Create your own problem using physical quantities on inverse variation a i Design a formula using the concepts of variation I and solve that problem and state conclusion]

Answer:

Assuming that those are fraction the simplest form would be --> 7/4

Hope this helps

Answer: (10.4, 0).

Step-by-step explanation: In order to determine the coordinate where the projectile would land, we need to find the parabolic path followed by the projectile. Since the catapult is launched from the origin (0,0), and has a maximum height at (5.2, 6.3), we can find the vertex of the parabola, which is also (5.2, 6.3).

The standard form of a parabola is y = a(x - h)^2 + k, where (h,k) is the vertex of the parabola. In this case, h = 5.2 and k = 6.3.

We can use the information from the origin to determine the value of 'a'. Substituting x = 0 and y = 0, we get:

0 = a(0 - 5.2)^2 + 6.3

Solving for 'a':

a(5.2)^2 = -6.3

a = -6.3 / (5.2)^2

Now that we have 'a', we can rewrite the equation of the parabola:

y = a(x - 5.2)^2 + 6.3

To find the x-coordinate where the projectile would land, we need to find the other x-intercept (the other point where y = 0). Since the parabola is symmetric, the other x-intercept will be equidistant from the vertex:

x = 5.2 * 2 = 10.4

Now, we can plug in x = 10.4 into the equation to find the y-coordinate:

y = a(10.4 - 5.2)^2 + 6.3

However, since we are looking for the landing coordinate, which is an x-intercept, we know the y-coordinate will be 0.

Thus, the coordinate where the projectile would land is (10.4, 0).

Answer:   4.

Step-by-step explanation: y = kx

36 = k(9)

k = 4

Answer:

Have a Nice Best Day : )

The number of ways of arrangement is 726, 485, 760 ways, 5040 ways, 210 ways resdpectively

Remember that Permutation is a mathematical technique that determines the number of possible arrangements in a set when the order of the arrangements matters.

How many ways can you choose 5 players from 14

= [tex]14^{} Px_{5}[/tex]

14!/(14-5)!

(14*13*12*11*10*9*8*7*6*5*4*3*2*1)/9*8*7*6*5*4*3*2*1

= 726, 485, 760 ways

3. How many ways can you line up 7 books on a shelf?

= 7!

= 5040 ways

4.  10!/(10-6)!6!

10*9*8*7*6*5*4*3*2*1/6*5*4*3*2*1*4*3*2*1

= 5040/24

= 210 ways

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Since the arrival times are independent and uniformly distributed, the probability that a single person arrives in the first hour is 1/10^6. Therefore, the number of people N that arrive in the first hour follows a binomial distribution with parameters n=10^6 and p=1/10^6.

The probability that exactly i people arrive in the first hour is then given by the binomial probability mass function:

P{N=i} = (10^6 choose i) * (1/10^6)^i * (1 - 1/10^6)^(10^6 - i)

Using the normal approximation to the binomial distribution, we can approximate this probability as:

P{N=i} ≈ φ((i+0.5 - np) / sqrt(np(1-p)))

where φ is the standard normal probability density function. Plugging in the values of n=10^6 and p=1/10^6, we get:

P{N=i} ≈ φ((i+0.5 - 1) / sqrt(1*0.999999)) = φ(i - 0.5)

Therefore, an approximation for P{N=i} is given by the standard normal density function evaluated at i-0.5.

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The calculated value of the expression when a = 3 and b = -2 is 1

The expression is given as

(StartFraction 3 a Superscript negative 2 Baseline b Superscript 6 Baseline Over 2 a Superscript negative 1 Baseline b Superscript 5 Baseline EndFraction) squared

Mathematically, this can be expressed as

(3a^-2b^6/2a^-1b^5)^2

The values of a and b are given as

a =3 and b = -2

substitute the known values in the above equation, so, we have the following representation

(3(3)^-2(-2)^6/2(3)^-1(-2)^5)^2

Evaluate

So, we have

1

Hence, the solution is 1

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The quantity that maximizes total revenue is 40. The correct option is d. 40.

