refer to the following table. first event second event a1 a2 a3 total b1 4 11 14 29
b2 7 24 10 41 total 11 35 24 70 a. determine p(a2). (round your answer to 2 decimal places.) b. determine p(b2 | a2). (round your answer to 2 decimal places.) c. determine p(a3 and b2). (round your answer to 2 decimal places.)

-13y + 7 - 11 - 4y can be written as -17y - 11.

Expression is a way of conveying emotions, thoughts, or ideas through verbal or nonverbal means. It is often used to communicate an individual's feelings, beliefs, or opinions, and can be expressed through language, art, music, dance, and other forms of communication. Expression can be used to convey a wide range of emotions, from joy and excitement to anger and sorrow. Expression is an essential part of communication, and without it, we would not be able to effectively convey our thoughts, feelings, and desires. The way we express ourselves can also be a reflection of our personality and can give people an insight into who we are. Expression is also important for self-expression, as it allows us to explore our own thoughts and emotions, and express them in a way that may be more comfortable for us.


This can further be written as -17y × (-1) - 11. Therefore, the expression can be written as a multiplication as -17y × (-1) - 11.

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Given that function f is defined and continuous on the closed interval [a,b], the correct statements are:

From the case we know that:

function f is on the closed interval [a,b]

c = relative maximum

A function reach its relative maximum when its first derivative function is equal to zero (0). Since c is the relative maximum of function f, then:

f'(c) = 0

The third statement is wrong because when c is the relative maximum point, then its second derivative function must be lower than zero (0), or:

f''(c) < 0

When the second derivative function is greater than zero (0), it indicates that the given critical point is the relative minimum.

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3(3)-4 = 5

Hope this helped

Answer: 90-99

Step-by-step explanation:

The top interval in the table would be 90-99. This is because the highest score (X = 96) falls within this interval, and the interval width is 10 points, which means that all scores between 90 and 99 would be included in this interval.

To calculate the distance traveled by the satellite, we need to find the circumference of the circle it traces in the sky. The circumference can be found using the formula:

C = 2 * π * r

where r is the radius of the circle (i.e., the distance from the center of the earth to the satellite), and π is pi, approximately equal to 3.14.

r = 36,000 km

C = 2 * π * 36,000 km

C = 72,000 * π km

Next, we need to find what fraction of the circumference the satellite has traveled in one hour. This is given by the fraction of the total circle the satellite has traced, which is equal to the angle it has formed divided by 360°.

fraction = 32° / 360°

Finally, we can multiply the circumference by the fraction to find the distance traveled by the satellite:

distance = C * fraction

distance = 72,000 * π * 32° / 360° km

Rounding the result to the nearest kilometer, the satellite has traveled approximately:

distance = 73,382 km

So, the satellite traveled approximately 73,382 km in one hour.

The difference between the highest guess and the lowest guess, in the whole number, is 120.

Subtraction is a mathematical operation. Which is used to remove terms or objects in the expression.

Given:

A table that shows the relationship between the guessed amount and the number of weeks.

Week          Amount

1                      120

2                      140

3                      160

4                      240

Here, the highest guess is 240.

And the lowest guess is 120.

The difference  between the highest guess and the lowest guess,

= 240 - 120

= 120

120 is a whole number.

Therefore, 120 is the required number.

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The complete question:

A table that shows the relationship between the guessed amount and the number of weeks.

Week          Amount

1                      120

2                      140

3                      160

4                      240

What was the difference between the highest guess and the lowest guess? Write your answer as a fraction, mixed number, or whole number

Answer:

See below for answers

Step-by-step explanation:

Part A

[tex]\displaystyle V=\pi\int\limits^d_c {(R^2-r^2)} \, dy\\=\pi\int\limits^3_0 {\biggr(\biggr(\frac{y^2}{9}\biggr)^2-\biggr(\frac{y}{3}\biggr)^2\biggr)} \, dy\\=\pi\int\limits^b_a {\biggr(\frac{y^4}{81}-\frac{y^2}{9}\biggr)} \, dy[/tex]

Make sure to write the functions in terms of y since you rotate on a vertical axis.

