Do couples get engaged or not? If they are engaged, how long did they date before becoming engaged? A poll of 1000 couples conducted by Bruskin and Goldring Research for Korbel Champagne Cellars gave Time Never Engaged Number of Couples 200 the following information: Time Number of Couples Never Engaged 200 Less than 1 year 240 1 to 2 years 210 More than 2 years 350 What is the sample space in this problem?

Answers

Answer 1

The sample space in this problem is the total number of couples surveyed, which is 1000.

The sample space in probability refers to the set of all possible outcomes of an experiment. In this case, the experiment is the survey conducted by Bruskin and Goldring Research, and the possible outcomes are the different categories of time taken by couples before getting engaged.

The given information provides the number of couples in each category, which can be added to find the total number of couples surveyed:

Sample space = Number of couples never engaged + Number of couples engaged for less than 1 year + Number of couples engaged for 1 to 2 years + Number of couples engaged for more than 2 years

Sample space = 200 + 240 + 210 + 350

Sample space = 1000

Therefore, the sample space in this problem is 1000, which represents the total number of couples surveyed by the research firm.

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Related Questions

Use cylindrical coordinates to evaluate the triple integral ∫∫∫√(x^2 + y^2) dV where E is the solid bounded by the
circular paraboloid z = 9 - (x^2 + y^2) and the xy-plane.

Answers

The value of the triple integral ∫∫∫[tex]E \sqrt{(x^2 + y^2)} dV[/tex] over the solid bounded by the circular paraboloid [tex]z = 9 - (x^2 + y^2)[/tex] and the xy-plane is 486π/5.

To evaluate the triple integral ∫∫∫[tex]E \sqrt{(x^2 + y^2)} dV[/tex], where E is the solid bounded by the circular paraboloid [tex]z = 9 - (x^2 + y^2)[/tex] and the xy-plane, we can use cylindrical coordinates. In cylindrical coordinates, the equation of the paraboloid becomes:

[tex]z = 9 - (r^2)[/tex]

The limits of integration are:

0 ≤ r ≤ 3 (since the paraboloid intersects the xy-plane at z = 0 when r = 3)

0 ≤ θ ≤ 2π

0 ≤ z ≤ 9 - (r^2)

The triple integral becomes:

∫∫∫[tex]E √(x^2 + y^2) dV = ∫0^3 ∫0^2π ∫0^(9-r^2) r√(r^2) dz dθ dr[/tex]

Simplifying, we get:

∫∫∫[tex]E √(x^2 + y^2) dV = ∫0^3 ∫0^2π ∫0^(9-r^2) r^2 dz dθ dr[/tex]

Evaluating the innermost integral, we get:

∫[tex]0^(9-r^2) r^2 dz = (9-r^2)r^2[/tex]

Substituting this back into the triple integral, we get:

∫∫∫[tex]E √(x^2 + y^2) dV = ∫0^3 ∫0^2π (9-r^2)r^2 dθ dr[/tex]

Evaluating the remaining integrals, we get:

∫∫∫[tex]E √(x^2 + y^2) dV = ∫0^3 (9r^2 - r^4) dθ[/tex]

= 2π [243/5]

= 486π/5

Therefore, the value of the triple integral ∫∫∫[tex]E \sqrt{(x^2 + y^2)} dV[/tex] dV over the solid bounded by the circular paraboloid [tex]z = 9 - (x^2 + y^2)[/tex] and the xy-plane is 486π/5.

