You want to estimate the proportion of students at your college or university who are employed for 10 or more hours per week while classes are in session. You plan to present your results by a 95% confidence interval. Using the guessed value p* = 0.33, find the sample size required if the interval is to have an approximate margin of error of m = 0.06.

Answer:

0.86 or 86%

Step-by-step explanation:

The data given represent 41% of people in a certain country like to cook and 68% of people in the country like to shop, while 14% enjoy both activities.

The probability that a randomly selected person in the country enjoys cooking or shopping or both.

People who like to cook P(C) = 41% = 0.40

People who like to shopping P(S) = 68% = 0.60

People who like cooking and shopping both P(C∩S) =  14% = 0.14

People who like cooking or shopping or both = P(C∪S)

                        = P(C) + P(S) - P(C∩S)

                        =  0.40 + 0.60 - 0.14

                        = 0.86

The probability that a randomly selected person in the country enjoys cooking or shopping or both is 0.86 or 86%

To calculate the probability that a selected person likes cooking or shopping or both, we add the probabilities of each individual event and subtract the overlapping probability. In this case, it's 94.1%.

You want to find the probability that a randomly selected person in the country enjoys cooking, shopping, or both. To calculate this probability, you can use the principle of inclusion-exclusion for two sets A and B, where:

The formula for the probability that a randomly selected person enjoys either cooking or shopping (or both) is:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Given:

Plug in the values:

P(A ∪ B) = 40.1% + 68% - 14%
= 94.1%

So, the probability that a randomly selected person enjoys cooking or shopping or both is 94.1%.

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Answer:

$8000 is invested for 7% interest and $2000 is invested for 9% interest

Step-by-step explanation:

Points to remember

Simple interest formula

I = PNR/100

P - Principle amount

N - Number of years

R - Rate of interest

To find the amount of investment

It is given that total amount = 10,000 and total interest = $740

Let 'x' be the amount invested at the rate of 7%

10,000 - x be the amount invested at the rate of 9%

I = PNR/100

740 = (x*1*7)/100 + (10000 - x)*1*9/100

740 = 7x/100 + 90000/100 - 9x/100

740 = 7x/100 + 900 - 9x/100

740-900 = -2x/100

-160 = -2x/100

x = 16000/2 = 8000

10000-8000 = 2000

Therefore $8000 is invested for 7% interest and $2000 is invested for 9% interest

Answer:

$8000 is invested for 7% interest and $2000 is invested for 9% interest

Step-by-step explanation:

Points to remember

Simple interest formula

I = PNR/100

P - Principle amount

N - Number of years

R - Rate of interest

To find the amount of investment

It is given that total amount = 10,000 and total interest = $740

Let 'x' be the amount invested at the rate of 7%

10,000 - x be the amount invested at the rate of 9%

I = PNR/100

740 = (x*1*7)/100 + (10000 - x)*1*9/100

740 = 7x/100 + 90000/100 - 9x/100

740 = 7x/100 + 900 - 9x/100

740-900 = -2x/100

-160 = -2x/100

x = 16000/2 = 8000

10000-8000 = 2000

Therefore $8000 is invested for 7% interest and $2000 is invested for 9% interest

I really hope it helped idiot noob die

Answer:

y = 2x − 1

Step-by-step explanation:

By eliminating the parameter, first solve for t:

x = 4 + ln(t)

x − 4 = ln(t)

e^(x − 4) = t

Substitute:

y = t² + 6

y = (e^(x − 4))² + 6

y = e^(2x − 8) + 6

Taking derivative using chain rule:

dy/dx = e^(2x − 8) (2)

dy/dx = 2 e^(2x − 8)