Part B

[tex]\displaystyle V=\pi\int\limits^b_a {(R^2-r^2)} \, dx\\=\pi\int\limits^1_0 {((3-3\sqrt{x})^2-(3-3x)^2)} \, dx\\=\pi\int\limits^1_0 {((9-18\sqrt{x}+9x)-(9-18x+9x^2))} \, dx\\=\pi\int\limits^1_0 {(-18\sqrt{x}-9x-9x^2)} \, dx[/tex]

Make sure to fix the radii so they start at y=3.

The informational text that is an expression of an opinion on a subject is an essay. That is option B

An essay is defined as the piece of writing or a literary write up that is used to represent the writer's opinion about a subject that is under study.

There are different types of easy written that includes the following:

Therefore, easy is an informational text that is an expression of an opinion on a subject.

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The statement that is true when comparing the two accounts after 10 years is: A: "Hannah's account has more money than Frank's account".

To compare the amount of money in the two accounts after 10 years, we can use the formula for continuous compound interest:

A = Pe^(rt)

where A is the amount of money in the account after t years, P is the principal (initial amount), r is the annual interest rate (expressed as a decimal), and e is the mathematical constant approximately equal to 2.71828.

For Frank's account, we have:

A = 2500e^(0.05410) = 4244.96 dollars (rounded to the nearest cent)

For Hannah's account, we have to take into account the quarterly compounding. The formula for quarterly compound interest is:

A = P(1 + r/n)^(nt)

where n is the number of compounding periods per year (4 for quarterly compounding), and t is the time in years. Thus, we have:

A = 3000*(1 + 0.032/4)^(4*10) = 4412.09 dollars (rounded to the nearest cent)

Therefore, the amount of money in Hannah's account after 10 years is greater than the amount in Frank's account.

"

Complete question

frank invests $2,500 in an account that has an annual interest rate of 5.4% compounded continuously. hannah invests $3,000 in an account that has an annual interest rate of 3.2% compounded quarterly. which of the following statements is true when comparing the amount of money in the two accounts after 10 years?

A: Hannah's account has more money than Frank's account

B: : Frank's account has more money than Hannah's account

"

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a

if you flip it the c point will be the opposite of the original coordinates

Note that the measure of ∠AED is 123°. (Option D) This is arrived at using the properties of a 4 sided polygon and angles on a straight light postulate.

A polygon is a planar figure characterized by a limited number of straight line segments joined to create a closed polygonal chain in geometry. A polygon is defined as a bounded planar region, a bounding circuit, or both. A polygonal circuit's segments are known as its edges or sides.

We know that the sum of angles in a polygon is 360°

Since ∠ABD = 46°

That is Sum of Angles in ΔBDC Less (79+57) =

180 - (101)

= 46°

And ∠EDB = 180-79 (angles on a straight line)

⇒ ∠EDB = 101°

And ∠BAE = 90° (Given)

Thus,

∠AED = 360 - 90-46-101; Thus

∠AED = 123°

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Answer:

20

Step-by-step explanation:

Easiest way to solve is to consider the formula for the area of the trapezoid, which is 0.5h(a+b), where h is the altitude and a and b are the two bases.

Here we know the area, the altitude and one base so let's sub these in to find the other base:

92=0.5*8*(3+b)

92=4(3+b)

23=4+b

b=20

An autonomous differential equation with equilibrium solutions at y=0 and y=3, and y' > 0 for 0 < y < 3 and y' < 0 for -inf < y < 0 and 3 < y < inf is dy/dx = 3y(1-y).

