Order the functions from the narrowest graph to the widest graph. f (x) 1/3 x^2, g (x) = 4x^2, h (x) = 2x^2

The probability of selecting two non-defective DVD players from a group of six is 2/5. This is based on the assumption that the selection is done without replacement.

We can use the formula for calculating probabilities of combinations:

P(not defective) = number of ways to choose 2 non-defective DVD players / total number of ways to choose 2 DVD players

Total number of ways to choose 2 DVD players out of 6 is:

C(6,2) = 6! / ([2!] [4!]) = 15

Number of ways to choose 2 non-defective DVD players out of 4 is:

C(4,2) = 4! / ([2!] [2!]) = 6

Therefore, the probability that Cecil chooses 2 non-defective DVD players is:

P(not defective) = 6/15 = 2/5

So the value of P(not defective) is 2/5.

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Answer:

Answer and Explanation: The symmetric bell shape of the normal curve implies that the skewness of the distribution of the data is 0, and most of the observation is located at the middle of the distribution. The shape of the normal distribution is not positive and negative skewed, the shape seems to be bell-shaped.

HOPE THIS HELPS!

The coordinates of point C are (4, 2), which is located in quadrant 1.

In a 30-60-90 degree triangle, the side opposite the 30-degree angle is half the length of the hypotenuse, and the side opposite the 60-degree angle is √3/2 times the length of the hypotenuse. Since the side opposite the 30-degree angle is the shortest side, it must be the distance between points A and B, which is 8 units.

Let's call point C (x, y). Since angle C is 90 degrees, side AC is perpendicular to side AB, which means it is a vertical line that passes through point A. Similarly, side BC is perpendicular to side AB, which means it is a horizontal line that passes through point B.

Therefore, point C must lie on both the vertical line passing through A and the horizontal line passing through B. The equation of the vertical line passing through A is x = -4, and the equation of the horizontal line passing through B is y = -2. So we have the system of equations

x = -4

y = -2

Solving this system gives us the coordinates of point C: (-4, -2). However, this point is not in quadrant 1, as we desired.

To find a point C in quadrant 1, we need to flip the signs of the x and y coordinates of point C. So the coordinates of point C are (4, 2).

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In this simplex tableau, x₄, x₅, and z are the basic variables, while x₁, x₂, x₃, x₆, x₇, and x₈ are the nonbasic variables. The values of all the variables associated with this basic feasible solution are x₁ = 0, x₂ = 0, x₃ = 0, x₄ = 620, x₅ = 12, x₆ = 0, x₇ = 0, x₈ = 0, and z = 620.

In this tableau, the basic variables are x₄, x₅, and z, while the nonbasic variables are x₁, x₂, x₃, x₆, x₇, and x₈.

The basic variable associated with row 1 is x₄, the basic variable associated with row 2 is x₅, and the basic variable associated with row 3 is z.

The values of all the variables associated with this basic feasible solution are:

x₁ = 0, x₂ = 0, x₃ = 0, x₄ = 620, x₅ = 12, x₆ = 0, x₇ = 0, x₈ = 0, z = 620.

Note that these values correspond to the entries in the tableau in the "BV" column.

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The missing tableau is in the image attached below

Answer:

[tex]60n - 97[/tex]

Step-by-step explanation:

[tex]3 - 5( - 3n + 5)3 - 5( - 3n + 5)[/tex]

[tex]3 + (15n - 25)3 + 15n - 25[/tex]

[tex]3 + 45n - 75 + 15n - 25[/tex]

[tex]60n - 97[/tex]

The roots of the given quadratic equation are x= 4, [tex]\frac{-3}{2}[/tex]

The definition of a quadratic as a second-degree polynomial equation demands that at least one squared term must be included. It also goes by the name quadratic equations. The quadratic equation has the following generic form:

ax² + bx + c = 0The roots of a quadratic equation are found using the quadratic formula. In place of the factorization method, this formula aids in evaluating the quadratic equations' solutions. The quadratic formula aids in identifying the problem's fictitious roots when a quadratic equation lacks actual roots. Shreedhara Acharya's formula is another name for the quadratic formula.

2x² - 5x​ -12 = 0

The Shridharacharya formula or quadratic formula -

[tex]x=\frac{-b\ +-\sqrt{b^2-4ac}}{2a}[/tex]

we have a=2, b=-5 and c=-12

[tex]x=\frac{5+-\sqrt{5^2-4*2*12}}{2*2}\\\\x=\frac{5+-\sqrt{25+96}}{4}\\\\x=\frac{5+\sqrt{121}}{4}\\\\x=\frac{5+-11}{4}\\now, \\x=\frac{5+11}{4}=\frac{16}{4}\\x=4\\x=\frac{5-11}{4}\\x=-6/4\\x=-3/2[/tex]

The mistake in your solution is we have the value of b=-5 so , when we put the value of b in a formula then -5 will be 5.