Evaluating at x = 4:

dy/dx = 2 e^(8 − 8)

dy/dx = 2

Writing equation of line using point-slope form:

y − 7 = 2 (x − 4)

y = 2x − 1

Now, without eliminating the parameter, take derivative with respect to t:

x = 4 + ln(t)

dx/dt = 1/t

y = t² + 6

dy/dt = 2t

Finding dy/dx:

dy/dx = (dy/dt) / (dx/dt)

dy/dx = (2t) / (1/t)

dy/dx = 2t²

At the point (4, 7), t = 1.  Evaluating the derivative:

dy/dx = 2(1)²

dy/dx = 2

Writing equation of line using point-slope form:

y − 7 = 2 (x − 4)

y = 2x − 1

To find the tangent to the curve represented by the parametric equations x = 4 + ln(t), y = t² + 6, both methods, eliminating and not eliminating the parameter t, yield the same result. The slope of the tangent line at the point (4, 7) is determined to be 2, thus the equation of the tangent is y - 7 = 2(x - 4).

To find the equation of the tangent to the given curve at the point (4, 7) with the parametric equations x = 4 + ln(t) and y = t² + 6, we can approach the problem in two ways: with and without eliminating the parameter t.

Firstly, without eliminating the parameter, we need to find the derivatives dx/dt and dy/dt, and then use them to find dy/dx which is the slope of the tangent at the given point. Since dx/dt = 1/t and dy/dt = 2t, at the point (4, 7), we have t = 1, making the slope dy/dx = (dy/dt)/(dx/dt) = 2 × 1 / (1/1) = 2. The equation of the tangent line can thus be written as y - y₁ = m(x - x₁), where m is the slope, and (x₁, y₁) is the point of tangency.

This gives us the equation y - 7 = 2(x - 4).

Answer:

The value of k is 6

Step-by-step explanation:

we need to find the value of k

Given : - [tex]y=e^{2x}[/tex] is the solution [tex]y''-5y'+ky=0[/tex]

[tex]y=e^{2x}[/tex]                               ........(1)                  

differentiate  [tex]y=e^{2x}[/tex] with respect to 'x'

[tex]\frac{dy}{dx}=\frac{d}{dx}e^{2x}[/tex]

Since, [tex]\frac{d}{dx}e^{x} =e^{x}\frac{d}{dx}(x)[/tex]

[tex]\frac{dy}{dx}=e^{2x}\frac{d}{dx}(2x)[/tex]

[tex]\frac{dy}{dx}=e^{2x}\times 2[/tex]

[tex]\frac{dy}{dx}=2e^{2x}[/tex]

so, [tex]y'=2e^{2x}[/tex]                     ..........(2)

Again differentiation above with respect to 'x'

[tex]\frac{d}{dx}\frac{dy}{dx}=\frac{d}{dx}2e^{2x}[/tex]

[tex]\frac{d^{2}y}{dx^{2}}=2e^{2x}\frac{d}{dx}(2x)[/tex]

[tex]\frac{d^{2}y}{dx^{2}}=2e^{2x}\times 2[/tex]

[tex]\frac{d^{2}y}{dx^{2}}=4e^{2x}[/tex]

so, [tex]y''=4e^{2x}[/tex]                         ........(3)

Now, put the value of [tex]y\ ,y' \ \text{and} \ y''[/tex] in [tex]y''-5y'+ky=0[/tex]

[tex]4e^{2x}-5(2e^{2x})+(e^{2x})k=0[/tex]

[tex]4e^{2x}-10e^{2x}+e^{2x}k=0[/tex]

[tex]-6e^{2x}+e^{2x}k=0[/tex]

add both the sides by [tex]6e^{2x}[/tex]

[tex]e^{2x}k=6e^{2x}[/tex]

Cancel out the same terms from left and right sides

[tex]k=6[/tex]

Hence, the value of k is 6

checking the vertex of this upside-down parabola, it has a vertex at (1000, 2000000), so that's the U-turn, when as the price "p" increases the revenue goes down.

[tex]\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{a}x^2\stackrel{\stackrel{b}{\downarrow }}{+b}x\stackrel{\stackrel{c}{\downarrow }}{+c} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)[/tex]

now, if we solve the quadratic using the value of 500000

[tex]\bf \stackrel{R(p)}{500000}=-2p^2+4000p\implies 250000=-p^2+2000p \\\\\\ p^2-2000p+250000=0[/tex]

and we run the quadratic formula on it, we get the values of x = 133.97 and x = 1866.03, one value is obviously when going upwards, the first one, and the other is when going downwards.

so we know that the R(p) is 500,000 at x = 133.97, and it keeps on going up, up to the vertex above at x = 1000, so we can say from x = [134, 1000] R(p) > 500000.