An autonomous differential equation is a type of ordinary differential equation that does not depend on any other variable but time. It has a single variable, usually denoted as y, and a single independent variable, usually denoted as t or x. An autonomous differential equation with equilibrium solutions at y=0 and y=3, and y' > 0 for 0 < y < 3 and y' < 0 for -inf < y < 0 and 3 < y < inf is dy/dx = 3y(1-y). This equation will have two equilibrium solutions at y=0 and y=3. The derivative of the equation, y’, is greater than 0 for 0 < y < 3, which means that the equation is increasing in this range. The derivative of the equation is less than 0 for -inf < y < 0 and 3 < y < inf, which means that the equation is decreasing in this range. This equation is a type of logistic equation and is often used to model population growth. It is also used to describe other phenomena such as the spread of an epidemic or the number of species in an ecosystem. Using the initial conditions y(0) = 2 and y(1) = 3, we can calculate the values of y at other time points using the equation dy/dx = 3y(1-y). At t=0.5, the value of y is y(0.5) = 2.5; at t=1.5, the value of y is y(1.5) = 2.875; and at t=2.5, the value of y is y(2.5) = 2.9375.

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5 is the measure of the value of x from the diagram

The given diagram is a triangle with the following parameters

Opposite = 5

Adjacent  = x

Using the trigonometry identity

tan theta = opp/adj

tan45 = 5/x

x = 5/tan45

x = 5/1

x = 5

Hence the measure of the value of x is 5.

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C = 45 + 0.25t. The 45 represents the baseline amount that she is charged per month. The t represents each additional text, and the 0.25 represents how much she is charged for each additional text.

The wall that is most stable when the same lateral force is applied is the wall that is anchored to the ground with the most support.

The wall that is anchored to the ground with the most support is the one that is most stable when the same lateral force is applied. This could include walls that are reinforced with steel beams or are built with heavier materials such as brick or concrete.

Lateral force is a force that acts perpendicular to the direction of motion of an object. It is usually caused by friction, gravity, or air resistance. The most common example of a lateral force is the force of friction between two objects when they are in contact.

This force can cause an object to move in a different direction than the one it was initially moving in. Lateral force can also cause an object to rotate or change its direction of motion entirely.

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Answer:1,283.803

Step-by-step explanation:

Answer:

Below

Step-by-step explanation:

Line DE grows to 5 as you move to the midpoints of  BC  and AC....

   go the the other half of the way to the end AB and it grows another 5  for a total of 10

The graph of the function f(x) = 3√(3 - x) is at the end of the answer.

We want to graph the function f(x) = 3√(3 - x)

To do so, we need to find some points on that line, to find the points we need to evaluate the function in some values of x

if x = -1

f(-1) = 3√(3 + 1) = 3*2 = 6

if x = 3

f(3) = 3√(3 - 3) = 3*0 = 0

if x = -6

f(-6) = 3*√(3 + 6) = 3*3 = 9

if x = -13

f(-13) =  3*√(3 + 13) = 3*4 = 12

Then the points are (-1, 6), (3, 0), (-6, 9), (-13, 12)

Now we can just graph these 4 points and connect them with a curve, the graph of the function is below.

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Short Answer:

Ruby's rate if reading the books is:

2 months per book.

Long Answer:

If you look at it this way, two months per book means that by the time 14 months come you should have 7 books read.

Easy check to verify:

7 (books) X( mutltiply)  2(# of books)  = 14 (total of months) - This shows that the number lead back to the original numbers

Hope this helps :)

The volume of a pyramid can be calculated using integration is [tex]V = (1/3)(7*282 - 2*282) = 224 cubic units..[/tex]

The formula for the volume of a pyramid with a rectangular base with dimensions l and w, and height h is given by V = (lwh)/3. Using this formula, the volume of a pyramid with a rectangular base with dimensions 2 and 7 and height 28 can be calculated to be 224 cubic units.

To calculate this using integration, we can use the formula V = (1/3)∫b a y2dx, where a and b are the lower and upper limits of the rectangular base, and y is the height of the pyramid. In this case, a = 2, b = 7, and y = 28. Therefore, V = (1/3)(7*282 - 2*282) = 224 cubic units.

