A wind turbine is located at the top of a hill where the wind blows steadily at 12 m/s, and stands 37 m tall. The air then exits the turbine at 9 m/s and the same elevation. Find the power generated by the wind if the mass flow rate is 137 kg/s. Report your answer in kW and to 2 decimal places.

Answer:

[tex]m(x)=\frac{5}{6(5+x)}[/tex]

Step-by-step explanation:

Slope of the secant line PQ:

P : (1, 1/6)

Q : (x, x/(5 + x))

[tex]m(x)=\frac{y_{Q}-y_{P}}{x_{Q}-x_{P}}=\frac{x/(5 + x)-1/6}{x-1}=\frac{5(x-1)}{6(5+x)(x-1)}[/tex]

The slope of the secant line PQ is 0.

To find the slope of the secant line PQ, we need to determine the coordinates of Q and P. Given that Q has the coordinates (x, x/(5+x)), we can substitute x=1 into the equation to find Q. So, Q is (1, 1/6).

The slope of a line passing through two points can be found using the formula:

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

Substituting the coordinates of P and Q into the formula:

m = (1/6 - 1/(5+1)) / (1 - 1) = (1/6 - 1/6)/(0) = 0

Therefore, the slope of the secant line PQ is 0.

Answer: [tex]p=0.610[/tex]

Step-by-step explanation:

Given : A market  research worker interviewed a random sample of 18 people about their  use of a certain product.

The results, in terms of Yes (Y) or No (N)  are as follows:

Y-N-N-Y-Y-Y-N-Y-N-Y-Y-Y-N-Y-N-Y-Y-N.

The number of people said "Yes" for the product= 11

Then, the sample proportion for the users of the product =[tex]\hat{p}=\dfrac{11}{18}0.611111111111\approx0.61[/tex]

We know that the sample proportion is the best estimate for the population proportion.

Thus the point estimate for population proportion : [tex]p=\hat{p}=0.610[/tex]

Step-by-step explanation:

A direct proof is a method that takes an statement p, which we assume to be true, and use it to show directly that another statement q is true. So this method has the following steps:

Fact that we need to use:

Every odd integer can be written in the form 2m + 1 for some unique other integer m

Let p be the statement a and b be odd integers and q be the statement that the product of a and b is odd.

Proposition if a and b are odd, then the product of a and b is odd

Proof: Assume that a and b are odd integers, the by definition a = 2m + 1 and b = 2n + 1 for some integers m and n. we will now use this to show that the product of a and b is odd.

[tex]a\cdot b= (2m+1) \cdot (2n+1)\\a\cdot b = 2m\cdot 2n+2m+2n+1\\a\cdot b =4mn+2m+2n+1\\a\cdot b = 2(2mn+2m+2n) +1\\\:If  \:k=2mn+2m+2n\\a\cdot b = 2k+1[/tex]

Hence we have shown that the product of a and b is odd since 2k + 1 is and odd integer. Therefore we have shown that p ⇒ q and so we have completed our proof.

Answer:

Step-by-step explanation:

The proof by the direct method that the product of two odd numbers is odd integer number, is the following:

Let [tex]z_1[/tex] and [tex]z_2[/tex] be two odd integers, then [tex]z_1 = 2a+1[/tex] and [tex]z_2 = 2b +1[/tex], for some integers a and b.

[tex]z_1z_2 = (2a + 1) (2b + 1)\\\\z_1z_2 = 4ab + 2a + 2b + 1\\\\z_1z_2 = 2 (2ab + a + b) +1\\\\z_1z_2 = 2n + 1[/tex]

where [tex]n = 2ab + a + b[/tex], which guarantees that [tex]n[/tex] is an integer number. In this way, [tex]z_1z_2[/tex] is an odd integer.

