PLEASE HELP MEEEE!!!!!!

Answer:

Sheridan Service reduce its debt $1,392, now its due is $1,200.

Step-by-step explanation:

Sheridan Service received a shipment on January 15, with an invoice dated January 14, terms 4/10 E.O.M.

Term written on the invoice means 4% discount if paid within 10 days or full amount is due for the payment at the End of the Month.

Invoice shipment having amount = $2,592

Partial payment of the invoice by check = $1,392

So amount due = 2,592 - 1,392 = $1,200

Sheridan Service reduce its debt $1,392, now its due is $1,200.

Answer:

C) x = 20.8

Step-by-step explanation:

Since the line with the length equal to 6 forms a right angle on line x then we know it is perpendicular and bisects it.

Plug in the values: a^2 + b^2 = c^2

a^2 + 6^2 = 12^2

a^2 + 36 = 144

a^2 + 36 - 36 = 144 - 36

a^2 = 108

√a^2 = √108

a = √108

a = 10.3923048454

a = 1/2 x

2a = x

2(10.3923048454) = x

x = 20.7846096908

x = 20.8

[tex]6^2+\left(\dfrac{x}{2} \right)^2=12^2\\36+\dfrac{x^2}{4}=144\\144+x^2=576\\x^2=432\\x=\sqrt{432}=12\sqrt3\approx20.8[/tex]

Answer:

8 is the number.

Step-by-step explanation:

We are given the following expression in words which we are to translate into mathematical expression and tell the number:

'eight times the sum of 5 and some number is 104'

Assuming the number to be [tex]x[/tex], we can write it as:

[tex] 8 ( 5 + x ) = 1 0 4 [/tex]

[tex] 5 + x = \frac { 1 0 4 } { 8 } [/tex]

[tex] 5 + x = 1 3 [/tex]

[tex]x=13-5[/tex]

x = 8

ANSWER

[tex]8[/tex]

EXPLANATION

Let the number be y.

Eight times the sum of the number and 5 is written as:

[tex]8(5 + y)[/tex]

From the question, this expression must give us 104.

This implies that:

[tex]8(5 + y) = 104[/tex]

Expand the parenthesis to get;

[tex]40 + 8y = 104[/tex]

Group similar terms to get:

[tex]8y = 104 - 40[/tex]

Simplify:

[tex]8y = 64[/tex]

[tex]y = \frac{64}{8} [/tex]

This finally evaluates to

[tex]y=8[/tex]

Hence the number is 8

The integral is path-independent if there is a scalar function [tex]f[/tex] whose gradient is

[tex]\nabla f=(2e^x\cos2y,-\sin2y)[/tex]

(at least, that's what it looks like the given integrand is)

Then

[tex]\dfrac{\partial f}{\partial x}=2e^x\cos 2y\implies f(x,y)=2e^x\cos2y+g(y)[/tex]

Differentiating both sides with respect to [tex]y[/tex] gives

[tex]\dfrac{\partial f}{\partial y}=-4e^x\sin 2y\neq-\sin2y[/tex]

so the line integral *is* dependent on the path. (again, assuming what I've written above actually reflects what the question is asking)

The question asks about path independence in vector calculus, which indicates a property of a vector field where the value of the line integral is the same regardless of the path taken, as long as the vector field is conservative.

The student's question is focused on the concept of path independence in the context of line integrals in vector calculus. The subject matter implies they are dealing with a conservative vector field, where the integral of a function along any path depends only on the endpoints of that path, not the specific route taken. The goal is to check if a given vector field is path independent and, if so, to perform the integration from a starting point (0, 0, 0) to an endpoint (a, b, c). To establish path independence, one common method is to verify if the curl of the vector field is zero throughout the domain of interest. If it is, the field is conservative, and the path independence principle applies.

A vector field is path independent if the line integral between two points is the same regardless of the path taken between those points. Path independence typically occurs in conservative fields, where there exists a potential function such that the original vector field is its gradient.

If a field is conservative and path independent, the integral of the field over any path from point P1 to point P2 will yield the same result as the integral over any other path from P1 to P2 in the field's domain.

Final answer:

The probability that a computer contains neither a virus nor a worm is found by subtracting the probability of having a virus or worm from 1. Based on the provided probabilities, this calculation results in a probability of 0.82 or 82% for a computer to be free of both.

