255 people.

Find the total amount of people who will be visiting on Saturday:

= 382 ( Youth Wing ) + 461 ( Social Issues ) + 355 ( Fiction and Literature )

= 1, 198

Then the proportion of those who went to Youth Wing :

= 382 / 1, 198

= 0. 319

The estimated people going for Youth Wing on Sunday :

= 0. 319 x 800

= 255 people

In conclusion, an estimated 255 people would be going to Youth Wing.

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The smaller number is 13

Let x and y represent the unknown number

x + y= 47........equation 1

x - y= 21.........equation 2

From equation 1

x + y= 47

x= 47-y

Substitute 47-y for x in equation 2

(47-y)-y= 21

47-y-y= 21

47-2y= 21

-2y= 21-47

-2y= -26

y= 26/2

y= 13

Substitute 13 for y in equation 1

x + y= 47

x + 13= 47

x= 47-13

x= 34

Hence the smaller number is 13

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9 5/2 in radical form is equal to 4√2 + √7.

We are able to rewrite 9 5/2 in radical form through first changing it to an improper fraction.

9 5/2 = (9 x 2 + 5) / 2 = 23/2

Now, we can express 23/2 as a mixed radical by using locating the largest perfect square that may be a aspect of 23. the largest perfect square which is much less than 23 is 16 (that's 4^2).

So, we are able to write:

23/2 = 16/2 + 7/2 = 8 + 7/2

Now, we will express 8 + 7/2 in radical form as:

8 + 7/2 = 4 x 2 + 7/2 = 4√2 + √7

Consequently, 9 5/2 in radical form is equal to 4√2 + √7.

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[tex] \Large{\boxed{\sf c = 21}} [/tex]

[tex] \\ [/tex]

Solving the equation for c means finding the value of that variable that makes the equality true.

[tex] \\ [/tex]

Given equation:

[tex] \sf 9c - 73 = 6c - 10[/tex]

[tex] \\ [/tex]

To isolate c, we will move the variables to the left member by subtracting 6c from both sides of the equation:

[tex] \sf 9c - 73 - 6c = 6c - 10 - 6c \\ \\ \sf3c - 73 = - 10[/tex]

[tex] \\ [/tex]

Then, we move the constants to the right member by adding 73 to both sides of the equation:

[tex] \sf 3c - 73+ 73 = - 10 + 73 \\ \\ \sf 3c = 63[/tex]

[tex] \\ [/tex]

Finally, divide both sides of the equation by the coefficient of the variable, 3:

[tex] \sf \dfrac{3c}{3} = \dfrac{63}{3} \\ \\ \implies \boxed{ \boxed{ \sf c = 21}}[/tex]

[tex] \\ \\ [/tex]

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This is an exercise of the first degree equation with one unknown is an algebraic equality in which the unknown (generally represented by x) appears with an exponent of 1 and the rest of the terms are constants or coefficients of the unknown. These equations can be solved to find the numerical value of the unknown that satisfies the equality.

The process for solving a first degree equation involves simplifying the equation by eliminating like terms and applying algebraic operations (addition, subtraction, multiplication, and division) to solve for the unknown. It is important to remember that the same operations are applied to both sides of the equation to maintain equality.

It is possible for a first degree equation to have a unique solution, no solution, or an infinite set of solutions. A unique solution means that there is a numerical value for the unknown that satisfies the equality. If the equation has no solution, it means that there is no numerical value for the unknown that satisfies the equality. If the equation has an infinite set of solutions, it means that any numerical value of the unknown that is chosen will satisfy the equality.

Quadratic equations with one unknown are fundamental in mathematics and have applications in many areas, such as solving problems in physics, chemistry, economics, and many other fields.

9c - 73 = 6c - 10

Solving a linear equation means finding the value of the variable that makes it true.

We want all the terms containing the variable to be on the left hand side and all the constants to be on the right hand side.

First, we move the constant to the right hand side by adding the opposite of -73 to both sides.

 9c - 73 + 73 = 6c - 10 + 73

Two opposite numbers add up to zero, so we remove it from the expression.

 9c = 6c - 10 + 73

We add the constants on the right hand side.

 9c = 6c + 63

Now, we move the variable to the left side by adding the opposite of 6c to both sides.

9c - 6c = +6c - 6c + 63

Let's remember! Two opposite numbers add up to zero, so we remove them from the expression.

9c - 6c = 63

We simplify the left hand side by adding like terms.

3c = 63

To isolate the variable c on the left hand side, we have to divide both sides by 3. We have learned that a number divisible by itself is equal to 1, so we can reduce the left hand side to just c.

c = 63/3

All we have to do now is simplify the final division equation.