Answer:

[tex]y(x)\ =\ \sqrt{x}[C_1cos\dfrac{\sqrt{3}}{2}logx+C_2sin\dfrac{\sqrt{3}}{2}logx][/tex]

Step-by-step explanation:

Given differential equation is

[tex]x^2y"-xy'+y=0[/tex]                     (1)

Let's assume that

[tex]x=e^t[/tex]

[tex]=>\ t\ =\ logx[/tex]

then,

[tex]\dfrac{dx}{dt}=e^t[/tex]

[tex]and\ \dfrac{d^2x}{dt^2}=e^t[/tex]

We can write,

[tex]\dfrac{dy}{dx}=\dfrac{dy}{dt}.\dfrac{dt}{dx}[/tex]

                      [tex]=e^{-t}\dfrac{dy}{dt}[/tex]

Similarly,

[tex]\dfrac{d^2y}{dt^2}=\dfrac{d^2y}{dt^2}.\dfrac{dt^2}{dx^2}[/tex]

                            [tex]=e^{-2t}.\dfrac{d^2y}{dt^2}[/tex]

Putting these values in equation (1), we will get

[tex]e^{2t}.e^{-2t}.\dfrac{d^y}{dt^2}-e^t.e^{-t}\dfrac{dy}{dt}+y=0[/tex]

[tex]=>\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}+y=0[/tex]

So, the characteristics equation can be given as

[tex]D^2-D+1=0[/tex]

[tex]=>D\ =\ \dfrac{1+\sqrt{1-4}}{2}\ or\ \dfrac{1-1\sqrt{1-4}}{2}[/tex]

[tex]=>D=\ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}\ or\ \dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}[/tex]

Hence, the general solution of the equation can be give by

[tex]y(t)\ =\ e^{\dfrac{t}{2}}[C_1cos\dfrac{\sqrt{3}}{2}t+C_2sin\dfrac{\sqrt{3}}{2}t][/tex]

Now, by putting the value of t in above solution, we will have

[tex]y(x)\ =\ e^{\dfrac{1}{2}logx}[C_1cos\dfrac{\sqrt{3}}{2}logx+C_2sin\dfrac{\sqrt{3}}{2}logx][/tex]

[tex]y(x)=\ \sqrt{x}[C_1cos\dfrac{\sqrt{3}}{2}logx+C_2sin\dfrac{\sqrt{3}}{2}logx][/tex]

Hence, the solution of above given differential equation can be given by

[tex]y(x)=\ \sqrt{x}[C_1cos\dfrac{\sqrt{3}}{2}logx+C_2sin\dfrac{\sqrt{3}}{2}logx][/tex]

Answer:

The shop’s profit if it sold 150 units of sunscreen yesterday is $1000

Step-by-step explanation:

The total daily cost associated with the sale of this item yesterday was $200.

Cost of 1 unit = $8

So, Cost of 150 units = [tex]8 \times 150[/tex]

                                  = [tex]1200[/tex]  

So, revenue = $1200

Cost = $200

So Profit = Revenue - Cost

Profit = $1200-$200

Profit = $1000

Hence the shop’s profit if it sold 150 units of sunscreen yesterday is $1000

Answer: 43 cm

Step-by-step explanation:

The perimeter of rectangle is given by :-

[tex]P=2(l+w)[/tex], where l is length and w is width of the rectangle.

Given : The length of rectangle is 17.5 cm and the width is 40 mm in cm.

Since , 1 cm = 10 mm

Then, 40 mm= [tex]\dfrac{40}{10}\ cm=4\ cm[/tex]

Then, the  perimeter of rectangle will be :-

[tex]P=2(17.5+4)\\\\\Rightarrow\ P=2(21.5)\\\\\Rightarrow\ P=43\ cm[/tex]

Hence, the perimeter of rectangle = 43 cm

Answer:

[tex]E(k)= \left[ \begin{array}{c} {6+0.10k\,\,if\,\,0\leq k \leq500} & 6+500\times0.10+0.11(k-500)\,\,if\,\,500<k\leq1000& 6+500\times0.10+500\times0.11+0.15(k-1000)\,\,if\,\,k>1000\end{array}[/tex]

Step-by-step explanation:

The cost function has 3 branches,

So the first branch the consumer pays 6 dollars plus .10cents for any additional kWh (k)

In the second, they pay the same as the first up to 500kWh, and after that they pay  0.11 for the additional kWh above 500: (k-500) but bellow 1000

In the third branch, for consumptions above 1000, they pay the fix amount, plus .10 for the first 500 ([tex]500\times0.10[/tex]) , .11 for the additional 500 ([tex]500\times0.11[/tex]) , and finally 0.15 for consumptions above 1000: [tex]0.15(k-1000)[/tex]

Final answer:

The monthly cost of electricity, E, as a function of the electricity used, x, is defined using a piecewise function with different rates for usage tiers, including a fixed base charge and variable rates for different consumption brackets.