To maximize total revenue, we need to find the quantity where marginal revenue (MR) equals zero.

Given the MR equation:

MR = 40 - Q

Set MR to zero and solve for Q:

0 = 40 - Q

Q = 40

So, the quantity that maximizes total revenue is 40. The correct option is d. 40.

Maximizing total revenue requires optimizing prices, bundling products, cross-selling/upselling, improving customer retention, expanding the customer base, and reducing costs. Make sure to regularly check on revenue performance and modify strategies accordingly.

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Suppose a monopolist has the following equations: P=40-0.5Q MR=40-Q What is the quantity that maximizes total revenue? a. 25 10 30 d.40 e. 20 T1e MC=10

There is a 90% probability that a resident has at least one of the two services, and a 62.5% probability that a resident has Internet service given that they already have television service.

To answer your question, we will use the formula for the probability of the union of two events: P(A ∪ B) = P(A) + P(B) - P(A ∩ B), where A represents having Internet service and B represents having television service.

The probability of having at least one of the two services is:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
= 0.60 (Internet) + 0.80 (television) - 0.50 (both)
= 1.40 - 0.50
= 0.90 or 90%

Now, to find the probability of having Internet service given that the resident already has television service, we'll use the conditional probability formula: P(A | B) = P(A ∩ B) / P(B)

P(Internet | Television) = P(Internet ∩ Television) / P(Television)
= 0.50 (both) / 0.80 (television)
= 0.625 or 62.5%

So, there is a 90% probability that a resident has at least one of the two services, and a 62.5% probability that a resident has Internet service given that they already have television service.

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The domain of this relation is the set of all non-negative real numbers that can be expressed as: A weight from 0 to 5 ounces A weight from 5 to 20 ounces that is a multiple of 0.2 ounces A weight greater than 20 ounces is a multiple of 0.15 ounces.

The domain of this relation is the set of all possible weights of packages that can be shipped using the postal service.

Since the postal service charges $2 for packages up to 5 ounces, the domain includes all weights from 0 to 5 ounces. For packages weighing between 5 and 20 ounces, the domain includes all weights from 5 to 20 ounces, with each weight being a multiple of 0.2 ounces.

For packages weighing more than 20 ounces, the domain includes all weights greater than 20 ounces, with each weight being a multiple of 0.15 ounces.

The domain does not include negative numbers or numbers that are not expressible using the above criteria.

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The time taken for the smokestack A is 22.5 hours.

The time taken for the smokestack B is 45 hours.

The time taken for the smokestack is calculated as follows;

Let's the rate at which smokestack B emits pollution = r

Then smokestack A = 2r

Their total rate of pollution combined;

= r + 2r

= 3r

The total amount of pollution they emitted after 15 hours;

= 3r x 15

= 45r pollution

The time taken for each to emit the same amount;

rate of B = r pollution/hr

time of B = 45r/r = 45 hours

rate of A = 2r

time of A = 45r/2r = 22.5 hours

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Answer:

29/16, or 1.81 in decimal form.

Step-by-step explanation:

To solve this, we have to find a common denominator between the 2 fractions.

We cannot simplify 15/16 any further without getting a decimal, so let's change 7/8.

A common denominator between the 2 fractions is 16, so multiply 7/8 by 2 to get 14/16.

Now, we have to find out how far she walked in total, so add 14/16 + 15/16:

29/16

Hope this helps! :)

The 95% confidence interval for the mean amount (in milligrams) of nicotine in the sampled brand of cigarettes is C.39.2 to 40.8

We can use a t-table or a calculator to calculate the t-score. The t-score for a 95% confidence interval with 22 degrees of freedom (n-1) is around 2.074.

When we plug in the values, we get:

CI = 40 ± 2.074 * 1.8/√23 = 40 ± 0.763 = (39.237, 40.763)

As a result, the 95% confidence interval for the mean nicotine content of the studied cigarette brand is (39.237, 40.763) mg.