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Answer:

6 and 10

Step-by-step explanation:

first find what 1 min is in meters

1/3 is 8 so 8x3 = 1m which is 24

now divide based on the next two numbers

1/4 of 24 = 6

6x 1 = 6

now the next

24/12 = 2

2x5 = 10

these two results, 6 and 10, are the answers to the missing meter values

As per the unitary method, Nikhil has 21 pencils and Cortez has 63 pencils.

Let's say Nikhil has x number of pencils. Then, according to the problem, Cortez has three times as many pencils as Nikhil. Therefore, Cortez has 3x number of pencils.

Together, they have a total of 84 pencils. So, we can write an equation based on the number of pencils owned by Nikhil and Cortez as follows:

x + 3x = 84

Simplifying the equation, we get:

4x = 84

Dividing both sides by 4, we get:

x = 21

So, Nikhil has 21 pencils. Using the fact that Cortez has three times as many pencils as Nikhil, we can find out how many pencils Cortez has:

Cortez has 3x = 3(21) = 63 pencils.

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Complete Question:

Lance has three times as many pencils as Nick, and they have 84 pencils together. How many pencils does each of them have?

The expression given as 11 + 8p²q² - 9pq - 8 - 20p²q² + 22pq when simplified is 3 +13pq - 12p²q²

Given that

11 + 8p²q² - 9pq - 8 - 20p²q² + 22pq

The given expression 11 + 8p²q² - 9pq - 8 - 20p²q² + 22pq contains several terms with different variables and exponents.

To simplify the expression, we need to combine the like terms.

So, we have

11 + 8p²q² - 9pq - 8 - 20p²q² + 22pq = 11 - 9pq - 8 - 12p²q² + 22pq

Similarly, we can combine the other terms

So, we have

11 + 8p²q² - 9pq - 8 - 20p²q² + 22pq = 3 +13pq - 12p²q²

So, the solution is 3 +13pq - 12p²q²

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Answer:  y = 2x + 8

Step-by-step explanation:

3y - 6x = 24

      3y = 6x + 24

        y = (6x + 24) / 3

        y = 2x + 8

Answer: No

Step-by-step explanation:

Because the length of the box is shorter than the length of the club

20in<26in

The width of the box is also shorter than the width of the club

13in<16in

The height of the box is also shorter than the height of the club

11in<16in

But what about putting it at an angle?

So we know  [tex]a^{2} +b^{2} =c^{2}[/tex]

so let's try [tex]20^{2} +13^{2} =x^{2}[/tex]

                               [tex]x^{2}[/tex]=569

                               [tex]x=\sqrt{159}[/tex]

x is near 23.85 in, but 23.85<26. So no.

The best type of inspection to use depends on the nature of the purchase. Each type of inspection has its own advantages and disadvantages, and should be chosen based on the requirements of the product and the level of risk associated with the inspection.

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BC will be congruent to AD under the C.P.C.T rule (corresponding parts of congruent triangles.)

Triangle congruence: Two triangles are said to be congruent if their three corresponding sides and their three corresponding angles are of identical size.

You can move, flip, twist, and turn these triangles to produce the same effect. When relocated, they are parallel to one another.

Two triangles are congruent if they satisfy all five conditions for congruence.

They include the right angle-hypotenuse-side (RAHS), angle-side-angle (ASA), angle-angle-side (AAS), side-side-side (SSS), and angle-side-angle (SSS) (RHS).

So, in the given △DAB and △DCB:

AC = AC = Common

∠DAC = ∠BAC = AC is the angle bisector

∠DCA = ∠BCA = AC is the angle bisector

Then, △DAB ≅ △DCB under the ASA congruency rule,

Then, BC will be congruent to AD under the C.P.C.T rule (corresponding parts of congruent triangles.)

Therefore, BC will be congruent to AD under the C.P.C.T rule (corresponding parts of congruent triangles.)

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Yes, we can say the shape of the sampling distribution for the sample mean if we know the population distribution.