Explanation:

To calculate the probability that a computer contains neither a virus (V) nor a worm (W), we use the principle of complementation. Given that P(V) = 0.17, P(W) = 0.05, and P(V AND W) = 0.04, we want to find P(neither V nor W). This is the same as finding 1 - P(V OR W). The probability of V OR W is given by P(V) + P(W) - P(V AND W), which simplifies to 0.17 + 0.05 - 0.04 = 0.18. Therefore, the probability of neither V nor W is 1 - 0.18 = 0.82.

So, the probability that the computer is free of both a virus and a worm is 0.82 or 82%.

Answer: 61,750 miles

Step-by-step explanation:

Given : The p-value of the tires to outlast the warranty = 0.96

The probability that corresponds to 0.96 from a Normal distribution table is 1.75.

Mean : [tex]\mu=60,000\text{ miles}[/tex]

Standard deviation : [tex]\sigma=1000\text{ miles}[/tex]

The formula for z-score is given by  : -

[tex]z=\dfrac{x-\mu}{\sigma}\\\\\Rightarrow\ 1.75=\dfrac{x-60000}{1000}\\\\\Rightarrow\ x-60000=1750\\\\\Rightarrow\ x=61750[/tex]

Hence, the tread life of tire should be 61,750 miles if they want 96% of the tires to outlast the warranty.

The company looking to ensure 96% of the tires outlast the warranty should use a mileage warranty of 58,250. This ensures a failure rate of only 4%, as calculated using statistics and the concept of normal distribution.

The question deals with the concept of normal distribution in statistics, specifically, with an application to a real-life situation - to decide the warranty for a product (in this case, a type of tire). The mean and standard deviation given represent the average life and variation in life of the tires respectively. If the company wants 96% of the tires to outlast the warranty, then they are looking to find the lifespan beyond which only 4% of the tires would fail.

The Z-score corresponding to the 96th percentile in a standard normal distribution table is roughly 1.75. Since standard deviation is 1,000 miles, that implies that 1.75 standard deviations below the average is acceptable for the warranty. Therefore, we calculate the warranty as Mean - 1.75*Standard Deviation, which results in 60,000 - 1.75*1000 = 58,250 miles. Thus, the company should use a warranty of 58,250 miles.

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Answer:

[tex]A'(t)=3\sqrt{t}+\frac{3}{2\sqrt{t}}[/tex]          

Step-by-step explanation:

Given : Length of rectangle = 2t+3

           Height of rectangle = [tex]\sqrt{t}[/tex]

To Find: Find the rate of the change of the area with respect to time.

Solution:

Area of rectangle = [tex]Length \times Width[/tex]

                              = [tex](2t+3) \times \sqrt{t}[/tex]

                              = [tex]2t^{\frac{3}{2}}+3t^{\frac{1}{2}}[/tex]

                              = [tex]2t^{\frac{3}{2}}+3t^{\frac{1}{2}}[/tex]

So, [tex]A(t)=2t^{\frac{3}{2}}+3t^{\frac{1}{2}}[/tex]

[tex]\frac{d}{dx} (x^n)=nx^{n-1}[/tex]

[tex]A'(t)=\frac{3}{2} \times 2t^{\frac{3}{2}-1}+\frac{1}{2} \times 3t^{\frac{1}{2}-1}[/tex]

[tex]A'(t)=3t^{\frac{1}{2}}+\frac{3}{2}t^{\frac{-1}{2}}[/tex]

[tex]A'(t)=3t^{\frac{1}{2}}+\frac{3}{2}t^{\frac{-1}{2}}[/tex]          

[tex]A'(t)=3\sqrt{t}+\frac{3}{2\sqrt{t}}[/tex]              

Hence the rate of the change of the area with respect to time is    [tex]A'(t)=3\sqrt{t}+\frac{3}{2\sqrt{t}}[/tex]                                                                                  

To find the rate of change of the area with respect to time, differentiate the area function and substitute the given expressions for the dimensions. The rate of change of the area with respect to time is 2√t + 3/√t.

To find the rate of change of the area of a rectangle with respect to time, we need to differentiate the area function and then substitute the given expressions for the dimensions. The formula for the area of a rectangle is A = length × height. So, A(t) = (2t + 3)(√t). To find the derivative of A(t), which represents the rate of change of the area, we can use the product rule of differentiation.

A'(t) = (2t + 3) × d/dt(√t) + (√t) × d/dt(2t + 3)

Let's differentiate each term of the product separately. Differentiating √t, we get 1/(2√t).