Explanation:

To express the monthly cost E as a function of the amount x of electricity used by the customer, we consider the following rates:

Therefore, the function is a piecewise function defined as:

E(x) =

This function helps calculate the monthly electricity bill based on the usage levels. Remember that the first bracket is for the initial 500 kWh, the second bracket is for 501 to 1000 kWh, and the last bracket applies to usage above 1000 kWh.

Answer:

1. Group A and B agree with each other.

2. Group A and C do not agree with each other.

Step-by-step explanation:

When we are analizing this problem, we will see what are the ranges of this measured times. Since we are taking into account the error we can see that :

Question 1 is about the overlapping response in Group A and Group B. And yes, we have an overlap between 7.35 to 7.39. Among this times both group A and B are in agree with each other within the experimental uncertainty.

Question 2 is now referring to Group A and Group C. And no, there isn't any common time where both groups agree with each other.

Answer:

100503

Step-by-step explanation:

Data provided in the question:

Fixed cost per month for the cell phone company = $1,000,000

Variable cost  per month per subscriber = $20

Charges for the customer per month = $29.95

Now,

the breakeven point is calculated as:

Breakeven point = [tex]\frac{\textup{Total fixed cost}}{\textup{Charges - variabel cost}}[/tex]

on substituting the respective values, we get

Breakeven point = [tex]\frac{\textup{1,000,000}}{\textup{29.95 - 20}}[/tex]

or

Breakeven point = 100502.51 ≈ 100503

Answer:

The area of the region between the curves y=6x^2 and y=4x is 8/27

Step-by-step explanation:

Use the diagram to visualize the problem, the area colored of blue is the one that needs to be found, let's do it in 3 parts:

Part 1: Find the intersection points of the curves

To do this we put both equations in one and solve it for x:

[tex]6x^2=4x[/tex]

[tex]6x^2-4x=0\\2x(3x-2)=0[/tex]

This equation has 2 possible solutions:

x=0 and x=2/3, so the interval for integration is 0 <= x <= 2/3

Part 2: Find the area below each curve

[tex]A_{blue}=\int\limits^0_{2/3} {6x^2} \, dx \\A_{blue}=2x^3[/tex], evaluate in 0 and 2/3

[tex]A_{blue}=\frac{16}{27}[/tex]

[tex]A_{red}=\int\limits^0_{2/3} {4x} \, dx \\A_{red}=2x^2[/tex], evaluate in 0 and 2/3

[tex]A_{red}=\frac{8}{9}[/tex]

Part 3: Substract the area of the blue curve (y=6x^2) to the area of the red curve (y=4x)

[tex]Area=\frac{8}{9}-\frac{16}{27}\\Area=\frac{8}{27}[/tex]

Answer:

If n is an odd integer, then 5n +3 is even.

Step-by-step explanation:

Below is the know-show table for the statement

Answer: 0.7258

Step-by-step explanation:

Given : A telemarketer makes a sale on 25% of his calls.

i.e. p=0.25

He makes 300 calls in a night, i.e. n=300

Let x be a random variable that represents the number of calls make in night.

To convert the given binomial distribution to normal distribution we have :-

[tex]\mu=np=300(0.25)=75[/tex]

[tex]\sigma=\sqrt{p(1-p)n}=\sqrt{(0.25)(1-0.25)(300)}\\\\=\sqrt{56.25}=7.5[/tex]

Now, using [tex]z=\dfrac{x-\mu}{\sigma}[/tex], the z-value corresponds to x= 70 :-

[tex]z=\dfrac{70-75}{7.5}\approx-0.67[/tex]

The z-value corresponds to x= 90 :-

[tex]z=\dfrac{90-75}{7.5}\approx2[/tex]

By using the standard normal distribution table for z, the probability that he will make more than 70 sales but less than 90 sales:-

[tex]P(-0.67<z<2)=P(z<2)-P(z<-0.67)\\\\=P(z<2)-(1-P(z<0.67))\\\\=0.9772-(1-0.7486)\\\\=0.9772-0.2514=0.7258[/tex]

Hence, the probability that he will make more than 70 sales but less than 90 sales= 0.7258

Answer:

They are 11.7475km away from Vancouver when they meet.