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A) No, P(A ∩ B) = 0.5, is not possible. In this case that P(A ∩ B) ≤ P(B). So, the correct choice is option (iii).

B) Probability that the selected student has at least one of these two types of cards is 0.8.

The probability of an event has different properties. If two events take place at the simultaneously, then we calculate their joint probabilities. The individual probabilities are called the marginal probabilities. Let us consider two events :

A : Event that the selected student has Visa card

B : Event that selected student has Master Card

Probability of occurrence of event A, P(A) =0.7

Probability of occurrence of event B, P(B) =0.4

A) P(A ∩ B) = 0.5 ,

No, this is not possible. Since, B is contain in the event (A ∩ B), it must be the case that

P( A ∩ B)≤ P(B). However, 0.5 > 0.4, violate this requirement. So, correct choice is option (iii).

B) Now, it is specific that the probability A and B is 0.3. The probability that the selected student has atleast one of these two types of cards will be, P(A∪B)

= P(A) + P(B) − P(A∩B)

= 0.7 + 0.4v− 0.3 = 0.8

So the probability is 0.8.

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Complete question:

Consider randomly selecting a student at a large university, and let A be the event that the selected student has a Visa card and B be the analogous event for MasterCard. Suppose that P(A) = 0.7 and P(B) = 0.4.

A) Could it be the case that P(A ∩ B) = 0.5? Pick one:

i. Yes, this is possible. Since B is contained in the event A ∩ B, it must be the case that P(B) ≤ P(A ∩ B) and 0.5 > 0.4 does not violate this requirement.

ii. Yes, this is possible. Since A ∩ B is contained in the event B, it must be the case that P(B) ≤ P(A ∩ B) and 0.5 > 0.4 does not violate this requirement.

iii. No, this is not possible. Since B is equal to A ∩ B, it must be the case that P(A ∩ B) = P(B). However 0.5 > 0.4 violates this requirement.

iiii. No, this is not possible. Since B is contained in the event A ∩ B, it must be the case that P(A ∩ B) ≤ P(B). However 0.5 > 0.4 violates this requirement.

v) No, this is not possible. Since A ∩ B is contained in the event B, it must be the case that P(A ∩ B) ≤ P(B). However 0.5 > 0.4 violates this requirement.

B) From now on, suppose that P(A ∩ B)

= 0.3. What is the probability that the selected student has at least one of these two types of cards?

The expected profit per week for David is $4.25, with a standard deviation of $4.28.

Based on the given information, David purchases 8 pounds of bananas every week at 10 cents per pound, which costs $0.80 per week. The demand for good bananas follows the same distribution as d given in the first question. Using the probabilities given in d, we can calculate the expected revenue from selling the bananas. The expected revenue from selling good bananas is (0.30.6 + 0.050.3) * 8 = $1.56. The expected revenue from selling bad bananas is 0.05 * 8 = $0.40.

Thus, the expected total revenue is $1.96. The expected profit is the difference between the total revenue and the cost of purchasing the bananas, which is $1.16. The standard deviation of the profit can be calculated using the formula for the standard deviation of a linear combination of random variables, which gives a standard deviation of $4.28.

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The equation for line 1 is y = x+6 and equation for line 2 is y= -x-4. The domain and range is as follows,

y = x+6 {(x,y)| x≤ -5 , y≤ 1}

y= -x-4  {(x,y) x≥-5, y≤1}

The line 1 passes through points (-6,0) and (-5,1).

Equation for a line is  y= Ax+B

Using two points we can formulate the following equations

-6A+B =0

-5A+B =1

Subtracting the two equations,

-A = -1, so A= 1

-6A+B = 0

-6 ×1 + B = 0

B= 6

So the equation for the line by substituting values of A and B is

y= x+6

The domain is all values ≤-5 and range is all values ≤1

The line 2 passes through (0,-4) and ( -5,1)

Formulating equations using points

B= -4

-5A+B =1

A = -1

Substituting in equation for line 2

y = -x-4

The domain is all values ≥ 5 and range is all values ≤1.