However, if we do not know the population distribution, we cannot determine the exact shape of the sampling distribution for the sample mean. In this case, we can make use of the Central Limit Theorem (CLT) to make some assumptions about the shape of the sampling distribution. According to CLT, as the sample size increases, the sampling distribution of the sample mean becomes approximately normal, regardless of the shape of the population distribution, provided that the sample size is sufficiently large. Therefore, if the sample size is 19 and the population distribution is unknown, we can assume that the sampling distribution of the sample mean is approximately normal if the sample data is not heavily skewed or contains extreme outliers.

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The inverse function is R(L) = (360 - L)/15. It will take 8 weeks to pay loan if student owes $240 and 24 weeks to pay off the whole loan.

A. Let's start by defining the function L(w) as the amount owed w weeks after the student borrows the money. The student borrowed $360 and agreed to pay $15 per week, so the amount owed after w weeks can be calculated as:

L(w) = $360 - $15w

B. To find the inverse of function L, we need to switch the roles of the input and output variables. Let's call the inverse function R, where R(L) is the number of weeks it takes to pay off the loan if the amount owed is L. We can solve the equation from part A for w:

L(w) = $360 - $15w

$15w = $360 - L

w = (360 - L)/15

Therefore, the inverse function R(L) is:

R(L) = (360 - L)/15

This function tells us how many weeks it will take to pay off the loan for a given amount owed. For example, if the student owes $240, we can plug that into the inverse function to find out how many weeks it will take to pay off the loan:

R($240) = (360 - 240)/15 = 8

So it will take 8 weeks to pay off the loan if the student owes $240.

C. To find out how many weeks it will take to pay off the loan, we need to find the value of w when L(w) = 0 (i.e., when the loan is fully paid off). We can set L(w) = 0 and solve for w:

L(w) = $360 - $15w = 0

$15w = $360

w = 24

So it will take 24 weeks to pay off the loan.

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After 6 weeks, the size of the bacteria is 1600, and after 15 weeks, the size of the bacteria is 12800

Let N₀ be the initial number of bacteria in a population and r be the growth rate of the bacteria.

The doubling time t doubles the population after each time interval t, hence:

Nt = N₀ × 2^(t/t) Nₜ

= N₀ × 2^(t/d)Nt/N₀

= 2^(t/d)ln(Nₜ/N₀)

= (t/d)ln2ln(Nₜ) - ln(N₀)

= (t/d)ln2ln(Nₜ/N₀)

= (t/d)ln2t/d

= ln(Nt/N₀) / ln2

Now, we can calculate the time required for the population to double in size as

t = d × ln2/ln(Nₜ/N₀)

Now, let's substitute the values given and solve:

a) After 6 weeks, t = 3 weeks (since doubling time is 3 weeks)

So, Nₜ/N₀ = 2^(t/d)

= 2^(6/3)

= 2^2 = 4

Nt = N₀ × 4

= 100 × 4

= 400.

After 6 weeks, the size of the bacteria is 400.

b) After 15 weeks,

t = d × ln2/ln(Nₜ/N₀)ₜ

= 3 × ln2/ln(128)

= 3 × 0.9983/7.1554

= 0.4186 weeks

So, Nₜ/N₀ = 2^(t/d)

= 2^(15/3)

= 2^5

= 32Nₜ

= N₀ × 32 = 100 × 32

= 3200

After 15 weeks, the size of the bacteria is 3200.

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The probability that a randomly selected adult weighs between 120 and 165 lbs is approximately 0.8186.

Since the weights of adults are normally distributed with a mean of 140 lbs and a standard deviation of 25 lbs, we can use the standard normal distribution to calculate the probability.

We first need to standardize the values using the formula: z = (x - μ) / σ, where x is the weight, μ is the mean, and σ is the standard deviation.

For x = 120 lbs, z = (120 - 140) / 25 = -0.8, and for x = 165 lbs, z = (165 - 140) / 25 = 1.0. We can then use a calculator to find the probability between -0.8 and 1.0, which is approximately 0.8186.

Thus, the chance of picking an adult at random who weighs between 120 and 165 lbs is roughly 0.8186.

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The bottom of the ladder is moving at a rate of 4/3 ft per second.

To solve the problem, we can use the Pythagorean Theorem:[tex]$x^2 + y^2 = 64$[/tex], where x is the distance from the wall to the bottom of the ladder and y is the length of the ladder. We differentiate this equation with respect to time t and use the chain rule to get [tex]$\frac{d}{dt} (x^2 + y^2) = \frac{d}{dt} 64$[/tex]

Simplifying, we get

[tex]$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$[/tex]

When the bottom of the ladder is 4 feet from the wall, we have x = 4 and y = 8, so we can substitute these values into our equation and solve for [tex]$\frac{dx}{dt}$[/tex]:

[tex]$2(4)\frac{dx}{dt} + 2(8)(-2) = 0$[/tex]

[tex]$\frac{dx}{dt} = \frac{16}{8} = \frac{4}{3}$[/tex]

Therefore, the bottom of the ladder is moving at a rate of [tex]$\frac{4}{3}$[/tex] ft/s.