Substituting the values back into the equation and simplifying, we have:

A'(t) = 2√t + 6/(2√t)

Our final answer for the rate of change of the area with respect to time is A'(t) = 2√t + 3/√t.

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Answer:

  x = 5

Step-by-step explanation:

The table and a graph of f(x) and g(x) are shown in the attachment. The solution is x=5. The table shows you f(5) = g(5) = 2, as does the graph.

It is generally convenient to make use of a graphing calculator or spreadsheet when repeated evaluation of a function is required.

Answer:

Since every kilo of apples cost $3.12, it is directly proportional to the total weight (which is not exactly the same as mass).

Step-by-step explanation:

Answer:

:p

Step-by-step explanation:

Answer:

[tex]600*10^{5}[/tex]

Step-by-step explanation:

All you have to do is multiply the original cost of the movie by 100.

The budget for the remake of a movie, which is calculated as 100 times the original movie budget of 6•10^5 dollars, would be 6•10^7 dollars.

The original budget for the movie is given as 6•10^5 dollars. If the budget for the remake of the movie is 100 times the original, we would multiply the original budget by 100. Written as a power, 100 is 10^2. To multiply numbers written in scientific notation, you can multiply the coefficients (numbers at the front) and add the exponents (the powers of 10).

So, 6•10^5 times 10^2 becomes (6 * 1)•10^(5+2), or 6•10^7.

Therefore, the budget for the remake of the movie would be 6•10^7 dollars.

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Answer:

whatever

Step-by-step explanation:

-789237

Answer:

Option D.  sine theta equals plus or minus square root of thirty-five over six, tangent theta equals plus or minus square root of thirty five

Step-by-step explanation:

we have that

[tex]cos(\theta)=\frac{1}{6}[/tex]

If the cosine is positive, then the angle theta lie on the first or fourth Quadrant

therefore

The sine of angle theta could be positive (I Quadrant) or negative (IV Quadrant) and the tangent of angle theta could be positive (I Quadrant) or negative (IV Quadrant)

step 1

Find [tex]sin(\theta)[/tex]

Remember that

[tex]sin^{2} (\theta)+cos^{2} (\theta)=1[/tex]

we have

[tex]cos(\theta)=\frac{1}{6}[/tex]

substitute

[tex]sin^{2} (\theta)+(\frac{1}{6})^{2}=1[/tex]

[tex]sin^{2} (\theta)+\frac{1}{36}=1[/tex]

[tex]sin^{2} (\theta)=1-\frac{1}{36}[/tex]

[tex]sin^{2} (\theta)=\frac{35}{36}[/tex]

[tex]sin(\theta)=(+/-)\frac{\sqrt{35}}{6}[/tex]

so

sine theta equals plus or minus square root of thirty-five over six

step 2

Find [tex]tan(\theta)[/tex]

Remember that

[tex]tan(\theta)=\frac{sin(\theta)}{cos(\theta)}[/tex]

we have

[tex]sin(\theta)=(+/-)\frac{\sqrt{35}}{6}[/tex]

[tex]cos(\theta)=\frac{1}{6}[/tex]

substitute

[tex]tan(\theta)=(+/-)\sqrt{35}[/tex]

tangent theta equals plus or minus square root of thirty five

Answer:

a) increasing: (-∞, -2)∪(1, ∞); decreasing: (-2, 1)

b) local maximum: (-2, 20); local minimum: (1, -7)

c) inflection point: (-0.5, 6.5); concave up: (-0.5, ∞); concave down: (-∞, 0.5)

Step-by-step explanation:

A graphing calculator can show you the local extremes. Everything else falls out from those.

a) The leading coefficient is positive, so the general shape of the graph is from lower left to upper right. The function will be increasing from -infinity to the local maximum (x=-2), decreasing from there to the local minimum (x=1), then increasing again to infinity.

__

b) See the attached. This is what we did first. If you want to do this by hand, you find where the derivative is zero:

  6x^2 +6x -12 = 0

  6(x+2)(x-1) = 0 . . . . . local maximum at x=-2; local minimum at x=1.

You know the left-most zero of the derivative is the local maximum because of the nature of the curve (increasing, then decreasing, then increasing again).