So the equation, domain and range of line 1 is y = x+6 {(x,y)| x≤ -5 , y≤ 1} and for line 2 is y= -x-4  {(x,y) x≥-5, y≤1}

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The complete question is included as image

By the quotient rule the derivative of all functions can be found in an easy way,

(1) [tex]f'(x)=\frac{x^6+128x^3-3x^2}{(x^3+32)^2}[/tex]    

(2)  [tex]f'(x)=\frac{5}{(2+x)^2}[/tex]

(3)  [tex]f'(t)=\frac{4t}{(t^2+1)^2}[/tex]

The Quotient Rule in calculus is a technique for figuring out a function's derivative (differentiation) as the ratio of two differentiable functions. It is a formal rule applied to issues involving differentiation where one function is split in half by another function. The definition of the limit of the derivative is followed by the quotient rule. Keep in mind that the bottom function is where the quotient rule starts and where it finishes.

The quotient rule is similar to the product rule. A Quotient Rule is stated as the ratio of the quantity of the denominator times the derivative of the numerator function minus the numerator times the derivative of the denominator function to the square of the denominator function. In short, the quotient rule is a way of differentiating the division of functions or the quotients

We have the functions

(1) [tex]f(x)=\frac{x^4+1}{x^3+32}[/tex]

derivative with respect to x by the quotient rule.

[tex]f'(x)=\frac{(x^3+32)(4x^3)-(x^4+1)(3x^2)}{(x^3+32)^2}\\\\f'(x)=\frac{4x^6+128x^3-3x^6-3x^2}{(x^3+32)^2}\\\\f'(x)=\frac{x^6+128x^3-3x^2}{(x^3+32)^2}[/tex]

2) derivative with respect to x of the below function

[tex]f(x)=\frac{3x+1}{2+x}\\\\f'(x)=\frac{(x+2)(3)-((3x+1))}{(2+x)^2}\\\\f'(x)=\frac{5}{(2+x)^2}[/tex]

3)

[tex]f(t)=\frac{t^2-1}{t^2+1}[/tex]

derivative with respect to t,

[tex]f'(t)=\frac{(t^2+1)(2t)-(t^2-1)(2t)}{(t^2+1)}\\\\f'(t)=(2t^3+2t-2t^3+2t)/(t^2+1)^2\\\\f'(t)=4t/(t^2+1)^2[/tex]

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The volume of the pyramid is mn, where m and n are positive integers, and n is not divisible by the square of any prime then m+n is 803.

Let the triangular base be ΔABC with

AB = 24, Altitude to the side AB is 16

Area of ΔABC is 24x16/2 = 192

Let fourth vertex of Tetrahedron be P and let the midpoint of AB be M

ΔABC will pass through circumcenter of Δ ABC which we will call O

OM + OC = 16

OM = d

d = √d² + 144 = 16

2d² + 2d√d² + 144 = 112

2d(d + √d² + 144) = 112

d = 7/2

Now, we find √d² + 144 = 25/2

Theorem , on the triangle AOP, BOP, COP

25² = h² + (25/2)²

625 = h² + 625/4

h = 25√3/2

Finally the formula for volume of Pyramid from

V = Ah/3

= 192 (25√3/2) / 3

= 800√3

m + n = 800 + 3

m + n = 803

Therefore, the value of m+n is 803.

The volume of a aggregate depends upon the type of aggregate’s base, whether it's a triangle, square or cube. A aggregate is a polyhedron figure which has only one base.

The base of the aggregate is a poly sided figure. Hence, the formula to find not only volume but also the face area of a aggregate will be grounded on the structure of its base and height of the aggregate.