[tex]\bf \stackrel{height}{h(t)}=-16t^2+32t+48\implies \stackrel{48~ft}{~~\begin{matrix} 48 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}=-16t^2+32t~~\begin{matrix} +48 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix} \\\\\\ 0=-16t^2+32t\implies 16t^2-32t=0\implies 16t(t-2)=0\implies t= \begin{cases} 0\\ 2 \end{cases}[/tex]

t = 0 seconds, when the ball first took off, and t = 2, 2 seconds later.

Answer: [tex]t_1=0\\t_2=2[/tex]

Step-by-step explanation:

We know that the function [tex]h(t)=-16t^2+32t+48[/tex] models the height  "h" of the ball above the ground as a function of time "t".

Then, to find the times in which the ball will be 48 feet above the ground, we need to substitute [tex]h=48[/tex] into the function and solve fot "t":

[tex]48=-16t^2+32t+48\\0=-16t^2+32t+48-48\\0=-16t^2+32t[/tex]

Factorizing, we get:

[tex]0=-16t(t-2)\\t_1=0\\t_2=2[/tex]

Answer:

Step-by-step explanation:

So there is a 3% probability that an athlete is using EPO .

The probability of showing positive on a test when you've used it is 0.99.

3% x 0.99= 2.97%

The probability of a positive result without EPO is 0.1

97% x 0,1 = 9,7 %

My guess is that 2.97% + 9,7% = 12.67% or 0.1267.

I don't know i may be wrong because you've put as an answer 0.0297 but if you like you may take only the first part of the answer.

There is a 0.1267 = 12.67% probability that a randomly selected athlete tests positive for EPO.

A positive test can happen in two cases:

Then, adding these probabilities:

[tex]p = 0.03(0.99) + 0.97(0.1) = 0.1267[/tex]

0.1267 = 12.67% probability that a randomly selected athlete tests positive for EPO.

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Answer:

{[tex]z|z<-\frac{1}{2}[/tex]}U{[tex]z|z>\frac{1}{2}[/tex]}

Step-by-step explanation:

Given the inequality [tex]|z| > \frac{1}{2}[/tex] you need to set up two posibilities:

FIRST POSIBILITY : [tex]z>\frac{1}{2}[/tex]

SECOND POSIBILTY: [tex]z<-\frac{1}{2}[/tex]

Therefore, you got that:

[tex]z<-\frac{1}{2}\ or\ z>\frac{1}{2}[/tex]

Knowing this, you can write the solution obtained in Set notation. This is:

Solution: {[tex]z|z<-\frac{1}{2}[/tex]}U{[tex]z|z>\frac{1}{2}[/tex]}

Answer:

{z|z < (-1/2) ∪ z > (1/2)}

Step-by-step explanation:

I got it right. I would explain if I were better at making sense lol

Answer:

Width of box= 2inches

Length of box= 4inches

Height of box= 6inches.

Step-by-step explanation:

Let width of box=x inches

Length of box = twice of width=[tex]2\times x[/tex]=[tex]2x[/tex]

Height of box= 4 inches greater than width= [tex]x+4[/tex]

Volume of box= 48 cubic inches

We know that the formula of volume of cuboid= [tex] length\times breadth\times height[/tex]

Apply the formula

Volume of box= [tex]x\times 2x\times (x+4)[/tex]

Volume of cube = [tex]2x^2(x+4)[/tex]

[tex]2x^2(x+4)=48[/tex]

[tex]x^2(x+4)=24[/tex]

[tex]x^3+x^2-24[/tex]

Apply inspection method to solve the equation

Put [tex]x=0[/tex]

Then we get [tex]-24\neq0 [/tex]

Hence, x=0 is not the solution of x

Put x=1 in the equation then we get

[tex]-22\neq 0[/tex]

Hence x=1 is not the solution of equation.