C = 21

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That we are 95% confident that the true population mean falls between 4.68 and 5.96. Based on this interval, it is likely that the true population mean is greater than 4.

a. The null hypothesis is that the average score on the Attributional Complexity scale is equal to or less than 4. The alternative hypothesis is that the average score is greater than 4. Symbolically:

H0: µ ≤ 4

Ha: µ > 4

b. We need to conduct a one-sample t-test, since we are comparing a sample mean to a known population mean (4). We will use a significance level of α = 0.05.

c. Using the information given, we can calculate the t-value as:

t = (x - µ) / (s / √n) = (5.32 - 4) / (0.54 / √10) = 5.04

where x is the sample mean, µ is the population mean, s is the sample standard deviation, and n is the sample size. The degrees of freedom (df) is n - 1 = 9.

At a significance level of α = 0.05 and with 9 degrees of freedom, the critical t-value is 1.833 (obtained from a t-table or calculator). Since our calculated t-value (5.04) is greater than the critical t-value (1.833), we can reject the null hypothesis.

Based on these sample data, we can say that there is evidence to suggest that the average score on the Attributional Complexity scale is greater than 4.

The p-value associated with the sample mean is less than 0.001. This means that there is less than a 0.1% chance of obtaining a sample mean of 5.32 (or higher) if the null hypothesis is true.

If the null hypothesis is true, we would expect the sample mean to be around 4. Therefore, the large difference between the sample mean (5.32) and the null hypothesis value (4) suggests that the null hypothesis is not true.

d. The 95% confidence interval can be calculated as:

CI =x ± t*(s / √n) = 5.32 ± 2.306*(0.54 / √10) = (4.68, 5.96)

This means that we are 95% confident that the true population mean falls between 4.68 and 5.96. Based on this interval, it is likely that the true population mean is greater than 4.

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Answer:

8.5 ounces.

Step-by-step explanation:

The volume of the conical cup is given by the formula V = (1/3)πr²h, where r is the radius and h is the height. Since the diameter of the cup is 3 inches, the radius is 1.5 inches. Thus, the volume of the conical cup is:

V = (1/3)π(1.5²)(3) = 7.07 cubic inches

The volume of the cylindrical cup is given by the formula V = πr²h. Since the radius is 1.5 inches and the height is 3 inches, the volume of the cylindrical cup is:

V = π(1.5²)(3) = 21.2 cubic inches

To find the weight of the water in each cup, we need to multiply the volume of each cup by the weight of 1 cubic inch of water:

Weight of water in conical cup = 7.07 × 0.6 = 4.24 ounces

Weight of water in cylindrical cup = 21.2 × 0.6 = 12.72 ounces

Therefore, the water in the cylindrical cup weighs 12.72 - 4.24 = 8.48 more ounces than the water in the conical cup. Rounded to the nearest tenth, this is 8.5 ounces.

a. the population standard deviation is not given, we cannot calculate the standard error of the mean.

b. b. The expected shape of the distribution would still be positively skewed, as the skewness of the population does not change with the sample size.

c. the probability of selecting a sample with a mean of at least $117,000 is 1 - 0.9772 = 0.0228, or about 2.28%.

d. the probability of selecting a sample with a mean of more than $107,000 is 1 - 0.0427 = 0.9573, or about 95.73%.

e. the probability of selecting a sample with a mean of more than $107,000 but less than $117,000 is the difference between these probabilities, which is 0.9772 - 0.0427 = 0.9345, or about 93.45%.

What is standard deviation?

Standard deviation is a measure of the amount of variation or dispersion in a set of data values. It shows how much the data deviates from the mean or average value.

a. The standard error of the mean is given by the formula:

SE = σ/√n

where σ is the population standard deviation, n is the sample size, and √n denotes the square root of n.

Since the population standard deviation is not given, we cannot calculate the standard error of the mean.

b. The expected shape of the distribution would still be positively skewed, as the skewness of the population does not change with the sample size.

c. To calculate the probability of selecting a sample with a mean of at least $117,000, we need to find the z-score corresponding to this sample mean:

z = (x - μ) / (σ / √n)

z = (117000 - 113000) / (35000 / √50)

z = 2.02

From Appendix B.1, we can find that the probability of a z-score being less than or equal to 2.02 is 0.9772. Therefore, the probability of selecting a sample with a mean of at least $117,000 is 1 - 0.9772 = 0.0228, or about 2.28%.