Because 39.2 to 40.8 is the closest response choice, the answer is 39.2 to 40.8.

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There are 8 successful trials (trials 2, 4, 5, 7, 8, 10, 11, and 12).

The probability that exactly two kittens will be chocolate brown is 1/12.

The probability that at least two kittens will be chocolate brown is 7/12.

Priya can increase the accuracy of the simulation by increasing the number of trials.

We have,

To determine if a trial was a success, we need to count the number of chocolate brown kittens in each trial.

If a trial has at least two chocolate brown kittens, it is considered a success.

Now,

Using the table provided, we can count the number of chocolate brown kittens in each trial:

trial number outcome count of chocolate brown kittens

1 ggggg 0

2 gggwg 1

3 wgwgw 0

4 gwggg 1

5 gggwg 1

6 wwggg 0

7 gwggg 1

8 ggwgw 1

9 wwwgg 0

10 ggggw 2

11 wggwg 1

12 gggwg 1

So,

There are 8 successful trials (trials 2, 4, 5, 7, 8, 10, 11, and 12).

To estimate the probability that exactly two kittens will be chocolate brown, we need to count the number of trials where exactly two chocolate brown kittens were born and divide it by the total number of trials.

From the table, we can see that there is only one trial where exactly two chocolate brown kittens were born (trial 10).

The estimated probability.

=  1/12

= 0.0833.

To estimate the probability that at least two kittens will be chocolate brown, we need to count the number of trials where at least two chocolate brown kittens were born and divide it by the total number of trials.

From the table, we can see that there are 7 successful trials.

The estimated probability.

= 7/12

= 0.5833.

Another question Priya could answer using this simulation is:

Question:

What is the probability that all five kittens will be white?

Answer:

We need to count the number of trials where all five cubes drawn were white (trial 6 and trial 9) and divide it by the total number of trials.

The estimated probability.

= 2/12

= 0.1667.

To increase the accuracy of the simulation, Priya could increase the number of trials conducted.

The more trials conducted, the more accurate the estimated probabilities will be.

Thus,

There are 8 successful trials (trials 2, 4, 5, 7, 8, 10, 11, and 12).

The probability that exactly two kittens will be chocolate brown is 1/12.

The probability that at least two kittens will be chocolate brown is 7/12.

Priya can increase the accuracy of the simulation by increasing the number of trials.

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All Trigonometric Expressions:

a. ∫5x * [tex](5x^{2} - 2)^{(3/2)[/tex]dx = ∫2sin³θ cos²θ dθ

b. ∫[tex]x^{2} dx/(4x^{2} + 6)[/tex]= ∫tan²θ sec²θ dθ

c. ∫x√(5x + 50)/(x + 118)dx = ∫(5tan²θ – 25)tanθ sec³θ dθ

d. ∫(19 – 50x)/(119 – 5x² + 50x)dx = -2∫dθ/(25tan²θ + 94)

a. The integral ∫5x * [tex](5x^{2} - 2)^{(3/2)[/tex]dx, we can use the substitution x = (2/5)sinθ. This gives dx = (2/5)cosθ dθ and 5x² – 2 = 5(2/5 sinθ)² – 2 = 2cos²θ. Substituting these expressions into the integral, we get:

∫5x * [tex](5x^{2} - 2)^{(3/2)[/tex]dx  

= ∫2sin³θ cos²θ dθ

b. For the integral ∫x²dx/(4x² + 6), we can use the substitution x = tanθ. This gives dx = sec²θ dθ and 4x² + 6 = 4tan²θ + 6 = 2sec²θ. Substituting these expressions into the integral, we get:

∫x²dx/(4x² + 6) = ∫tan²θ sec²θ dθ

c. For the integral ∫x√(5x + 50)/(x + 118)dx, we can use the substitution

x + 25 = 5tan²θ.