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"Consider the following function: f(x,y)=y^5 ln(−2x^4+3y^5) find fx and fy"

Answers

From the function f(x,y)=y⁵ ln(−2x⁴+3y⁵).  The value of  fx = -10x³y⁵ / (-2x⁴ + 3y⁵) and

fy = y⁴ * ln(-2x⁴ + 3y⁵) * d/dy [(-2x⁴ + 3y⁵)]

To find fx, we differentiate f(x,y) with respect to x, treating y as a constant:

fx = d/dx [y⁵ ln(-2x⁴ + 3y⁵)]

Using the chain rule and the derivative of ln u = 1/u, we have:

fx = y⁵ * 1/(-2x⁴ + 3y⁵) * d/dx [-2x⁴ + 3y⁵]

Simplifying and applying the power rule of differentiation, we get:

fx = -10x³y⁵ / (-2x⁴ + 3y⁵)

Similarly, to find fy, we differentiate f(x,y) with respect to y, treating x as a constant:

fy = d/dy [y⁵ ln(-2x⁴ + 3y⁵)]

Using the chain rule and the derivative of ln u = 1/u, we have:

fy = y⁴ * ln(-2x⁴ + 3y⁵) * d/dy [(-2x⁴ + 3y⁵)]

Applying the power rule of differentiation and simplifying, we get:

fy = 15y⁴ ln(-2x⁴ + 3y⁵)

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I need someone to do this for me rq

Answers

Answer:

Step-by-step explanation:

The triangle area= 1/2 * the perpendicular height * breath

                           = 1/2*2*(3/4)

                          =0.75

The diameter of a wheel is 3 feet witch of the following is closest to the area of the whee

Answers

The area of the wheel is approximately 7.07 square feet.

The area of a circle is calculated using the formula A = πr^2, where A is the area and r is the radius of the circle. In this case, the diameter of the wheel is given as 3 feet, so the radius is half of that, which is 1.5 feet.

Substituting the value of the radius into the formula, we get A = π(1.5)^2. Simplifying this expression gives us approximately 7.07 square feet. Therefore, the closest answer to the area of the wheel is 7.07 square feet.

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Choose the correct symbol to compare the expressions. Do not multiply. 7×

2

10

?7

Answers

The correct symbol to compare the expressions is < (less than).

7 × (2/10) is equivalent to 1.4, which is less than 7. Therefore, 7 is greater than 1.4, and we can write 7 × (2/10) < 7 as the comparison between the expressions.

To compare the two expressions, we can analyze their values without actually multiplying them. The expressions are:

1. 7 × (2/10)
2. 7

Now let's simplify the first expression without multiplying:

7 × (2/10) = 7 × (1/5) (since 2 and 10 have a common factor of 2)

Now let's compare:

7 × (1/5) ? 7

Since we're multiplying 7 by a fraction that is less than 1 (1/5), the result will be smaller than 7. Therefore, the correct comparison symbol is "<":

7 × (1/5) < 7

The correct expression so formed is 7 × (2/10) < 7.

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In 1990, Jane became a real estate agent. Eight years later, she sold a house for $144,000. Eleven years later, she sold the same house for $245,000. Write an equation that represents the value of the house (V) related to the number of years (t) since Jane became a real estate agent. (Hint: Be careful! The second sale of the house was 11 years after the first sale which was 8 years after she became a real estate agent! That means the second sale took place 19 years after she became an agent!)

Answers

Let's break down the information given in the problem:

- Jane became a real estate agent in 1990.

- She sold a house 8 years later (in 1998) for $144,000.

- She sold the same house 11 years after that sale (in 2009), which is 19 years after she became an agent, for $245,000.

To write an equation that represents the value of the house (V) related to the number of years (t) since Jane became a real estate agent, we can use the information from the two sales to find the rate of change in the value of the house over time. We can use this rate of change to write an equation in point-slope form:

V - V1 = m(t - t1)

where V1 is the value of the house at time t1, m is the rate of change in the value of the house, and t is the time since Jane became a real estate agent.

Using the two sales, we can find the rate of change in the value of the house as follows:

m = (V2 - V1) / (t2 - t1)

where V2 is the value of the house at the second sale, t2 is the time of the second sale (19 years after Jane became an agent), V1 is the value of the house at the first sale, and t1 is the time of the first sale (8 years after Jane became an agent).