The function values at those points are easily found by evaluating the function written in Horner form:

  ((2x +3)x -12)x

at x=-2, this is ((2(-2)+3)(-2) -12)(-2) = (2-12)(-2) = 20 . . . . (-2, 20)

at x = 1, this is 2 +3 -12 = -7 . . . . . . . . . . . . . . . . . . . . . . . . . (1, -7)

__

c) The point of inflection of a cubic is the midpoint between the local extremes: ((-2, 20) +(1, -7))/2 = (-1, 13)/2 = (-0.5, 6.5)

A cubic curve is symmetrical about the point of inflection. When you consider the derivative is a parabola symmetric about the vertical line through its vertex, perhaps you can see why. The local extremes of the cubic are the zeros of the parabola, which are symmetric about that line of symmetry. Of course the vertex of the derivative (parabola) is the place where its slope is zero, hence the second derivative of the cubic is zero--the point of inflection.

__

The cubic is concave down to the left of the point of inflection; concave up to the right of that point. The interval of downward concavity corresponds to the interval on which the first derivative (parabola) is decreasing or the second derivative is negative.

Answer: 0.9660

Step-by-step explanation:

Given: Mean : [tex]\mu =0[/tex]

Variance : [tex]\sigma^2=2.00[/tex]

⇒ Standard deviation : [tex]\sigma = \sqrt{2}[/tex]

The formula to calculate z is given by :-

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x= -3

[tex]z=\dfrac{-3-0}{\sqrt{2}}=-2.12132034356\approx-2.12[/tex]

The P Value =[tex]P(z<-2.12)=0.017003[/tex]

For x= 3

The P Value =[tex]P(z<2.12)=0.9829969[/tex]

[tex]\text{Now, }P(-3<X<3)=P(X<3)-P(X<-3)\\\\=P(z<2.12)-P(z<-2.12)\\\\=0.9829969-0.017003=0.9659939\approx 0.9660[/tex]

Hence, the probability that a given error will be between -3 and 3=0.9660

[tex]|\Omega|=27^{39}\\|A|=1\\\\P(A)=\dfrac{1}{27^{39}}[/tex]

In case you want it as a fraction

[tex]P(A)=\dfrac{1}{66555937033867822607895549241096482953017615834735226163}[/tex]

:D

Answer:

  C)  They are parallel, but not opposites

Step-by-step explanation:

They would be opposites if they had the same magnitude. As is, they are simply parallel with opposite directions.

Answer: 0.2

Step-by-step explanation:

Given: The commuting time on a particular train is uniformly distributed over the interval (42,52).

∴ The probability density function of X will be :-

[tex]f(x)=\dfrac{1}{b-a}\\\\=\dfrac{1}{52-42}=\dfrac{1}{10}, 42<x<52[/tex]

Thus, the required probability :-

[tex]P(X<44)=\int^{44}_{42}f(x)\ dx\\\\=\int^{44}_{42}\dfrac{1}{10}\ dx\\\\=\dfrac{1}{10}[x]^{44}_{42}=\dfrac{1}{10}(44-42)=\dfrac{1}{5}=0.2[/tex]

Hence, the  probability that the commuting time will be less than 44 ​minutes= 0.2

the probability that the commuting time will be less than 44 minutes is 0.2 or 20%.

To calculate the probability that the commuting time will be less than 44 minutes, given that it is uniformly distributed between 42 and 52 minutes, we use the formula for the probability of a continuous uniform distribution:

P(a < X < b) = (b - a) / (max - min)

Here, min is 42, max is 52, a is 42 (since that's the minimum time), and b is 44 (the time we are interested in). So the probability that the commuting time is less than 44 minutes will be:

P(42 < X < 44) = (44 - 42) / (52 - 42) = 2 / 10

This simplifies to 1/5 or 0.2.

Therefore, the probability that the commuting time will be less than 44 minutes is 0.2 or 20%.

Answer:

  (a) 120 square units, underestimate

  (b) 248 square units, overestimate

Step-by-step explanation:

(a) left sum

The left sum is the sum of the areas of the rectangles whose width is the total interval width (8-0) divided by the number of divisions (n=4). The height of each rectangle is the function value at its left edge.

We can compute the sum by adding the function values and multiplying that total by the width of the rectangles:

  left sum = (1 + 5 + 17 + 37)×2 = 60×2 = 120 . . . square units

The curve is increasing throughout the interval of interest, so the left sum underestimates the area under the curve.