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A candidate's final multiple score main answer can have a maximum of three cumulative Pass Not Advanced (PNA) points applied to it.

PNA points are a form of assessment for a candidate's answers to multiple-choice questions.

They are awarded when the candidate selects an answer that is not necessarily wrong, but does not go into sufficient depth or demonstrate the required level of understanding.

Each PNA point is worth a fraction of a mark, with a total of three PNA points making up one mark of the candidate's final multiple score main answer.

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If i have four boxes arranged in a [tex]2 \times 2[/tex] grid, in 6 distinct ways can i place the digits 1, 2, and 3 in the boxes, using each digit exactly once, such that each box contains at most one digit

If a student has four boxes arranged in a 2 × 2 grid, the distinct ways to place the digits 1, 2, and 3 in the boxes, using each digit exactly once, such that each box contains at most one digit are six in number.

There are two possibilities for which box is left blank, so let's consider them separately:

Case 1: The top left box is left blank. In this case, the other three boxes must contain the digits 1, 2, and 3. There are three choices for what digit goes in the top right box, and then two choices for what digit goes in the bottom left box, and then one choice for what digit goes in the bottom right box.

This gives a total of 3·2·1 = 6 ways to place the digits in the boxes when the top left box is left blank.

Case 2: A different box is left blank. In this case, one of the digits must be left out. There are three choices for which digit is left out, and then three choices for which box is left blank. Once the digit and the blank box have been chosen, the remaining two digits can be placed in the other two boxes in any order, giving 2 ways.

This gives a total of 3·3·2 = 18 ways to place the digits in the boxes when a different box is left blank.

Therefore, the distinct ways to place the digits 1, 2, and 3 in the boxes, using each digit exactly once, such that each box contains at most one digit are six in number.

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Answer: It should be -1,7

Step-by-step explanation: x=1-2=-1

y=8-1=7

Answer:

(-1, 7)

Step-by-step explanation:

A left translation is a negative number affecting the x-coordinate.

1 - 2 = -1

A down translation is a negative number affecting the y-coordinate.

8 - 1 = 7

(1, 8) --------> (-1, 7)

a. The probability that all four units are good is 0.2067 (approx),

b. The probability that exactly two units are good is 0.0226 (approx).

Given a shipment of 13 microwave ovens contains four defective units and a vending company purchases four units at random, we need to calculate the probability of the following events:

(a) all four units are good.

(b) exactly two units are good.

(a) What is the probability that all four units are good?

To solve this, we need to use the formula for the probability of an intersection of independent events.

Since the probability of getting a good unit is 9/13, then the probability of getting 4 good units in a row is calculated as follows:

P(All 4 units are good) = P(Good unit) × P(Good unit) × P(Good unit) × P(Good unit) = 9/13 × 9/13 × 9/13 × 9/13 = 47829609/232044048 = 0.2067 (approx)

(b) What is the probability that exactly two units are good?

Here, we need to use the binomial probability formula since the number of good units follows a binomial distribution. We need to find the probability of getting exactly 2 good units, given that we are purchasing 4 units.

P(exactly 2 units are good) = C(4,2) × P(Good unit)² × P(Defective unit)²

= 6 × (9/13)² × (4/13)²

= 52488/2320440

= 0.0226 (approx)

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The most reasonable domain of the function C = 4ⁿ is all positive integers, which is answer choice (c) i.e. all whole numbers .

A function is a mathematical object that takes one or more inputs, performs a specified operation on them, and produces an output.

The inputs to a function are called the domain, and the outputs are called the range.

An equation, a graph, or a table can all be used to depict a function.

The function that represents the number of new coins the jester receives on the n-th day after January 1 is given by C = 4ⁿ.

Since n represents the number of days after January 1, it is most reasonable to assume that n is a positive integer, because it doesn't make sense to talk about a negative or fractional number of days in this context.

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The complete question is given below.