Substituting the given values, we get:

m = ($245,000 - $144,000) / (19 - 8) = $10,100 per year

Now we can use the point-slope form equation to find the value of the house at any time t since Jane became a real estate agent. Let's choose 1990 as our initial time (t1), so V1 = $0:

V - 0 = $10,100 (t - 0)

Simplifying, we get:

V = $10,100t

Therefore, the equation that represents the value of the house (V) related to the number of years (t) since Jane became a real estate agent is V = $10,100t. Note that this equation assumes a constant rate of change in the value of the house over time, which may not be accurate in real life.

The local regional transit authority of a large city was interested in determining the mean commuting time for workers who drove to work. They selected a random sample of 125 residents of the metropolitan region and asked them how long they spent commuting to work (in minutes). A 95% confidence interval was constructed and reported as (27. 74, 30. 06). Interpret the interval in the context of this problem. 2. A long distance telephone company recently conducted research into the length of calls (in minutes) made by customers. In a random sample of 45 calls, the sample mean was minutes and the standard deviation was s 5. 2 minutes. (a) Find a 95% confidence interval for the true mean length of long distance telephone calls made by customers of this company. X 1. 68

Answers

For the first problem, we can interpret the confidence interval as follows:

We are 95% confident that the true mean commuting time for workers who drive to work is between 27.74 and 30.06 minutes.

This means that if we were to repeat the sampling process many times and construct a 95% confidence interval each time, about 95% of those intervals would contain the true mean commuting time.

For the second problem, we can use the following formula to find a 95% confidence interval for the true mean length of long distance telephone calls:

[tex]CI = X ± t*(s/sqrt(n))[/tex]

Where X is the sample mean, s is the sample standard deviation, n is the sample size, and t is the t-value from the t-distribution with n-1 degrees of freedom for a 95% confidence interval.

Plugging in the values given, we get:

[tex]CI = 1.68 ± t*(5.2/sqrt(45))[/tex]

To find the value of t, we can look it up in a t-distribution table or use a calculator. For a 95% confidence interval with 44 degrees of freedom, we get t = 2.015.

Plugging this value in, we get:

[tex]CI = 1.68 ± 2.015*(5.2/sqrt(45)) = (0.86, 2.50)[/tex]

So we can interpret the interval as follows:

We are 95% confident that the true mean length of long distance telephone calls made by customers of this company is between 0.86 and 2.50 minutes longer or shorter than the sample mean of 1.68 minutes.

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helppppppppppppplppppppooo

Answers

Answer:

B, A, C

Step-by-step explanation:

The rate is the another name for the slope.

A:

Change in y over the change in x.  You find the change by subtracting

[tex]\frac{7-3}{5-3}[/tex] = [tex]\frac{4}{2}[/tex] = 2

The rate is 2.

B:

Change in y over the change in x. You find the change by subtracting.

[tex]\frac{0-3}{-5-0}[/tex] = [tex]\frac{-3}{-5}[/tex] = [tex]\frac{3}{5}[/tex]

The rate is [tex]\frac{3}{5}[/tex].

C:

The rate is the number before the x in the equation.

The rate is 3.

Helping in the name of Jesus.

A right triangle is shown. The length of the hypotenuse is 4 centimeters and the lengths of the other 2 sides are congruent.
The hypotenuse of a 45°-45°-90° triangle measures 4 cm. What is the length of one leg of the triangle?

2 cm
2 StartRoot 2 EndRoot cm
4 cm
4 StartRoot 2 EndRoot cm

Answers

Answer:

The length of one leg of this right triangle is 4/√2 = 2√2 cm.

Indetify the mononial, binomail or trinomial

4x2 - y + oz4

Answers

The given expression is a trinomial because it consists of three terms: 4x²-y+oz⁴

A trinomial is a polynomial with three terms. It is a type of algebraic expression that consists of three monomials connected by addition or subtraction. The general form of a trinomial is:

ax^2 + bx + c

A monomial is an algebraic expression that consists of a single term. It is a polynomial with only one term. A term is a combination of a coefficient and one or more variables raised to non-negative integer exponents.  The general form of a monomial is:c * xᵃ,  yᵇ, zⁿ....

where 'c' represents the coefficient (a constant), and 'x', 'y', 'z', etc., represent variables, each raised to a non-negative exponent (a, b, n, etc.).

example of monomials:  5x² - This monomial has a coefficient of 5 and a single variable 'x' raised to the power of 2.