Step-by-step explanation:

The first step to solve this problem is modeling the position of each ferry. The position can be modeled by a first order equation in the following format:

[tex]S(t) = S_{0} + vt[/tex], in which [tex]S_{0}[/tex] is the initial position of the ferry, t is the time in hours and v is the speed in km/h.

I am going to say that the positive direction is from Nanaimo to Vancouver, and that Nanaimo is the position 0 and Vancouver the position 22.

First ferry:

Leaves Nanaimo, so [tex]S_{0} = 0[/tex]. It is 2km/h faster than the second ferry, so i am going to say that [tex]v = v + 2[/tex]. It moves in the positive direction, so v is positive. The equation of the position of this train is modeled as:

[tex]S_{1}(t) = 0 + (v+2)t[/tex],

Second ferry:

Leaves Vancouver, so [tex]S_{0} = 22[/tex]. It has a speed of v, that is negative, since it moves in the negative direction. So

[tex]S_{2}(t) = 22 - vt[/tex]

The problem states that they meet in 45 minutes. Here we have to pay attention. Since the speed is in km/h, the time needs to be in h. So 45 minutes = 0.75h.

They meet in 0.75h. It means that

[tex]S_{1}(0.75) = S_{2}(0.75)[/tex]

With this we find the value o v, and replace in the equation of [tex]S_{2}[/tex] to see how far they are from Vancouver when they meet.

[tex]S_{1}(0.75) = S_{2}(0.75)[/tex]

[tex]0.75(v+2) = 22 - 0.75v[/tex]

[tex]0.75v + 1.50 = 22 - 0.75v[/tex]

[tex]1.50v = 20.50[/tex]

[tex]v = \frac{20.50}{1.50}[/tex]

[tex]v = 13.67[/tex]km/h.

[tex]S_{2}(t) = 22 - vt[/tex]

[tex]S_{2}(0.75) = 22 - 13.67*0.75 = 11.7475[/tex]

They are 11.7475km away from Vancouver when they meet.

The ferries are 10.25 km away from Vancouver when they meet after traveling for 45 minutes, with the ferry from Nanaimo moving at 15.67 km/h and the one from Vancouver at 13.67 km/h.

Distance Between Ferries

Let's denote the speed of the slower ferry as s km/h. Therefore, the speed of the faster ferry leaving Nanaimo will be s + 2 km/h. As they both start at the same time, we can express the distance they travel in terms of their speeds multiplied by the time, which is 45 minutes (or 0.75 hours when converted to hours).

Now, let's set up the equation for the distance each ferry has traveled when they meet: the distance traveled by the slower ferry plus the distance traveled by the faster ferry equals the total distance between Nanaimo and Vancouver, which is 22 km.

Distance of slower ferry: s (0.75) km

Distance of faster ferry: (s + 2)(0.75) km

The sum of both distances is 22 km, so we have:

s (0.75) + (s + 2)(0.75) = 22

Rearranging terms and solving for s gives us:

1.5s + 1.5 = 22

s = (22 - 1.5)/1.5

Calculating the value of s and consequently the distance each ferry traveled before meeting will give us the answer.

After solving, we can determine that the speed of the slower ferry is 13.67 km/h and the faster ferry is 15.67 km/h. To find the distance from Vancouver to the meeting point, we only need to calculate the distance traveled by the slower ferry:

Distance from Vancouver = (13.67 km/h) (0.75 h) = 10.25 km

Therefore, the ferries are 10.25 km away from Vancouver when they meet.

Answer:

First one: Its place value is 600,000.

Second one: Its place value is 600.

Step-by-step explanation:

Answer: a) FALSE b) FALSE

Step-by-step explanation:

a) For the given proposition p ∧ ¬(q ∨ s) you can solve first (q v s)

q v s is true if either q is true or s is true or both. It is only false if both q and s are false. So, the proposition (q v s) is true because q is true.

Now you can solve the negation: ¬(q ∨ s)

As we know, (q v s) is true then its negation ¬(q ∨ s) is false.

p ∧ ¬(q ∨ s) should be true when both p and ¬(q ∨ s) are true, and false otherwise. So, the proposition is false because p is true and ¬(q ∨ s) is false.

b) For the given proposition  ¬(q ∧ p ∧ ¬s)

You can rewrite the expression as: ¬[q ∧ (p ∧ ¬s) ]  to solve first each part of the propositions in parenthesis.