__

(b) right sum

The rectangles whose area is the right sum are shown in the attachment, along with the table of function values. The right sum is computed the same way as the left sum, but using the function value on the right side of each subinterval.

  right sum = (5 + 17 + 37 + 65)×2 = 124×2 = 248 . . . square units

The curve is increasing throughout the interval of interest, so the right sum overestimates the area under the curve.

_____

The actual area under the curve on the interval [0, 8] is 178 2/3, just slightly less than the average of the left- and right- sums.

The left-hand and right-hand sums are common methods to estimate the area under a curve. Using n=4, the left-hand sum tends to be an underestimate, while the right-hand sum is usually an overestimate of the actual area.

The integral 8(x²+1)dx from 0 to a is the area under the curve defined by the function 8(x²+1). A common way to approximate this area is by using left-hand and right-hand sums.

(a) Left-Hand Sum: Using n=4, we divide the interval [0, a] into 4 equal subintervals. For each subinterval, we find the left endpoint and plug it into our function, then multiply by the length of the subinterval. Sum these values to get the left-hand sum. This sum is usually an underestimate of the true area as it leaves out the area under the curve that lies to the right of the left-hand rectangle in each subinterval.

(b) Right-Hand Sum: The process is similar to the left-hand sum, but we use the right endpoint of each subinterval. The right-hand sum tends to be an overestimate of the true area as it includes the area above the curve that lies to the right of the rectangle in each subinterval.

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Answer:

Step-by-step explanation:

Here's the gameplan for this.  First of all we need a general formula, then we will define the variables for each.

The general formula for all of these is the same:

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

where A(t) is the amount after the compounding, P is the initial investment, n is the number of compoundings per year, r is the interest rate in decimal form, and t is time in years.  

Then after we find the amount after the compounding, we will subtract the initial amount from that, because the amount at the end of the compounding is greater than the initial amount.  It's greater because it represents the initial amount PLUS the interest earned.  The difference between the initial amount and the amount at the end is the interest earned.

For A:

A(t) = ?

P = 450

n = 1

r = .06

t = 7

[tex]A(t)=450(1+\frac{.06}{1})^{(1)(7)}[/tex]

Simplifying gives us

[tex]A(t)=450(1.06)^7[/tex]

Raise 1.06 to the 7th power and then multiply in the 450 to get that

A(t) = 676.63 and

I = 676.63 - 450

I = 226.63

For B:

A(t) = ?

P = 450

n = 4 (there are 4 quarters in a year)

r = .06

t = 7

[tex]A(t)=450(1+\frac{.06}{4})^{(4)(7)}[/tex]

Simplifying inside the parenthesis and multiplying the exponents together gives us

[tex]A(t)=450(1.015)^{28}[/tex]

Raising 1.015 to the 28th power and then multiplying in the 450 gives us that

A(t) = 682.45

I = 682.45 - 450

I = 232.75

For C:

A(t) = ?

P = 450

n = 12 (there are 12 months in a year)

r = .06

t = 7

[tex]A(t)=450(1+\frac{.06}{12})^{(12)(7)}[/tex]

Simplifying the parenthesis and the exponents:

[tex]A(t)=450(1+.005)^{84}[/tex]

Adding inside the parenthesis and raising to the 84th power and multiplying in 450 gives you that

A(t) = 684.17

I = 684.17 - 450

I = 234.17

Answer:

See below.

Step-by-step explanation:

C1  = {0, 1, 2} and C2 = {2, 3, 4}.

C1 ∪ C2 = {0, 1, 2, 3, 4}       (Note: the 2 is not repeated in the result}.

C1 ∩ C2 = {2}.

The table contains the values from (A)

A limit tells us the value that a function approaches as that function's inputs get closer and closer to some number. The idea of a limit is the basis of all calculus.

Given equation:

limx→2 (x^2+8x−2)

Table A  does a nice job getting closer and closer to x=2, equals, 2 from both sides.

Table D does not make it clear which x-value we are approaching,

Table B only approaches x=1, equals, 1 from both side.

Table C is approaching only the positive side.

Learn more about limit here:

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Answer:

I don't know if it's low enough,but i can give you another answer.

Step-by-step explanation:

So the probability of finding bacteria in a public swimming are is 0.005.

But if you combine the samples then the probability of finding bacteria is 0.01 because you add the probabilities of finding bacteria of the two swimming pools.