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if the area of a circle is 153.86m find diamiter and perimeter

Answers

Answer:

the diameter is 14 and the perimeter is 43.97

can someone did this step by step correctly and not give the wrong answer
A cylinder has the net shown.

net of a cylinder with diameter of each circle labeled 3.8 inches and a rectangle with a height labeled 3 inches

What is the surface area of the cylinder in terms of π?

40.28π in2

22.80π in2

18.62π in2

15.01π in2

Answers

40.28 in2 hope it helps

Pls help me find the exponent!

Answers

Answer:

1.6×10^-12..............

17 Question Evaluate y=−4(5)x y = − 4 ( 5 ) x for x

Answers

The equation y=−4(5)x y = − 4 ( 5 ) x can be simplified to y = -20x. This equation is a linear function with a slope of -20. As x increases, y decreases at a rate of 20 units for every one unit increase in x. Therefore, the value of y will decrease rapidly as x increases.

A large apartment complex has 1,500 units, which are filling up at a rate of 10% per month. If the


apartment complex starts with 15 occupied units, how many months will pass before the complex


has 800 occupied units? (Assume logistic growth). Round to the nearest tenth.


0. 58 months


15. 7 months


1. 3 months


47. 3 months

Answers

After about 15.7 months, the apartment complex will have 800 occupied units.

Logistic growth is a type of growth in which the growth rate of a population decreases as the population size approaches its maximum value. In this case, the apartment complex has a maximum capacity of 1500 units.

Starting with 15 occupied units and growing at a rate of 10% per month, the number of occupied units can be modeled by a logistic function.

To find the number of months it takes to reach 800 occupied units, we need to solve for the time when the logistic function equals 800.

Let P(t) be the number of occupied units at time t (in months), then we have:

P(t) = 1500 / (1 + 1485[tex]e^{(-0.1t)}[/tex])

We want to find t such that P(t) = 800. Solving for t, we get:

t = -10 ln(1 - 4/37) ≈ 15.7 months

This means that after about 15.7 months, the apartment complex will have 800 occupied units.

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A ball is dropped from a window at a height of 36 feet. the function h(x) = -16x2 + 36 represents the height (in feet) of the ball after x seconds. round
to the nearest tenth.
how long does it take for the ball to hit the ground?

Answers

It takes about 1.5 seconds for the ball to hit the ground.

How to calculate the time for ball to hit the ground?

To find how long it takes for the ball to hit the ground, we need to find the value of x when h(x) = 0, since the height of the ball is 0 when it hits the ground. We can set -16x²+36 = 0 and solve for x:

-16x²+ 36 = 0

Dividing both sides by -16:

x² - 2.25 = 0

Adding 2.25 to both sides:

x²= 2.25

Taking the square root of both sides (we can ignore the negative root since time cannot be negative):

x = √(2.25) ≈ 1.5

Therefore, it takes about 1.5 seconds for the ball to hit the ground.

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Use the normal approximation to find the indicated probability. the sample size is n, the population proportion of successes is p, and x is the number of successes in the sample.
n = 81, p = 0.5: p(x ≥ 46)
group of answer choices

0.1210

0.1335

0.8790

0.1446

Answers

We know that the indicated probability is approximately 0.1210.

To use the normal approximation, we need to check if the conditions for a normal approximation are met. In this case, we have:

np = 81 * 0.5 = 40.5 ≥ 10
n(1-p) = 81 * 0.5 = 40.5 ≥ 10

Since both conditions are met, we can use the normal approximation to find the probability.