Put x=2 then we get

[tex](2)^3+(4)^2-24[/tex]

8+16-24=0

Hence, x=2 is the solution of equation .

[tex] (x-2)(x^2+6x+12)[/tex]=0

Now substitute equation [tex]x^2+6x+12[/tex]=0

Sum roots =6

Product of roots=12

When sum of roots  is greater than zero and product of roots is greater than zero then value of roots of equation is imaginary.

Hence, the roots of equation [tex]x^2+6x+12=0[/tex] are imaginary.

Lenght , widht and height are dimensions of box therefore, imaginary value are not possible.

Hence,[tex] x=2 [/tex] is the only real values of root of equation .Therefore, it is possible and other two imaginary value of roots are not possible .

Widht of box=2 inches

Length of box = [tex]2\times2[/tex]=4inches

Height of box=[tex]x+4[/tex]=2+4=6 inches

The dimensions of the box are: length = 4 inches, width = 2 inches, and height = 6 inches.

Let's use the given information to solve for the dimensions of the box:

Let the width of the box be represented by x inches.

The length of the box is twice as long as the width, so the length is 2x inches.

The height is 4 inches greater than the width, so the height is (x + 4) inches.

The volume of a box can be calculated by multiplying the length, width, and height. Since the volume of the box is given as 48 cubic inches, we can set up the equation: 2x * x * (x + 4) = 48.

Simplifying the equation, we get 2x^3 + 8x^2 - 48 = 0.

Factoring the equation, we find that (x - 2)(x + 4)(x + 6) = 0.

The possible solutions are x = 2, x = -4, or x = -6.

Since we are dealing with dimensions, the width cannot be negative, so we can disregard the negative solutions. The width, therefore, is 2 inches.

The length is twice as long as the width, so the length is 2 * 2 = 4 inches.

The height is 4 inches greater than the width, so the height is 2 + 4 = 6 inches.

Therefore, the dimensions of the box are: length = 4 inches, width = 2 inches, and height = 6 inches.

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Answer:

103.62 cm

Step-by-step explanation:

Given : Diameter, D = 33 cm

circumference = πD = 3.14 x 33 = 103.62 cm

Hello, I believe your answer is C.

You can find your answer by plugging in the diameter into the circumference formula: 2πr² (You must divide your diameter in half to get the radius).

Approximately 85.87 mg of Bismuth-210 would remain after 8 days.

To calculate the amount of Bismuth-210 remaining after 8 days, we need to apply the concept of radioactive decay. Bismuth-210 decays by about 13% each day, meaning that 13% of the remaining Bismuth-210 transforms into another atom (Polonium-210) each day.

Let's calculate the amount remaining:

Therefore, after 8 days, approximately 85.87 mg of Bismuth-210 would remain.

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After 8 days, approximately 76.69 mg of Bismuth-210 remains out of the initial 233 mg.

To calculate the amount of Bismuth-210 remaining after 8 days of radioactive decay, follow these steps:

Step 1:

Understand the decay rate.

Each day, 13% of the remaining Bismuth-210 decays into Polonium-210. This means that 87% of the Bismuth-210 remains each day.

Step 2:

Calculate the remaining amount each day.

Start with the initial amount of 233 mg of Bismuth-210.

After the first day: [tex]\( 233 \text{ mg} \times 0.87 = 202.71 \text{ mg} \)[/tex]

After the second day: [tex]\( 202.71 \text{ mg} \times 0.87 = 176.43 \text{ mg} \)[/tex]

Continue this process for 8 days.

Step 3:

Perform the calculations for 8 days.