The negation of s: ¬s  is true because s is false

Now, you can solve (p ∧ ¬s) which is true because both p and ¬s are true.

To continue, you have to solve (q ∧ p ∧ ¬s) which is true because both q and (p ∧ ¬s) are true.

To finish, the negation: ¬(q ∧ p ∧ ¬s) is false.

Answer:

The desviation is 8 4/7 or 8.571

Step-by-step explanation:

The conversion for any variable X to a standard z is

[tex]Z=\frac{X-\[Mu]}\\{\[Sigma]}[/tex]

where mu is the mean and sigma de desviation

You can find the value of Z whith the tables of the accumulated probability function. The accumulated probability for the mean is 0.5 .Remenber that the accumlated probability function represent the area at the left of an abscissa. Then

0.5+0.3531=0.8531

Acording to the accumulated probability function table, a Z=1.05 has an area of 0.8531 at its left.

Now it is only solving the equation

[tex]\[Sigma]=\frac{X-\[Mu]}{Z}[/tex]

σ=[tex]\frac{34-25}{1.05}[/tex]

σ=8.571

Answer:

There are 8 teams that have nicknames without a color and don't end in "s.

Step-by-step explanation:

This can be solved by treating each value as a set, and building the Venn Diagram of this.

-I am going to say that set A are the teams that have nicknames that end in S.

-Set B are those whose nicknames involve a color.

-Set C are those who have nicknames without a color and don't end in "s.

We have that:

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

In which a are those that have nickname ending in "s", but no color, and [tex]A \cap B[/tex] are those whose nickname involves a color and and in "s".

By the same logic, we have

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

In which b are those that nicknames involves a color but does not end in s.

We have the following subsets:

[tex]a,b, (A \cap B), C[/tex]

There are 129 schools, so:

[tex]a + b + (A \cap B) + C = 129[/tex]

Lets find the values, starting from the intersection.

The problem states that:

13 nicknames involve both a color and end in "s". So:

[tex]A \cap B = 13[/tex]

19 have nicknames that involve a color. So:

[tex]B = 19[/tex]

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

[tex]b + 13 = 19[/tex]

[tex]b = 6[/tex]

115 have nicknames that end in "s". So:

[tex]A = 115[/tex]

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

[tex]a + 13 = 115[/tex]

[tex]a = 102[/tex]

Now, we just have to find the value of C, in the following equation:

[tex]a + b + (A \cap B) + C = 129[/tex]

[tex]102 + 6 + 13 + C = 129[/tex]

[tex]C = 129 - 121[/tex]

[tex]C = 8[/tex]

There are 8 teams that have nicknames without a color and don't end in "s.

Answer:

y = 26

Step-by-step explanation:

9 + 22 = x + y

9 + 22 = 31

31 = x + y

x = 5

31 = 5 + y

31 - 5 = 26

y = 26

Hey!

-----------------------------------------------

Solution:

9 + 22 = 5 + y

31 = 5 + y

31 - y = 5 + y - y

31 - y = 5

31 - y - 31 = 31 - 5

y = 26

-----------------------------------------------

Answer:

y = 26

-----------------------------------------------

Hope This Helped! Good Luck!

Answer: -2.12°C

Step-by-step explanation:

Let x denotes the reading of the thermometers .

We assume that the readings on the thermometers are normally distributed.

Let a be the reading that separates the rejected thermometers from the others.

Given: Population mean : [tex]\mu=0[/tex]

Standard deviation: [tex]\sigma= 1[/tex]

Also, [tex]P(x<a)=0.017[/tex]

By using the z-table , the z-value corresponds to the p-value (one -tailed)0.017 is [tex]\pm2.12[/tex].

Now, [tex]z=\dfrac{a-\mu}{\sigma}[/tex]

i.e. [tex]\pm2.12=\dfrac{a-0}{1}[/tex]

i.e. [tex]\pm2.12=a[/tex]

For left tailed , [tex]a=-2.12[/tex]

It means the reading that separates the rejected thermometers from the others = -2.12°C.

Answer: The plot will be linear as described by the equation of the problem

Step-by-step explanation:

The first thing to do is to determine values of X and place them in one column. The following column should include all values solved by the equation in terms of y. Once you plot the graph, you should have something as the graph attached. To save it as a pdf, simply scan your graph paper to pdf and follow the instructions for assignment submission.