First, we need to find the mean and standard deviation of the sampling distribution of sample proportions:

mean = np = 81 * 0.5 = 40.5
standard deviation = sqrt(np(1-p)) = sqrt(81 * 0.5 * 0.5) = 4.5

Next, we need to standardize the value of x:

z = (x - mean) / standard deviation
z = (46 - 40.5) / 4.5 = 1.22

Finally, we can use a standard normal table or calculator to find the probability:

P(z ≥ 1.22) = 0.1118

Therefore, the answer is approximately 0.1210.

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Make a table of values in a graph for fabian's income inexpenses the expenses e to make n cakes per month is given by the equation E = 825 + 3.25n the income I for selling n Cakes given by the equation I equals 8.20 n also graphic and make a table

Answers

The table of values in a graph for fabian's income in expenses is

n            E(n)

0           825

1            828.25

2           831.5

4          838

Making a table of values in a graph for fabian's income inexpenses

From the question, we have the following parameters that can be used in our computation:

The expenses E to make n cakes per month is given by the equation

E = 825 + 3.25n

Next, we assume values for n and calculate E

Using the above as a guide, we have the following:

E = 825 + 3.25(0) = 825

E = 825 + 3.25(1) = 828.25

E = 825 + 3.25(2) = 831.5

E = 825 + 3.25(4) = 838

So, we have

n            E(n)

0           825

1            828.25

2           831.5

4          838

This represents the table of values

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b) In a certain group of 200 persons, 110 can speak Nepali, 85 can speak Maithili and 60 can speak both the languages. Find, (i) how many of them can talk in either of these languages? (ii) how many of them can talk in neither of these languages?​

Answers

Answer:

(i) 135, (ii) 65

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

Given:

Total number in the group - 200 persons,Nepali speakers - 110,Maithili speakers - 85,Both - 60.

(i) We know 60 out of 110 can speak both languages, so as 60 out of 85. The number 60 is counted twice if we add them together.

Find the number of those speak either language:

Either = sum of each - bothEither = 110 + 85 - 60 = 135

(ii) Find the number of thise who can talk neither of these languages:

Neither = total - eitherNeither = 200 - 135 = 65

Solve for X pleaseee!!!

Answers

The answer
X=10
If you want Another way

6x-7=10x-47
We move all terms to the left:
6x-7-(10x-47 =0
We get rid of parentheses
6x-10X+47-7=0
We add all the numbers together, and all t
-4X+40=0
We move all terms containing x to the left
-4X=-40
x=-40/-4
X=+10

A gardener has a rectangular vegetable garden that is 2 feet longer than it is wide. The area of the garden is at


least 120 square feet.


Enter an inequality that represents all possible widths, w, in feet of the garden

Answers

This is the inequality that represents all possible widths, w, in feet of the garden is W^2 + 2W - 120 ≥ 0

The area of a rectangle is given by the formula A = L x W, where A is the area, L is the length, and W is the width. In this problem, we are given that the garden is rectangular and that the length is 2 feet longer than the width, so we can write L = W + 2.

We are also told that the area of the garden is at least 120 square feet, so we can write:
A = L x W ≥ 120
Substituting L = W + 2, we get:
(W + 2) x W ≥ 120

expanding the left side, we get:
W^2 + 2W ≥ 120
Rearranging, we get:
W^2 + 2W - 120 ≥ 0

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In which type of statistical study is the population influenced by researchers?

Answers

The type of statistical study in which the population is influenced by researchers is known as an experimental study.

In an experimental study, researchers manipulate one or more variables to observe the effect on another variable. The population in an experimental study is usually a sample that is randomly selected to represent the larger population.

The researchers intentionally intervene in the study, which can impact the behavior or responses of the participants. This can be seen as a form of bias since the researchers are influencing the population. However, in some cases, this is necessary to determine causality or to test a hypothesis.

To minimize bias, experimental studies often use control groups. The control group is used to provide a baseline for comparison with the group that is exposed to the manipulated variable. This helps to determine if any observed effects are due to the intervention or if they are due to other factors.