[tex]\[ \text{Day 1: } 233 \text{ mg} \times 0.87 = 202.71 \text{ mg} \][/tex]

[tex]\[ \text{Day 2: } 202.71 \text{ mg} \times 0.87 = 176.43 \text{ mg} \][/tex]

[tex]\[ \text{Day 3: } 176.43 \text{ mg} \times 0.87 = 153.62 \text{ mg} \][/tex]

[tex]\[ \text{Day 4: } 153.62 \text{ mg} \times 0.87 = 133.67 \text{ mg} \][/tex]

[tex]\[ \text{Day 5: } 133.67 \text{ mg} \times 0.87 = 116.33 \text{ mg} \][/tex]

[tex]\[ \text{Day 6: } 116.33 \text{ mg} \times 0.87 = 101.28 \text{ mg} \][/tex]

[tex]\[ \text{Day 7: } 101.28 \text{ mg} \times 0.87 = 88.10 \text{ mg} \][/tex]

[tex]\[ \text{Day 8: } 88.10 \text{ mg} \times 0.87 = 76.69 \text{ mg} \][/tex]

Step 4:

Interpret the result.

After 8 days, approximately 76.69 mg of Bismuth-210 remains.

So, after 8 days, approximately 76.69 mg of Bismuth-210 remains.

Answer:

[tex]1000±100\sqrt{55}[/tex]

Step-by-step explanation:

To simplify that expression, first we need to find the largest common of the expression inside the radical, in this case: 2.200.000.

We know that 2.200.000 = 2 · 2 · 2 · 2 · 2 · 2 · 5 · 5 · 5 · 5 · 5 · 11 = [tex]2^{6}[/tex] ×[tex]5^{5}[/tex]× [tex]11[/tex]

Now, [tex]\sqrt{2^{6}5^{5}11} = 200\sqrt{55}[/tex].

Now we have: [tex]\frac{2000±200\sqrt{55}}{2}[/tex]

Dividing by 2: [tex]1000±100\sqrt{55}[/tex]

So the simplified expression is: [tex]1000±100\sqrt{55}[/tex]

Answer:

the awnser is a -2.75

Step-by-step explanation:

Answer:

129 468  

Step-by-step explanation:

I'm not sure if you're supposed to reuse the same numbers just in a different Three digit number. Or if you're supposed to use the number one time and one time only. But if not here's there's some numbers that you could use

He basically all You have to do Is take the numbers and turning them into three digit numbers Without repetition.

Hope this helps you!

Also it's saying that 664 is not allowed Because it they are reusing the six When there's no Extra six to use. So remind you not to use the same number twice!

Answer:

The answer is in the attachment.

Step-by-step explanation:

Look at the picture.

The rectangle that has a vertex of C and has one of it's sides on line CD, with the stated lengths is constructed as shown in the image attached below (see attachment).

A rectangle can be described as a 4-sided polygon having all its four interior angles measuring 90 degrees each and has two pairs of opposite equal sides.

Thus, the rectangle that has a vertex of C and has one of it's sides on line CD, with the stated lengths is constructed as shown in the image attached below (see attachment).

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Answer: Hence, a) No, they are not independent

b) 0.193

c) 0.60

Step-by-step explanation:

Since we have given that

Probability that a child's gestational age is less than 37 weeks say P(A)= 0.142

Probability that his or her birth weight is less than 2500 grams say P(B) = 0.051

P(A∩B) = 0.031

We need to check whether it is independent or not.

Since ,

[tex]P(A).P(B)=0.142\times 0.051=0.0072[/tex]

and

[tex]P(A\cap B)=0.051[/tex]

So, we can see that

[tex]P(A).P(B)\neq P(A\cap B)[/tex]

So, it is not independent.

a) Hence, A and B are not independent.

b) P(A∪B) is given by

[tex]P(A\ or B\ or\ both)=P(A)+P(B)\\\\P(A\ or\ B\ or\ both)=0.142+0.051=0.193[/tex]

c) P(A|B) is given by

[tex]P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.031}{0.051}=0.60[/tex]

Hence, a) No,

b) 0.193

c) 0.60

Answer:

120/s^6

Step-by-step explanation:

There is an easy formula for this...