Answer:

Option (c) is correct

Step-by-step explanation:

Case (a)

A = 3 + 5i = (3, 5)

B = 2 + 2i = (2, 2)

C = 5i = (0, 5)

Use the distance formula to find the distance between two points

[tex]AB = \sqrt{(2-3)^{2}+(2-5)^{2}}=\sqrt{10}[/tex]

[tex]BC = \sqrt{(0-2)^{2}+(5-2)^{2}}=\sqrt{13}[/tex]

[tex]CA = \sqrt{(0-3)^{2}+(5-5)^{2}}=\sqrt{9}[/tex]

For the triangle to be right angles triangle

[tex]BC^{2}=AB^{2}+CA^{2}[/tex]

Here, it is not valid, so these are not the points of a right angled triangle.

Case (b)

A = 2i = (0, 2)

B = 3 + 5i = (3, 5)

C = 4 + i = (4, 1)

Use the distance formula to find the distance between two points

[tex]AB = \sqrt{(3-0)^{2}+(5-2)^{2}}=\sqrt{18}[/tex]

[tex]BC = \sqrt{(4-3)^{2}+(1-5)^{2}}=\sqrt{17}[/tex]

[tex]CA = \sqrt{(4-0)^{2}+(1-2)^{2}}=\sqrt{17}[/tex]

For the triangle to be right angles triangle

[tex]AB^{2}=BC^{2}+CA^{2}[/tex]

Here, it is not valid, so these are not the points of a right angled triangle.

Case (c)

A = 6 + 4i = (6, 4)

B = 7 + 5i = (7, 5)

C = 8 + 4i = (8, 4)

Use the distance formula to find the distance between two points

[tex]AB = \sqrt{(7-6)^{2}+(5-4)^{2}}=\sqrt{2}[/tex]

[tex]BC = \sqrt{(8-7)^{2}+(4-5)^{2}}=\sqrt{2}[/tex]

[tex]CA = \sqrt{(8-6)^{2}+(4-4)^{2}}=\sqrt{4}[/tex]

For the triangle to be right angles triangle

[tex]CA^{2}=BC^{2}+AB^{2}[/tex]

Here, it is valid, so these are the points of a right angled triangle.

Answer:

  x ∈ {-4, 14}

Step-by-step explanation:

Take the square root and add 5.

  (x -5) = ±√81 = ±9

  x = 5 ± 9

  x ∈ {-4, 14} . . . . . . x may be either of -4 or 14

Answer:

4.08 + 2 = 6.08 years

Step-by-step explanation:

we know that

Simple Interest(S.I.) = (P × R × T) ÷ 100

where, P = Principal = 750

R = Rate = 6%

T = unknown

⇒ S.I. = (750 × 6 × t)÷ 100

⇒ S.I. = 45t

Also, Amount = S.I + Principal

⇒ Amount = 750 + 45t

Now Formula for Compound Interest is:

[tex]A = P(1+\frac{r}{100})^{t}[/tex]

where A = Amount

=1000

P = Principle

r = rate

t = total number of year

Here, P = 750 + 45t, r = 3.5% , and t = 2.

Putting all these values in above formula:

[tex]1000 = (750 + 45t)(1+\frac{3.5}{100})^{2}[/tex]

⇒ [tex]1000 = (750 + 45t)(1.071)[/tex]

⇒ t = 4.08

Hence, total time required will be 2 + 4.08 = 6.08 years.

Answer:

Option d. 12.40963855% is the answer.

Step-by-step explanation:

Monthly salary of a sales representative = $1,100

He was paid for the month = $2,300

His total sales were = $8,300

His commission for the month of sales = Total payment - monthly salary

                                                                 = 2,300 - 1,100

                                                                 = $1,030

He got $1,030 as commission on the sales of $8,300.

The percentage of the commission = [tex]\frac{1030}{8300}\times 100[/tex]

                                                           = 12.40963855421687%

Option d. 12.40963855% is the answer.

Answer:

(a) The confidence interval is: 0.0304 ≤ π ≤ 0.0830.