In summary, an experimental study is the type of statistical study in which the population is influenced by researchers. While this can introduce bias, the use of control groups and other measures can help to minimize the impact of this bias on the results.

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Tom Jones, a mechanic at Golden Muffler Shop, is able to install new mufflers at an average rate of 3 per hour (exponential distribution). Customers seeking this service, arrive at the rate of 2 per hour (Poisson distribution). They are served first-in, first-out basis and come from a large (infinite population). Tom only has one service bay.



a. Find the probability that there are no cars in the system.


b. Find the average number of cars in the system.


c. Find the average time spent in the system.


d. Find the probability that there are exactly two cars in the system

Answers

a. To find the probability that there are no cars in the system, we need to use the formula for the steady-state probability distribution of the M/M/1 queue:
P(0) = (1 - λ/μ)
where λ is the arrival rate (2 per hour) and μ is the service rate (3 per hour).
P(0) = (1 - 2/3) = 1/3 or 0.3333
Therefore, the probability that there are no cars in the system is 0.3333.

b. To find the average number of cars in the system, we can use Little's Law:
L = λW
where L is the average number of cars in the system, λ is the arrival rate (2 per hour), and W is the average time spent in the system.
We can solve for W by using the formula:
W = 1/(μ - λ)
W = 1/(3 - 2) = 1 hour
Therefore, the average number of cars in the system is:
L = λW = 2 x 1 = 2 cars

c. To find the average time spent in the system, we already calculated W in part b:
W = 1 hour

d. To find the probability that there are exactly two cars in the system, we need to use the formula for the steady-state probability distribution:
P(n) = P(0) * (λ/μ)^n / n!
where n is the number of cars in the system.
P(2) = P(0) * (λ/μ)^2 / 2!
P(2) = 0.3333 * (2/3)^2 / 2
P(2) = 0.1111 or 11.11%
Therefore, the probability that there are exactly two cars in the system is 11.11%.

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Solve the optimization problem. Maximize P= xy with x + 2y = 26.
P=

Answers

The optimization problem has a maximum value of P when x = 13 and y = 6.5. The maximum value of P = 13 * 6.5 = 84.5.

To solve the optimization problem and maximize P = xy with the constraint x + 2y = 26, follow these steps:

1. Express one variable in terms of the other using the constraint: x = 26 - 2y

2. Substitute the expression for x into the objective function P: P = (26 - 2y)y

3. Differentiate P with respect to y to find the critical points: dP/dy = 26 - 4y

4. Set the derivative equal to zero and solve for y: 26 - 4y = 0 => y = 6.5

5. Plug the value of y back into the expression for x: x = 26 - 2(6.5) => x = 13

6. Check the second derivative to confirm it's a maximum: d²P/dy² = -4 (since it's a constant negative, this confirms it's a maximum)

Thus, the optimization problem has a maximum value of P when x = 13 and y = 6.5. The maximum value of P = 13 * 6.5 = 84.5.

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When angela and walker first started working for the​ supermarket, their weekly salaries totaled ​$550. now during the last 25 years walker has seen his weekly salary triple angela has seen her weekly salary become four times larger. together their weekly salaries now total ​$2000. write an algebraic equation for the problem. how much did they each make 25 years​ ago?

Answers

Angela made $350 per week 25 years ago and Walker made $200 per week 25 years ago.

Let's assign variables to represent Angela and Walker's salaries 25 years ago. Let A be Angela's salary 25 years ago and W be Walker's salary 25 years ago.

Using the information given in the problem, we can set up two equations:

A + W = 550 (their total salary 25 years ago)
4A + 3W = 2000 (their total current salary)

To solve for A and W, we can use substitution or elimination. Let's use substitution.

From the first equation, we can rearrange to solve for A:

A = 550 - W

Substitute this into the second equation:

4(550 - W) + 3W = 2000

Distribute the 4:

2200 - 4W + 3W = 2000

Simplify:

W = 800

Now that we know Walker's salary 25 years ago was $800, we can plug that into the first equation to solve for Angela's salary:

A + 800 = 550

A = -250

Uh oh, a negative salary doesn't make sense in this context. We made a mistake somewhere.