This gives x = 5tan²θ – 25 and dx = 10tanθ sec²θ dθ, and

5x + 50 = 25sec²θ. Substituting these expressions into the integral, we get:

∫x√(5x + 50)/(x + 118)dx

= ∫(5tan²θ – 25)tanθ sec³θ dθ

d. For the integral:

∫(19 – 50x)/(119 – 5x² + 50x)dx,

we can use the substitution

5x – 5 = √(50x – 5)tanθ.

This gives x = (1/10)[(tanθ)² + 1] and

dx = (1/5)(tanθ sec²θ) dθ, and 119 – 5x² + 50x

= (25tan²θ + 94)².

Substituting these expressions into the integral, we get:

∫(19 – 50x)/(119 – 5x² + 50x)dx

= -2∫dθ/(25tan²θ + 94)

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Correct Question:

For each of the following integrals find an appropriate trigonometric substitution of the form x = f(t) to simplify the integral.

a. ∫5x * ∫5x * [tex](5x^{2} - 2)^{(3/2)[/tex]dx

b. ∫[tex]x^{2} dx/(4x^{2} + 6)[/tex]

c. ∫x√(5x + 50)/(x + 118)dx

d. ∫(19 – 50x)/(119 – 5x² + 50x)dx

The given set of ordered pairs represents a function because each output has exactly one corresponding input. So, the correct answer is A) Yes, because for each output there is exactly one input.

A relation between two variables is a set of ordered pairs, where the first element in each pair corresponds to the input or independent variable (usually denoted by x), and the second element corresponds to the output or dependent variable (usually denoted by y).

In the given set of ordered pairs, each output has exactly one corresponding input, and therefore the relation satisfies the definition of a function. For example, the input of 3/4 is associated with only one output of -2, and the output of -7 is associated with only one input of -2. Hence, the relation represents a function.

So, the correct answer is A).

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The value of the iterated integral is [tex]7/3[/tex] √2 in the given case

To convert to polar coordinates, we need to express the integrand and the limits of integration in terms of polar coordinates. Let's start by finding the limits of integration:

0 ≤ y ≤ √2 - y[tex]^2[/tex]

0 ≤ x ≤ 1

The first inequality can be rewritten as [tex]y^2 + x^2[/tex] ≤ 2, which is the equation of a circle centered at the origin with a radius √of 2. Therefore, the limits of integration in polar coordinates are:

0 ≤ r ≤ √2

0 ≤ θ ≤ π/2

Now, let's express the integrand in polar coordinates:

7xy = 7r cos(θ) sin(θ)

And the differential area element in polar coordinates is:

dA = r dr dθ

Therefore, the integral becomes:

= [tex]7/3[/tex] √2

Therefore, the value of the iterated integral is [tex]7/3[/tex] √2.

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(a) The joint probability mass function of X and Y x1=0 x2=2 x3=4

y1=0 1/8 0 PY(y1)=1/2

y2=1 1/8 1/4 PY(y2)=1/4

y3=2 0 0 PY(y3)=1/4

(b) X and Y are uncorrelated. (c) The random variables X and Y are not independent with each other.

(a) We know that P(X=x2,Y=y1) = 0, since there are no entries in the table where X=x2 and Y=y1. Therefore,

x1=0 x2=2 x3=4

y1=0 1/8 0 PY(y1)=1/2

y2=1 1/8 1/4 PY(y2)=1/4

y3=2 0 0 PY(y3)=1/4

(b) The covariance of X and Y is :

Cov(X,Y) = E[XY] - E[X]E[Y]

where E[XY] is the expected value of the product XY,

E[X] = x1P(X=x1) + x2P(X=x2) + x3P(X=x3) = 0(1/4) + 2(1/2) + 4(1/4) = 2

E[Y] = y1P(Y=y1) + y2P(Y=y2) + y3P(Y=y3) = 0(1/2) + 1(1/4) + 2(1/4) = 1

Now,

E[XY] = x1y1P(X=x1,Y=y1) + x2y1P(X=x2,Y=y1) + x2y2P(X=x2,Y=y2) + x3y2P(X=x3,Y=y2) = 0(1/8) + 2(0) + 2(1/8) + 4(1/4) = 1.5

Therefore,

Cov(X,Y) = E[XY] - E[X]E[Y] = 1.5 - 2(1) = -0.5

Since the covariance is negative, hence X and Y are negatively correlated.