(b) Upper confidence bound = 0.0787

Step-by-step explanation:

(a) The confidence interval for p (proportion) can be calculated as

[tex]p \pm z*\sigma_{p}[/tex]

[tex]\sigma=\sqrt{\frac{\pi*(1-\pi)}{N} }\approx\sqrt{\frac{p(1-p)}{N} }[/tex]

NOTE: π is the proportion ot the population, but it is unknown. It can be estimated as p.

[tex]p=17/300=0.0567\\\\\sigma=\sqrt{\frac{p(1-p)}{N} }=\sqrt{\frac{0.0567(1-0.0567)}{300} }=0.0134[/tex]

For a 95% two-sided confidence interval, z=±1.96, so

[tex]\\LL = p-z*\sigma=0.0567 - (1.96)(0.0134) = 0.0304\\UL =p+z*\sigma= 0.0567 + (1.96)(0.0134) = 0.0830\\\\[/tex]

The confidence interval is: 0.0304 ≤ π ≤ 0.0830.

(b) The confidence interval now has only an upper limit, so z is now 1.64.

[tex]UL =p+z*\sigma= 0.0567 + (1.64)(0.0134) = 0.0787[/tex]

The confidence interval is: -∞ ≤ π ≤ 0.0787.

Final answer:

To calculate a confidence interval for the fraction of defective circuits, use the formula for the confidence interval of a proportion. The 95% two-sided confidence interval for the fraction of defective circuits is 0.0182 ≤ p ≤ 0.0951. The 95% upper confidence bound on the fraction of defective circuits is 0.0951.

Explanation:

To calculate a confidence interval for the fraction of defective circuits, we can use the formula for the confidence interval of a proportion. Let p-hat be the proportion of defectives in the sample, which is equal to 17/300 = 0.0567. We can calculate the standard error as [tex]\sqrt{((p-hat*(1-p-hat))/n)[/tex], where n is the sample size.

Using a 95% confidence level, we can find the critical value from the standard normal distribution, which is approximately 1.96. The lower bound of the confidence interval is given by p-hat - z*(standard error), and the upper bound is given by p-hat + z*(standard error).

Therefore, the 95% two-sided confidence interval for the fraction of defective circuits is 0.0182 ≤ p ≤ 0.0951. The 95% upper confidence bound on the fraction of defective circuits is 0.0951.

Answer:

A. -0.87

Step-by-step explanation:

The correlation coefficient is a measure of the strength and nature of the linear relationship between two variables, one dependent and one independent. This coefficient takes values between -1 and 1, indicating with its sign whether the relationship is direct or inverse between the variables involved and with its absolute value indicates the strength of the linear relationship between them. A coefficient with absolute value close to 1 indicates great strength and better fit.

Conclusion: The best coefficient, of the propuetso in the problem, is that of -0.87, which indicates a strong relationship between the variables, a good fit and an inverse relationship between them.

Answer:

£ 121.57

Step-by-step explanation:

As given in question,

Total amount inherited by Tim = £1000

He spends each month = 10 % of remaining money

savings reduced each month = 0.9 of remaining amount

So, the amount of money after one month = 0.9 x £1000

 amount of money remained after 2 months = 0.9 x 0.9 x £1000

amount of money remained after 3 months = 0.9 x 0.9 x 0.9 x £1000

Hence, the amount of money remained after the n months can be given by,

[tex]E\ =\ 0.9^n\times 1000[/tex]

Hence, amount of money remained after 20 months can be given by,

[tex]E_{20}\ =\ 0.9^{20}\times1000[/tex]

            = £ 121.57

So, the amount of money remained after 20 months will be £ 121.57.

Answer:

Use induction for the prove

Step-by-step explanation:

Mathemathical induction is an useful method to prove things over natural numbers, you check for the first case, supose for the n and prove using your hypothesis for n+1

there says any integer bigger than 12 can be written as 4r+5s

so first number n can be is 13.

we can check n=13  =  4*2+5*1   r=2 and s=1 give 13.

Now we suppose n can be written as 4r+5s

and we can check if n+1=4r'+5s'  with  r' and s' integers.

we replace n as 4r+5s because that is our hypotesis

n+1=4r+5s+1

if we write that 1 as 5-4

4r+5s+1

4r+5s+5-4

then we can write

4(r-1)+5(s+1)   , we got n+1= 4 (r-1) +5(s+1)  where r-1 and s+1 are non negative integers. because r and s were no negative integers ( if r is not 0)

what if r=0?

if r is 0 , n is a multiple of 5   and n+1 can be written as 5s+1

first multiple of 5 we can write is 15 since n is bigger than 12 , then smaller s is 3.

for any n+1 we can write

n+1=5s+1=5 (s-3) +3*5+1=5(s-3)+4*4,   s-3 is 0 or bigger.