Let's go back to our original equations and try elimination instead:

A + W = 550
4A + 3W = 2000

Multiplying the first equation by 4, we get:

4A + 4W = 2200

Subtracting the second equation from this, we get:

W = 200

Now we can plug this into either equation to solve for A:

A + 200 = 550

A = 350

So Angela made $350 per week 25 years ago and Walker made $200 per week 25 years ago.

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Find the inverse for each relation: 4 points each


1. {(1,‐2), (2, 3),(3, ‐3),(4, 2)}


2. {(4,2),(5,1),(6,0),(7,‐1)}


Find an equation for the inverse for each of the following relations.


3. Y=-8x+3


4. Y=2/3x-5


5. Y=1/2x+10


6. Y=(x-3)^2


Verify that f and g are inverse functions.


7. F(x)=5x+2;g(x)=(x-2)/5


8. F(x)=1/2x-7;g(x)=2x+14

Answers

The inverse for each relation:

1. {(1,‐2), (2, 3),(3, ‐3),(4, 2)} - {(-2, 1), (3, 2), (-3, 3), (2, 4)}

2. {(4,2),(5,1),(6,0),(7,‐1)} - {(2, 4), (1, 5), (0, 6), (-1, 7)}

3. Inverse equation: y=(-1/8)x+3/8

4. Inverse equation: y=3/2x+15/2

5. Inverse equation: y=2x-20

6. Inverse equation: y=[tex]x^{(1/2)}+3[/tex]

7. Since fog(x) = gof(x) = x, f and g are inverse functions.

8. Since fog(x) = gof(x) = x, f and g are inverse functions.

1. To find the inverse of the relation, we need to swap the positions of x and y for each point and then solve for y.

{(1, -2), (2, 3), (3, -3), (4, 2)}

Inverse: {(-2, 1), (3, 2), (-3, 3), (2, 4)}

2. Again, we swap x and y and solve for y.

{(4, 2), (5, 1), (6, 0), (7, -1)}

Inverse: {(2, 4), (1, 5), (0, 6), (-1, 7)}

3. To find the inverse equation for y=-8x+3, we swap x and y and solve for y.

x=-8y+3

x-3=-8y

y=(x-3)/-8

Inverse equation: y=(-1/8)x+3/8

4. To find the inverse equation for y=2/3x-5, we swap x and y and solve for y.

x=2/3y-5

x+5=2/3y

y=3/2(x+5)

Inverse equation: y=3/2x+15/2

5. To find the inverse equation for y=1/2x+10, we swap x and y and solve for y.

x=1/2y+10

x-10=1/2y

y=2(x-10)

Inverse equation: y=2x-20

6. To find the inverse equation for y=(x-3)², we swap x and y and solve for y.

x=(y-3)²

[tex]x^{(1/2)}=y-3[/tex]

[tex]y=x^{(1/2)}+3[/tex]

Inverse equation: [tex]y=x^{(1/2)}+3[/tex]

7. To verify that f(x)=5x+2 and g(x)=(x-2)/5 are inverse functions, we need to show that fog(x)=gof(x)=x for all x in the domain of f and g.

fog(x) = f(g(x)) = f((x-2)/5) = 5((x-2)/5) + 2 = x

gof(x) = g(f(x)) = g(5x+2) = ((5x+2)-2)/5 = x/5

Since fog(x) = gof(x) = x, f and g are inverse functions.

8. To verify that f(x)=1/2x-7 and g(x)=2x+14 are inverse functions, we need to show that fog(x)=gof(x)=x for all x in the domain of f and g.

fog(x) = f(g(x)) = f(2x+14) = 1/2(2x+14) - 7 = x

gof(x) = g(f(x)) = g(1/2x-7) = 2(1/2x-7) + 14 = x

Since fog(x) = gof(x) = x, f and g are inverse functions.