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The signal, (a) x(t) = 3cos(4t) is periodic with period T = pi/2. To see this, we can compute:

x (t + T) = 3cos (4(t + pi/2)) = 3cos (4t + 2pi) = 3cos(4t)

which shows that the signal repeats itself every T = pi/2 seconds.

[tex](b) \times (t) = e^{( - t)} [/tex]

This signal is not periodic.

[tex]e^{(-t)} = e^{(-(t+T))} [/tex]

for all t.

However, this is not possible since e^(-t) approaches 0 as t approaches infinity. Therefore, the signal is not periodic.

[tex](c) \times (t) = [cos (2t - \frac{\pi}{3}] ^2[/tex]

This signal is periodic with period T = pi. To see this, we can compute:

[tex]x (t + T) = [cos (2(t + \pi) - \pi/3)] ^2 = [cos (2t + 5\pi/3)] ^2 = [cos (2t - pi/3)] ^2 = x(t) \\

(d) x(t) = cos(4t) u(t) + sin(4t) u(-t)[/tex]

This signal is not periodic. To see this, suppose that there exists a period T such that x(t) = x (t + T) for all t. Then we must have:

cos(4t) u(t) + sin(4t) u(-t) = cos(4(t+T)) u(t+T) + sin(4(t+T)) u(-(t+T)) for all t.

However, this is not possible since the two terms on the left-hand side have different signs for t < 0, whereas the two terms on the right-hand side have the same sign for t < -T.

Therefore, the signal is not periodic.

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We have also determined when the amount of salt in tank 2 reaches its maximum, which is when[tex]e^{t/180} = 6, \text{or}[/tex] t ≈ 1,080 minutes.

A mixture is a combination of two or more substances that are physically intermingled with each other, but not chemically bonded.

To determine when the amount of salt in tank 2 reaches its maximum, we can set the derivative of x2(t) with respect to time to zero and solve for t.

[tex]dx_2(t)/dt = (x_1(t)/500) * 500 * (1/180) * e^{t/180} - (x_2(t)/300) * 500 * (1/180) * e^{t/180} = 0[/tex]

Simplifying, we get:

[tex]x_1(t) * e^{-t/180} - (300/5) = x_2(t)[/tex]

Substituting the expression for x2(t) in terms of x1(t), we get:

[tex]x_1(t) * e^{-t/180} - (300/5) = (x_1(t) - (300/5)) * e^{t/18} + (300/5)[/tex]

Simplifying, we get:

[tex]2x_1(t) * e^{-t/180} = (600/5) * e^{t/180}[/tex]

Dividing by [tex]e^{t/180}[/tex] on both sides, we get:

[tex]2x_1(t) * e^{-t/180 - t/180} = (600/5)[/tex]

Simplifying, we get:

[tex]x_1(t) = (300/5) * e^{t/180}[/tex]

Substituting this value of x₁(t) in the expression for x₂(t), we get:

[tex]x_2(t) = (300/5) * (e^{t/180} - 1)[/tex]

This expression for x₂(t) gives us the amount of salt in tank 2 at any time t, and we can see that it increases with time until it reaches its maximum value of 300 oz. This occurs when [tex]e^{t/180}=6,[/tex], or t ≈ 1,080 minutes (or 18 hours), which is independent of the initial concentration in tank 1.

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The length of PJ in the right triangle PNJ is determined as 24 inches.

The length of PJ is calculated by applying Pythagoras theorem as follows;

PJ² = PN² + NJ²

The given parameters;

NJ = 10 in

The length of PN = The length of PL = The length of PM

PL = 3x + 4

PM = 6x - 14

3x + 4 = 6x - 14

3x - 6x = -14 - 4

-3x = -18

x = 18/3

x = 6

Length PN = 3x + 4

PN = 3(6) + 4

PN = 22

Length PJ is calculated as follows;

PJ² = PN² + NJ²

PJ² = 22² + 10²

PJ² = 584

PJ = √584

PJ = 24.17 in

PJ ≈ 24 in

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