(check 3*5+1 is 16, the same as 4*4)

Final answer:

Any integer greater than 12 can be expressed as the sum 4r + 5s with nonnegative integers r and s by providing concrete examples for n values from 13 to 17 and then proving for any n > 17 using number theory.

Explanation:

To show that any integer n greater than 12 can be written as the sum 4r + 5s for some nonnegative integers r and s, we can provide examples and create a general proof. We know that for n = 13 through n = 17, specific values of r and s can be found that satisfy the equation:

n = 13 = 4(1) + 5(1)

n = 14 = 4(3) + 5(0)

n = 15 = 4(0) + 5(3)

n = 16 = 4(4) + 5(0)

n = 17 = 4(1) + 5(3)

For n > 17, we can write n = 17 + k, where k is a nonnegative integer. Since every nonnegative integer can be expressed as a multiple of 4 plus an additional 0, 1, 2, or 3 (as k = 4q + r, with 0 ≤ r < 4), we can substitute into our equation for n to n = 17 + 4q + r. The key is to notice that for any additional 4 we add to the sum, we can simply increase r (our existing count of 4-cent stamps), ensuring that the equation 4r + 5s will always hold.

This proof shows that we can always add enough 4-cent stamps (or subtract them and add a 5-cent stamp) to reach any postage amount above 12 cents. This is a classic example of a problem that uses the concept of number theory and diophantine equations.

Answer:

The cost of producing 100 items is $723.35

Step-by-step explanation:

The marginal cost is the derivative of the total cost function, so we have

[tex]C^{'}(x)=1.65-0.002x[/tex]

To find the total cost function we need to do integration

[tex]C(x)= \int\, C^{'}(x)dx \\C(x)=\int\,(1.65-0.002x) dx[/tex]

Apply the sum rule to find the integral

[tex]\int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)[/tex]

[tex]\int \:1.65dx=1.65x\\\int \:0.002xdx=0.001x^2[/tex]

[tex]C(x)=\int\,(1.65-0.002x) dx = 1.65x-0.001x^2+D [/tex]

D is the constant of integration

We are given that C(1) = $570, we can use this to find the value of the constant in the total cost function

[tex]C(1)=570=1.65*(1)+0.001*(1)^2+D\\D=570-1.649=568.351[/tex]

So the total cost function is [tex]C(x)=1.65x-0.001x^2+568.351 [/tex] and the cost of producing 100 items is

x=100

[tex]C(100)=1.65*(100)-0.001*(100)^2+568.351 = 723.35[/tex]

Answer:

Case 1

None of the sides is yellow.

Probability of rolling a yellow on your next toss: zero %

Case 2

All the sides are yellow.

Probability of rolling a yellow on your next toss: 100 %

Case 3  

At least one of the sides is yellow.

[tex]\frac{q}{N}\times100\%[/tex]

where  

N = total number of tosses  in one hour.

q = number of tosses you rolled a yellow.

Step-by-step explanation:

Case 1

None of the sides is yellow.

Probability of rolling a yellow on your next toss: zero %

Case 2

All the sides are yellow.

Probability of rolling a yellow on your next toss: 100 %

Case 3  

At least one of the sides is yellow.

As you do not know if the die is fair or not, the only way to approximate a probability of rolling a yellow is by making a table of frequencies and record the times you have rolled yellow.

If the number of tosses made in an hour is big enough as to draw a conclusion, then according to the Law of Large Numbers, the probability of rolling a yellow in one toss of the die should be

[tex]\frac{q}{N}\times100\%[/tex]

where  

N = total number of tosses

q = number of tosses you rolled a yellow.

Now, suppose you want to roll the dice once more.  

As the event of rolling a die is independent of the previous tosses, this means that the probable outcome of the event does not depend on the previous results. So, the probability remains the same.  

Probability of tossing a yellow on your next toss of the die

[tex]\frac{q}{N}\times100\%[/tex]

where  

N = total number of previous tosses

q = number of tosses you rolled a yellow.