L(t^n)=n!/(s^(n+1))

Your n=5 here

L(t^5)=5!/(s^6)

L(t^5)=120/s^6

[tex]\mathcal{L}\{t^n\}=\dfrac{n!}{s^{n+1}}[/tex]

So

[tex]\mathcal{L}\{f(t)\}=\dfrac{5!}{s^{5+1}}=\dfrac{120}{s^6}[/tex]

Since there is a right angle, you can use Pythagoras' Theorem:

So x = √(24² + 7²) = 25

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

Answer:

25

Answer: Simple random sampling

Step-by-step explanation:

Given: A market researcher obtains a list of all streets in a town. She randomly samples 10 street names from the list, and then administers survey questions to every family living on those 10 streets.

Since she randomly samples street names , therefore the type of sampling is simple random sampling.

Answer:

Faster cyclist: 6x

Slower: x

Step-by-step explanation:

x = mph

The speed of the faster cyclist is expressed as v + 6 mph, where v represents the speed of the slower cyclist.

To express the speed of the faster cyclist in terms of the speed of the slower cyclist, let's denote the speed of the slower cyclist as v mph. The problem states that the faster cyclist travels at a speed that is six miles per hour faster than the slower cyclist. Therefore, the speed of the faster cyclist can be expressed as v + 6 mph.

For example, if the slower cyclist is travelling at a speed of 9 mph, the faster cyclist would be traveling at 9 mph + 6 mph = 15 mph.

Answer:  Probability that he will receive a grade of B or better in both courses is 0.30.

Step-by-step explanation:

Since we have given that

Probability that he will get a grade of B or better in introduction to Finance say P(A) = 0.6

Probability that he will get a grade of B or better in introduction to Accounting say P(B) = 0.5

Since A and B are independent events.

We need to find the probability that he will receive a grade of B or better in both the courses.

So, it becomes,

[tex]P(A\cap B)=P(A).P(B)\\\\P(A\cap B)=0.6\times 0.5\\\\P(A\cap B)=0.30[/tex]

Hence, Probability that he will receive a grade of B or better in both courses is 0.30.

The probability that you will receive a grade of B or better in both courses is 0.3.

The probability that you will receive a grade of B or better in both Introduction to Finance and Introduction to Accounting, given that the events are independent, is calculated by multiplying the individual probabilities of each event.

[tex]\( P(F) = 0.6 \) \( P(A) = 0.5 \)[/tex]

Since the events are independent, the probability of both events occurring is given by the product of their individual probabilities:

[tex]\( P(F \text{ and } A) = P(F) \times P(A) \)[/tex]

Substituting the given probabilities:

[tex]\( P(F \text{ and } A) = 0.6 \times 0.5 \) \( P(F \text{ and } A) = 0.3 \)[/tex]

The final answer is [tex]\(\boxed{0.3}\).[/tex]

Answer:

[tex]\large\boxed{(-7x+4)(-7x-4)}[/tex]

Step-by-step explanation:

[tex]\text{The difference of squares:}\ a^2-b^2=(a-b)(a+b)\\\\(-7x+4)(-7x-4)=(-7x)^2-4^2=49x^2-16[/tex]

(-7x + 4) (-7x - 4) can be written as a difference of squares.

Option C is the correct answer.

An expression contains one or more terms with addition, subtraction, multiplication, and division.

We always combine the like terms in an expression when we simplify.

We also keep all the like terms on one side of the expression if we are dealing with two sides of an expression.

Example:

1 + 3x + 4y = 7 is an expression.com

3 + 4 is an expression.

2 x 4 + 6 x 7 – 9 is an expression.

33 + 77 – 88 is an expression.

We have,

The difference of squares is a special algebraic form that occurs when we multiply two binomials of the form (a + b)(a - b).

This results in the product of two terms:

The square of the first term minus the square of the second term.

In other words, we have (a + b)(a - b) = a² - b².

In the given options, only (-7x + 4) (-7x - 4) can be written as a difference of squares, by applying the above formula.

We can rewrite it as:

(-7x + 4) (-7x - 4) = (-7x)² - 4² = 49x² - 16

The other options do not follow this pattern and cannot be written as a difference of squares.