(c) To determine whether X and Y are independent, check whether:

P(X=x,Y=y) = P(X=x)P(Y=y)

for all possible values of x and y.

Using the joint probability mass function we determined in part (a), we can check this condition:

P(X=0,Y=0) = 1/8 ≠ (1/4)(1/2) = P(X=0)P(Y=0)

P(X=2,Y=1) = 1/8 ≠ (1/2)(1/4) = P(X=2)P(Y=1)

P(X=4,Y=2) = 1/4 ≠ (1/4)(1/4) = P(X=4)P(Y=2)

Therefore, X and Y are not independent.

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The equation that can be used to find the value of x is 120 = (x + 5) × (120/x - 1/3).

Carmen's first trip was 120 miles, and she traveled at an average of x miles per hour. We can use the formula:

distance = rate × time, which can be written as:
120 miles = x miles/hour × time

where, time is the time for outgoing.

For the return trip, Carmen traveled at a rate that was 5 miles per hour faster, so her speed was (x + 5) miles/hour. The time for the return trip was one-third of an hour less than the time for the outgoing trip, so we can represent the return trip time as (time - 1/3) hours. Using the distance formula again for the return trip:
120 miles = (x + 5) miles/hour × (time - 1/3) hours

Now, let's express both times in terms of x. From the first equation, we can find the time for the outgoing trip as:
time = 120 miles / x miles/hour

Substitute this expression for time in the return trip equation:
120 miles = (x + 5) miles/hour × (120/x - 1/3) hours

Now you have an equation that can be used to find the value of x:
120 = (x + 5) × (120/x - 1/3)

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The amount of money in the account after 16 years is approximately $17,526.64.

Using the formula for continuous compounding, we have:

V = Pe[tex]^(rt)[/tex]

where V is the value of the account after t years, P is the principal initially invested, e is the base of the natural logarithm, r is the annual interest rate, and t is the time in years.

Substituting the given values, we get:

V = 8430e[tex]^(0.038*16)[/tex]

Simplifying this expression, we have:

V = 8430[tex]e^0.608[/tex]

Using a calculator, we get:

V ≈ $17,526.64

Therefore, the amount of money in the account after 16 years is approximately $17,526.64.

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The number of miles is 7 while the number of minutes for this trip is 21.

Let x represent the number of minutes and y represent the number of miles. According to the given information, we have two equations:

1) 0.20x + 1.40y = $14
2) x = 3y

First, we will solve equation (2) for x:

x = 3y

Next, substitute this value of x into equation (1):

0.20(3y) + 1.40y = $14

Now, simplify and solve for y:

0.60y + 1.40y = $14
2.00y = $14
y = 7

Now that we have the value for y (number of miles), we can find the value for x (number of minutes) using equation (2):

x = 3y
x = 3(7)
x = 21

So, the number of minutes for this trip is 21, and the number of miles is 7.

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Santiago used 185 cell phone minutes over the course of a month.

To determine the total number of cell phone minutes that Santiago used over the course of a month, you would use the operation of addition.

You would add up the number of minutes that Santiago spoke with his mom, dad, and friends:

Total cell phone minutes = 45 minutes + 30 minutes + 110 minutes

Total cell phone minutes = 185 minutes

Therefore, Santiago used 185 cell phone minutes over the course of a month.

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The sample space for choosing one card is, S={I, L, L, M, P, T, U, Y}

Since, We know that;

A sample space is a set of potential results from a random experiment. The letter "S" is used to denote the sample space. Events are the subset of possible experiment results. Depending on the experiment, a sample area could contain a variety of results.

Given that,

A set of eight cards were labeled with M, U, L, T, I, P, L, Y.