Answer:

The length of arc AC is [tex]15\pi\ cm[/tex]

Step-by-step explanation:

step 1

Find the circumference of the circle

The circumference is equal to

[tex]C=\pi D[/tex]

we have

[tex]D=36\ cm[/tex]

substitute

[tex]C=\pi (36)[/tex]

[tex]C=36\pi\ cm[/tex]

step 2

Find the length of arc AC

Remember that the circumference subtends a central angle of 360 degrees

The measure of arc AC is equal to

arc AC+30°=180° ----> because the diameter divide the circle into two equal parts

arc AC=180°-30°=150°

using proportion

[tex]\frac{36\pi}{360}=\frac{x}{150}\\ \\x=36\pi*150/360\\ \\x=15\pi\ cm[/tex]

Answer:

  5.655 m³/min

Step-by-step explanation:

Halfway from the bottom of the tank, the radius is half that at the top, so is 1 m. That means the surface area of the water at that point is ...

  A = πr² = π(1 m)² = π m²

The rate of flow of water into the tank is the product of this area and the rate of change of depth:

  flow rate = area × (depth rate of change)

  = (π m²) × (1.8 m/min) = 1.8π m³/min

  flow rate ≈ 5.655 m³/min

Expressing the radius in terms of the height allows for finding the relationship between the changing volume and the rising water level.

We must use the concept of related rates, which involves finding the relationship between different rates of change within a geometrical context, particularly using the volume formula for a cone V = (1/3)\pi r^2 h and the fact that dv/dt represents the rate of change of volume with respect to time, which is equivalent to the water flow rate we are looking for.

Given the dimensions of the conical tank (height 3 m and radius 2 m), and the current water level at 1.5 m from the bottom, we know that the water level is rising at 1.8 m/min. By expressing the tank's radius as a function of its height, we can relate the changing volume to the changing height of the water level using the chain rule, eventually finding the rate at which water is flowing into the tank in m³/min, which is the same as the rate at which volume is increasing.

Answer: 0.345

Step-by-step explanation:

Given: The the setup for this hypothesis test:

[tex]H_0:p=0.65\n\nH_a:p < 0.65[/tex]

Since , the alternative hypothesis is left tail , so the test is left tail test.

[tex]P_0=0.65\n\nQ_0=1-0.65=0.35\n\np=(126)/(200)=0.63[/tex]

The test static for population proportion is given by :-

[tex]z=\frac{p-P_0}{\sqrt{(P_0Q_0)/(n)}}\n\n\Rightarrow\ z=\frac{0.63-0.65}{\sqrt{(0.65-0.35)/(126)}}=−0.409878030638\approx-0.4[/tex]

The p-value of [tex]z = P(z < -0.4)= 0.3445783\approx0.345[/tex]

Answer:

A. a false generalization

Step-by-step explanation:

All people who work do so in an office, at a computer. Bill works, so he works in an office, at a computer.

This is false generalization.

A false generalization is a mistake that we do when we generalize something without considering all the points or variables.

Like here, if people work, that does not generally means they always work in offices and work in front of computers.

There are many professions like various sports, civil engineering, police patrolling etc that cannot be done while sitting in an office and in front of a computer.

Hence, option A is the answer.

Yes.

1467-776=691

Which means 776 believed the country was protected and 691 believed the country wasn't.

To determine if a majority of surveyed citizens in 1945 believed the US could develop protection against atomic bombs, a one-sample proportion test must be conducted. If the test results in a p-value less than the 0.01 level of significance, we can conclude that a majority did hold this belief. This involves concepts in hypothesis testing and statistics.

The question you're asking requires a basic understanding of hypothesis testing in statistics. Let's denote the proportion of those who answered 'yes' as p and the total number of citizens surveyed as n.

In this case, p = 776 / 1467 = 0.529, meaning 52.9% of the sampled citizens responded yes leading to the claim that a majority believe the US could develop protection against atomic bombs.

However, to determine if this is truly a majority at a 0.01 level of significance, we need to conduct a one-sample proportion test. Our null hypothesis (H0) would state that the population proportion is 0.5 (no majority), while our alternative hypothesis (H1) would be that the population proportion is greater than 0.5 (there is a majority).

If we find a p-value less than the 0.01 level of significance from the test, we reject H0 and conclude that a majority of Americans did believe in the capacity for atomic bomb protection. If the p-value is greater than 0.01, we would fail to reject H0, concluding that a majority belief can't be inferred.