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Circle A is located at (6, 5) and has a radius of 4 units. What is the equation of a line that is tangent to circle A from point C (2, 8)? x = 2 y = −0. 75x + 9. 5 y = 1. 33x + 1. 66 x = 8

Answers

We can use the point-slope form of the equation of a line to find the equation of the tangent line.

How to find the equation of the line that is tangent to circle A from point C (2, 8)?

To find the equation of the line that is tangent to circle A from point C (2, 8), we need to first find the point of tangency, which is the point where the line intersects the circle.

Point C (2, 8) is outside the circle, so the tangent line will be perpendicular to the line connecting the center of the circle to point C and will pass through point C.

Step 1: Find the center of the circle

The center of the circle A is at (6, 5).

Step 2: Find the slope of the line connecting the center of the circle to point C

The slope of the line connecting the center of the circle (6, 5) and point C (2, 8) is:

m = (8 - 5) / (2 - 6) = -3/4

Step 3: Find the equation of the line perpendicular to the line from Step 2 passing through point C

The slope of the line perpendicular to the line from step 2 is the negative reciprocal of the slope:

m_perp = -1 / (-3/4) = 4/3

Now we can use the point-slope form of the equation of a line to find the equation of the tangent line:

y - 8 = (4/3)(x - 2)

Simplifying, we get:

y = (4/3)x + 4.67

So the equation of the line that is tangent to circle A from point C (2, 8) is y = (4/3)x + 4.67.

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To celebrate halloween, lacey's class is making candy necklaces. lacey is helping pass out string from a 50-yard-spool. she gives 30 inches of string to each student. if there are 24 students in her class, how many yards of string will be leftover?

Answers

The class will use 20 yards of the 50-yard spool, leaving 30 yards of string leftover.

This leftover string could be used for future projects or saved for another occasion.

Lacey's class will use a total of 720 inches (30 inches per student x 24 students) of string for the candy necklaces.

To convert this to yards, we divide by 36 (since there are 36 inches in a yard). 720 inches ÷ 36 = 20 yards

It's important to note that when working with different units of measurement, it's necessary to convert them to the same unit before performing calculations.

In this case, we converted inches to yards in order to determine the amount of string used by the class. By doing so, we were able to determine how much string was leftover in yards, which is a more appropriate unit of measurement for a spool of string.

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In the shown figure, DE←→
is parallel to side BC¯¯¯¯¯¯¯¯
in triangle ABC
. If m∠B=52
°, what is m∠DAB
?


m∠DAB
=
°

Answers

Answer:

In triangle ABC, m∠BAC = 50°. If m∠ACB = 30°, then the triangle is triangle. If m∠ABC = 40°, then the triangle is triangle. If triangle ABC is isosceles, and AB = 6 and BC = 4, then AC =

Answer:

52 degrees

Step-by-step explanation: because i looked and they looked the same so i put 52 and it was right

You buy a sheet with 10 stamps. Some are 45 cents and some are 30 cents. If it cost $4. 20 how many of each did you get

Answers

You bought 8 stamps that cost 45 cents each and 2 stamps that cost 30 cents each.

Let's assume you bought x stamps that cost 45 cents each and y stamps that cost 30 cents each.

From the given information, we can create two equations:

The total number of stamps is 10: x + y = 10.

The total cost is $4.20: 45x + 30y = 420 (since the cost is given in cents).

Now we can solve this system of equations to find the values of x and y.

We can multiply the first equation by 30 to eliminate y:

30x + 30y = 300.

Now we have a system of equations:

30x + 30y = 300,

45x + 30y = 420.

Subtracting the first equation from the second equation, we get:

45x + 30y - (30x + 30y) = 420 - 300,

15x = 120,

x = 120/15,

x = 8.

Substituting the value of x back into the first equation:

8 + y = 10,

y = 10 - 8,

y = 2.

Therefore, you bought 8 stamps that cost 45 cents each and 2 stamps that cost 30 cents each.

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