Here, the sample space is;

{ S, U, B, T, R, A, C, T}

Now, Elements in order is,

⇒ S = {I, L, L, M, P, T, U, Y}

Therefore, the sample space for the given cards is,

S = {I, L, L, M, P, T, U, Y}

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The number of people that are expected to go to the youth wing on Sunday would be = 543.

For Saturday;

The number of people that went to the youth wing = 497

The number of people that went to social issues = 369

The number of people that went to Fiction and Literature = 416

The total number of people that went to the central library = 497+369+416 = 1,282

For Sunday, the number of people that would visit the youth wing ;

= 497/1282× 1400/1

= 695800/1282

= 543 (to the nearest whole number)

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A graph that represents the inequality y > x² - 3 include the following: A. graph A.

In Mathematics and Geometry, the general form of a quadratic function can be modeled and represented by using the following quadratic equation;

y = ax² + bx + c

Where:

Since the leading coefficient (value of a) in the given quadratic function  y > x² - 3 is positive 1, we can logically deduce that the parabola would open upward. Also, the value of the quadratic function f(x) would be minimum at -3.

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The sample mean of the population is 3/4 and the variance is 3/80. Using the central limit theorem, P( > 0.8) can be simplified as 0.003.

The mean of the population can be computed as follows:

µ = ∫x f(x) dx from 0 to 1

  = ∫x (3x²) dx from 0 to 1

  = 3/4

The variance of the population can be computed as follows:

σ² = ∫(x-µ)² f(x) dx from 0 to 1

    = ∫(x-(3/4))² (3x²) dx from 0 to 1

    = 3/80

By the Central Limit Theorem, as the sample size n = 80 is large, the distribution of the sample mean  can be approximated by a normal distribution with mean µ and variance σ²/n.

Therefore, P( > 0.8) can be approximated by P(Z >0.8- 0.75)/(sqrt(3/80)/(sqrt(80))), where Z is a standard normal random variable.

Simplifying, we get P( > 0.8) ≈ P(Z > 2.73) ≈ 0.003.

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The probability that she will get a white button and a blue button is 18/30 * 3/29 = 9/145 or approximately 0.062.

The total number of buttons is 18 + 9 + 3 = 30. The probability of getting a white button on the first draw is 18/30. After drawing a white button, there are 29 buttons left, including 3 blue buttons, so the probability of getting a blue button on the second draw is 3/29.

To find the probability of both events happening, we multiply the probabilities:

18/30 * 3/29 = 9/145

Therefore, the probability that she will get a white button and a blue button is 9/145 or approximately 0.062.

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a. The test is a left-tailed test. b. What is the test statistic is -1.73.

a. The test is a left-tailed test because we are testing the claim that less than 10% of treated subjects experienced headaches.

b. To find the test statistic, we'll use the normal distribution as an approximation to the binomial distribution. Here are the steps:

Step 1: Determine the null and alternative hypotheses.
H0: p = 0.10 (The proportion of treated subjects experiencing headaches is equal to 10%.)
H1: p < 0.10 (The proportion of treated subjects experiencing headaches is less than 10%.)

Step 2: Calculate the sample proportion (p-hat).
p-hat = number of subjects with headaches / total subjects = 19 / 276 ≈ 0.0688

Step 3: Determine the standard error.
SE = sqrt((p * (1 - p)) / n) = sqrt((0.10 * (1 - 0.10)) / 276) ≈ 0.0180

Step 4: Calculate the test statistic (z-score).
z = (p-hat - p) / SE = (0.0688 - 0.10) / 0.0180 ≈ -1.73

So, the test statistic is -1.73 (rounded to two decimal places).

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The answer to the first question is Full Factorial Design, and the answer to the second question is Fractional Factorial Designs.

Full Factorial Design involves testing all possible combinations of the factors in an experiment at a number of levels.

Fractional Factorial Designs are used for screening experiments to identify critical factors. These designs are a subset of the full factorial design, and they only test a fraction of the possible combinations of the factors in an experiment. This allows for a more efficient use of resources when conducting experiments.

Therefore, the answer to the first question is Full Factorial Design, and the answer to the second question is Fractional Factorial Designs.

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