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a. We're looking for a scalar function [tex]f(x,y)[/tex] such that [tex]\vec F(x,y)=\nabla f(x,y)[/tex]. This means

[tex]\dfrac{\partial f}{\partial x}=y[/tex]

[tex]\dfrac{\partial f}{\partial y}=x[/tex]

Integrate both sides of the first PDE with respect to [tex]x[/tex]:

[tex]\displaystyle\int\frac{\partial f}{\partial x}\,\mathrm dx=\int y\,\mathrm dx\implies f(x,y)=xy+g(y)[/tex]

Differentating both sides with respect to [tex]y[/tex] gives

[tex]\dfrac{\partial f}{\partial y}=x=x+\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=C[/tex]

so that [tex]\boxed{f(x,y)=xy+C}[/tex]. A potential function exists, so the fundamental theorem does apply.

Green's theorem also applies because [tex]C[/tex] is a simple and smooth curve.

b. Now with (and I'm guessing as to what [tex]\vec G[/tex] is supposed to be)

[tex]\vec G(x,y)=\dfrac y{x^2+y^2}\,\vec\imath-\dfrac x{x^2+y^2}\,\vec\jmath[/tex]

we want to find [tex]g[/tex] such that

[tex]\dfrac{\partial g}{\partial x}=\dfrac y{x^2+y^2}[/tex]

[tex]\dfrac{\partial g}{\partial y}=-\dfrac x{x^2+y^2}[/tex]

Same procedure as in (a): integrating the first PDE wrt [tex]x[/tex] gives

[tex]g(x,y)=\tan^{-1}\dfrac xy+h(y)[/tex]

Differentiating wrt [tex]y[/tex] gives

[tex]-\dfrac x{x^2+y^2}=-\dfrac x{x^2+y^2}+\dfrac{\mathrm dh}{\mathrm dy}\implies h(y)=C[/tex]

so that [tex]\boxed{g(x,y)=\tan^{-1}\dfrac xy+C}[/tex], which is undefined whenever [tex]y=0[/tex], and the fundamental theorem applies, and Green's theorem also applies for the same reason as in (a).

c. Same as (b) with slight changes. Again, I'm assuming the same format for [tex]\vec H[/tex] as I did for [tex]\vec G[/tex], i.e.

[tex]\vec H(x,y)=\dfrac x{x^2+y^2}\,\vec\imath+\dfrac y{x^2+y^2}\,\vec\jmath[/tex]

Now

[tex]\dfrac{\partial h}{\partial x}=\dfrac x{x^2+y^2}\implies h(x,y)=\dfrac12\ln(x^2+y^2)+i(y)[/tex]

[tex]\dfrac{\partial h}{\partial x}=\dfrac y{x^2+y^2}=\dfrac y{x^2+y^2}+\dfrac{\mathrm di}{\mathrm dy}\implies i(y)=C[/tex]

[tex]\implies\boxed{h(x,y)=\dfrac12\ln(x^2+y^2)+C}[/tex]

which is undefined at the point (0, 0). Again, both the fundamental theorem and Green's theorem apply.

The potential functions for the given vector fields F, G and H are f(x,y) = xy, g(x,y) = (1/2)xy² - arctan(x/y), and h(x,y) = (1/2)(x² + y²). Also, Green's theorem and the Fundamental Theorem of Line Integrals apply under certain conditions and constraints on the fields.

The potential function for F is indeed f(x,y) = xy as the partial derivatives of this function are equal to the components of F. The Fundamental Theorem of Line Integrals applies to F • dr C because F is the gradient of the scalar potential f and C is a closed curve. Green's Theorem applies as well since both F = y i + x j and C satisfy the appropriate conditions of continuity and differentiability.

The potential function for G is g(x,y)= (1/2)xy² - arctan(x/y) when y !=0. It is not defined where y=0. The Fundamental Theorem of Line Integrals does not apply because G is not conservative, meaning it doesn't have a potential function with continuous derivatives everywhere. Green's Theorem does apply to G • dr C because both the vector field G and C satisfy the conditions it requires.

The potential function for H is h(x,y)= (1/2)(x² + y²) and it is defined everywhere except at the origin (0,0) where it becomes undefined because of division by zero in its components. The Fundamental Theorem of Line Integrals does apply to H · dr C but only for curves which avoid the origin, because H is not conservative everywhere. Green's Theorem does not apply to H · dr C because H's components are not differentiable at the origin.

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