Thus,

(-7x + 4)(-7x - 4) = 49x² - 16

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You can use the law sines, which states that in a triangle the ratio between one side length and the sine of the opposite angle is constant.

So, we have

[tex]\dfrac{PR}{\sin(Q)}=\dfrac{QR}{\sin(P)}=\dfrac{PQ}{\sin(R)}[/tex]

In particular, we can use

[tex]\dfrac{PR}{\sin(Q)}=\dfrac{QR}{\sin(P)}[/tex]

to write

[tex]\dfrac{68}{\sin(73)}=\dfrac{47.6}{\sin(P)} \iff \sin(P) = \dfrac{47.6\sin(73)}{68}\approx 0.66[/tex]

Which means

[tex]P\approx \arcsin(0.66)\approx 42[/tex]

[tex]\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec s=\int_0^1(t^2+t^4,4t^3)\cdot(1,2t)\,\mathrm dt=\int_0^1(t^2+9t^4)\,\mathrm dt=\boxed{\frac{32}{15}}[/tex]

[tex]\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec s=\int_0^1(2t^2,4t^2)\cdot(1,1)\,\mathrm dt=\int_0^16t^2\,\mathrm dt=\boxed{2}[/tex]

The value of the line integral depends on the path, so [tex]\vec F[/tex] is not a gradient vector field.

The line integrals ∫c1F⋅ds and ∫c2F⋅ds are computed by replacing x and y with the parametric representations, calculating ds, completing the dot product, and conducting the integration. If the results are identical, F is a gradient vector field.

To compute the line integrals, ∫c1F⋅ds and ∫c2F⋅ds, where c1(t)=(t, t2) and c2(t)=(t,t) for 0≤t≤1 of the vector field F=(x2+y2,4xy), we can reduce each of them to an integral over t, the parameter of the path. In the case of c1(t), replace x and y by t and t² correspondingly, for calculation. Similarly, in the case of c2(t), replace x and y by t in calculations.

Let's consider ∫c1F⋅ds. Here, F = (t²+t⁴,4t³) and ds can be calculated using the Pythagorean theorem leading to sqr(1+4t²). The dot product F.ds is then calculated and integrated from 0 to 1. Repeat the process for ∫c2F⋅ds.

A vector field F is said to be a gradient vector field if integral from one point to another remains the same regardless of the path chosen to get from one point to the other. Comparing the obtained results will determine the truth of this statement.

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Answer:

Step-by-step explanation:

This sort of problem is best worked by a tool such as a graphing calculator or spreadsheet.

Answer:

There is a 94.52% probability that a randomly selected adult has an IQ less than 132.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 20. This means that [tex]\mu = 100, \sigma = 20[/tex].

The probability that a randomly selected adult has an IQ less than 132 is?

This probability is the pvalue of Z when [tex]X = 132[/tex]. So:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{132 - 100}{20}[/tex]

[tex]Z = 1.6[/tex]

[tex]Z = 1.6[/tex] has a pvalue of 0.9452.

This means that there is a 94.52% probability that a randomly selected adult has an IQ less than 132.

Assume that adults have IQ scores that are normally distributed with a mean of mu equals 100 and a standard deviation sigma equals 20. The probability that a randomly selected adult has an IQ less than 132 is 0.9452 or 94.52%.

Let standardize the IQ value of 132 using the formula for standardization:

Z = (X - μ) / σ

Where:

Z= standardized value (Z-score)

X = IQ value

μ = mean= 100

σ = standard deviation =20

Let's calculate the Z-score for an IQ of 132:

Z = (132 - 100) / 20

Z = 32 / 20

Z = 1.6

Using a standard normal distribution table, the probability is 0.9452.

Therefore, the probability that a randomly selected adult has an IQ less than 132 is 0.9452 or 94.52%.

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Answer:

12 a. 4605 feet 12 b. 1,459,063 square feet

Step-by-step explanation:

For the perimeter, we simply add the lengths of each of the 5 sides together (or multiply 5 times one side length).

P = 5(921)

P = 4605 feet

For the area, we will use composition...add the area of the triangle to the area of the trapezoid.