d. To find the likelihood of selecting a sample with a mean of more than $107,000, we need to find the z-score corresponding to this sample mean:

z = (x - μ) / (σ / √n)

z = (107000 - 113000) / (35000 / √50)

z = -1.72

From Appendix B.1, we can find that the probability of a z-score being less than or equal to -1.72 is 0.0427. Therefore, the probability of selecting a sample with a mean of more than $107,000 is 1 - 0.0427 = 0.9573, or about 95.73%.

e. To find the likelihood of selecting a sample with a mean of more than $107,000 but less than $117,000, we need to find the z-scores corresponding to these sample means:

z1 = (107000 - 113000) / (35000 / √50)

z1 = -1.72

z2 = (117000 - 113000) / (35000 / √50)

z2 = 2.02

From Appendix B.1, we can find that the probability of a z-score being less than or equal to -1.72 is 0.0427, and the probability of a z-score being less than or equal to 2.02 is 0.9772. Therefore, the probability of selecting a sample with a mean of more than $107,000 but less than $117,000 is the difference between these probabilities, which is 0.9772 - 0.0427 = 0.9345, or about 93.45%.

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Answer:population: 39 million; parameter: 56%; sample: 98,422; statistic: 78%

Step-by-step explanation:

The total sample size, because each pair of data provides one less degree of freedom than the number of individuals in the pair.

The formula used to calculate the degrees of freedom for a test for dependent groups is:

df = n - 1

where n is the number of pairs of data in the sample.

In a dependent groups (or paired samples) test, the same group of individuals is measured twice, or two groups of individuals are matched in pairs. The difference between the two measurements or the two pairs of measurements is the data used in the test.

The degrees of freedom is a measure of the amount of information available to estimate a population parameter, such as the mean difference between the paired observations in the case of a dependent groups test. In general, a higher degrees of freedom value means more information and thus more precision in the estimate.

In the case of a dependent groups test, the degrees of freedom is determined by the number of pairs of data in the sample, rather than the total sample size, because each pair of data provides one less degree of freedom than the number of individuals in the pair.

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The PDF of T1 is:

f(T1) = d/dt [P(T1 ≤ t)] = d/dt [1 - P(T1 > t)] = d/dt [1 - P(N(t) = 0)] = d/dt [1 - e^(-Lt)] = L * e^(-Lt)

So the PDF of T1 is an exponential distribution with parameter L.

To find the PDF of T1, we need to find the probability that the first event occurs at time t given that it has not occurred before time t. That is, we need to find P(T1 = t | N(t) = 0).

Using Bayes' rule, we have:

P(T1 = t | N(t) = 0) = P(N(t) = 0 | T1 = t) * P(T1 = t) / P(N(t) = 0)

Since T1 is the arrival time of the first event, we know that there are no events in the interval [0, t), so we can write:

P(N(t) = 0 | T1 = t) = P(N(t - T1) = 0)

Since {N(t) : t ≥ 0} is a conditional Poisson process with rate L that is distributed as Exp(λ), we know that the number of events in any interval of length t has a Poisson distribution with mean L*t. Therefore, the probability that there are no events in an interval of length t is:

P(N(t) = 0) = e^(-L*t)

So we can write:

P(T1 = t | N(t) = 0) = P(N(t - T1) = 0) * P(T1 = t) / e^(-L*t)

The probability density function (PDF) of the exponential distribution with parameter λ is given by:

f(x) = λ * e^(-λ*x)

Therefore, the PDF of T1 is:

f(T1) = d/dt [P(T1 ≤ t)] = d/dt [1 - P(T1 > t)] = d/dt [1 - P(N(t) = 0)] = d/dt [1 - e^(-Lt)] = L * e^(-Lt)

So the PDF of T1 is an exponential distribution with parameter L.

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The probability that in a random sample of 100 TCS employees, the monthly expenditure lies between Rs. 38000 and Rs. 39000 is 0.1359 or approximately 13.59%.

To solve this problem, we need to standardize the random variable using the z-score formula:

z = (x - μ) / (σ / sqrt(n))

where:

x = 38000 and 39000

μ = 40000

σ = 20000

n = 100

For x = 38000:

z = (38000 - 40000) / (20000 / sqrt(100)) = -1

For x = 39000:

z = (39000 - 40000) / (20000 / sqrt(100)) = -0.5

Now, we need to find the probability that the z-score falls between -1 and -0.5. We can use a standard normal distribution table or calculator to find this probability.

Using a standard normal distribution table, we find that the probability of z falling between -1 and -0.5 is 0.1359.

Therefore, the probability that in a random sample of 100 TCS employees, the monthly expenditure lies between Rs. 38000 and Rs. 39000 is 0.1359 or approximately 13.59%.

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