For the area of the triangle, the formula is

[tex]A=\frac{1}{2}bh[/tex].

Filling in our values gives us

[tex]A=\frac{1}{2}(1490)(541)[/tex] and

A = 403,045 square feet.

Now for the trapezoid.  The formula for a trapezoid is

[tex]A=\frac{1}{2}(b_{1}+b_{2})(h)[/tex]

where the b's represent the bases and the h represents the height.  Filling in our values gives us

[tex]A=\frac{1}{2}(921+1490)(876)[/tex]

Work inside the parenthesis first:

[tex]A=\frac{1}{2}(2411)(876)[/tex] and

A = 1,056,018

Now we add those together to get that area of the Pentagon is 1,459,063 square feet

Answer:

A vertical line has an infinite or undefined slope since the denominator is zero.

Step-by-step explanation:

Parallel lines by definition refers to lines that never intersect or meet since they have identical slopes. The slope of line is defined as;

(change in y)/(change in x)

For a vertical line, the y values are changing while the x values remain constant. The slope of this line will thus have a zero value in the denominator implying that its slope will not defined or will be infinity.

Answer:

A vertical line has an infinite or undefined slope since the denominator is zero.

Step-by-step explanation:

a. The average value of [tex]f[/tex] on the given interval is

[tex]\displaystyle f_{\rm ave}=\frac1{\pi-0}\int_0^\pi(2\sin x-\sin2x)\,\mathrm dx=\boxed{\frac4\pi}[/tex]

b. Solve for [tex]c[/tex]:

[tex]\dfrac4\pi=2\sin c-\sin2c\implies\boxed{c\approx1.238\text{ or }c\approx2.808}[/tex]

The derivative of the function f(x) is:

                 [tex]f'(x)=-2x[/tex]

We are given a function f(x) as:

[tex]f(x)=5-x^2[/tex]

We have:

[tex]f(x+h)=5-(x+h)^2\\\\i.e.\\\\f(x+h)=5-(x^2+h^2+2xh)[/tex]

( Since,

[tex](a+b)^2=a^2+b^2+2ab[/tex] )

Hence, we get:

[tex]f(x+h)=5-x^2-h^2-2xh[/tex]

Also, by using the definition of f'(x) i.e.

[tex]f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}[/tex]

Hence, on putting the value in the formula:

[tex]f'(x)= \lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-(5-x^2)}{h}\\\\\\f'(x)=\lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-5+x^2}{h}\\\\i.e.\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2-2xh}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2}{h}+\dfrac{-2xh}{h}\\\\f'(x)=\lim_{h \to 0} -h-2x\\\\i.e.\ on\ putting\ the\ limit\ we\ obtain:\\\\f'(x)=-2x[/tex]

      Hence, the derivative of the function f(x) is:

          [tex]f'(x)=-2x[/tex]

Answer:

The derivative of given function is -2x.

Step-by-step explanation:

The first principle of differentiation is

[tex]f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}[/tex]

The given function is

[tex]f(x)=5-x^2[/tex]

[tex]f'(x)=lim_{h\rightarrow 0}\frac{5-(x+h)^2-(5-h^2}{h}[/tex]

[tex]f'(x)=lim_{h\rightarrow 0}\frac{5-(x^2+2xh+h^2)-5+h^2}{h}[/tex]

[tex]f'(x)=lim_{h\rightarrow 0}\frac{5-x^2-2xh-h^2-5+h^2}{h}[/tex]

[tex]f'(x)=lim_{h\rightarrow 0}\frac{-x^2-2xh}{h}[/tex]

[tex]f'(x)=lim_{h\rightarrow 0}\frac{-x^2}{h}-\frac{2xh}{h}[/tex]

[tex]f'(x)=lim_{h\rightarrow 0}\frac{-x^2}{h}-2x[/tex]

Apply limit.

[tex]f'(x)=\frac{-x^2}{0}-2x[/tex]

[tex]f'(x)=0-2x[/tex]

[tex]f'(x)=-2x[/tex]

Therefore, the derivative of given function is -2x.