Using the identity sin² 0 + cos² 0 = 1, the value of cos 0 is 0.951 (to the nearest hundredth)

Using the identity sin² 0 + cos² 0 = 1, we can solve for cos 0:

cos² 0 = 1 - sin² 0

cos² 0 = 1 - (-0.31)²

cos² 0 = 1 - 0.0961

cos² 0 = 0.9039

Taking the square root of both sides, we get:

cos 0 ≈ ±0.951

Since 0 is in the interval ³ < 0 < 2π, we know that cos 0 must be positive. Therefore, to the nearest hundredth, cos 0 ≈ 0.95.

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The perimeter of the updated blueprint will be 24 times the sum of the original length and width.

If Ezra wants to multiply the dimensions of the stage by a certain percentage to make the image 12 times larger than the original, he needs to find out what percentage that is.

To do this, he can divide the desired size of the new stage by the original size of the stage, and then multiply by 100 to get the percentage increase. So, if the original blueprint dimensions are x by y, and he wants to make the image 12 times larger, the new dimensions will be 12x by 12y.

To find the percentage increase, he can use the following formula:

Percentage increase = [(new size - original size) / original size] x 100

In this case, the new size is 12 times the original size, so the formula becomes:

Percentage increase = [(12x * 12y - x * y) / (x * y)] x 100

Simplifying this expression gives:

Percentage increase = [(144xy - x * y) / (x * y)] x 100 = 14300%

Therefore, Ezra needs to multiply the dimensions of the stage by 14300% to make the image 12 times larger than the original blueprint.

To find the perimeter of the updated blueprint, he can use the formula for the perimeter of a rectangle, which is: Perimeter = 2(length + width)

In this case, the length and width have been multiplied by 12, so the new perimeter becomes:

Perimeter = 2(12x + 12y) = 24(x + y)

Therefore, the perimeter of the updated blueprint will be 24 times the sum of the original length and width.

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The dimensions that maximize the enclosed area are L = 41.665 feet and W = 41.665 feet for each corral and the maximum area is 10868.09 square feet.

To find the dimensions that maximize the enclosed area, we need to use optimization techniques. Let's denote the length of each rectangular corral by L and the width by W. We can write the total enclosed area as A = 6LW.

The perimeter of each corral is given by P = 2L + 2W, and we have a total of 6 corrals, so the total length of fencing required is 6P = 12L + 12W.

We are given that we have 1,000 feet of fencing, so we can write 12L + 12W = 1000, or equivalently, L + W = 83.33 (rounded to two decimal places).

We can now use this equation to express one of the variables (say, W) in terms of the other: W = 83.33 - L.

Substituting this expression for W into the formula for the enclosed area, we get A = 6L(83.33 - L) = 499.98L - 6L^2.

To find the value of L that maximizes the area, we need to take the derivative of A with respect to L and set it equal to zero: dA/dL = 499.98 - 12L = 0. Solving for L, we get L = 41.665 (rounded to three decimal places).

Substituting this value back into the expression for W, we get W = 83.33 - L = 41.665.

The maximum area is A = 6LW = 10868.09 square feet (rounded to two decimal places).

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The solution to the system of equations is (x, y) = (2, 1).

A system of equations is a set of two or more equations that are to be solved simultaneously, meaning that the values of the variables that satisfy each equation in the system must be found. The solution to a system of equations is the set of values for the variables that satisfy all the equations in the system.

To solve the system of equations:

8x - 5y = 11 ...(1)

4x - 3y = 5 ...(2)

We can use the elimination method to eliminate one of the variables. We want to eliminate the variable "y", so we need to multiply equation (2) by -5/3, which will give us:

-5/3(4x - 3y) = -5/3(5)

-20x/3 + 5y = -25/3 ...(3)

Now we can add equations (1) and (3) to eliminate "y":

8x - 5y + (-20x/3 + 5y) = 11 - 25/3

Combining like terms, we get:

(24x - 15y - 20x + 15y)/3 = 8/3

Simplifying, we get:

4x/3 = 8/3

Multiplying both sides by 3, we get:

4x = 8

Dividing both sides by 4, we get:

x = 2

Now we can substitute x = 2 into equation (1) or (2) to find y. Let's use equation (1):

8x - 5y = 11

8(2) - 5y = 11

16 - 5y = 11

Subtracting 16 from both sides, we get:

-5y = -5

Dividing both sides by -5, we get:

y = 1

Therefore, the solution to the system of equations is (x, y) = (2, 1).

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Complete question:

Given the system of equation:

8x - 5y = 11

4x - 3y = 5

Find the value of x and y.

The probability that it will take at least 5 minutes until the student is served when he enters the restaurant today is 50%.

This is because there is an equal chance that it could take less than 5 minutes or more than 5 minutes until the student is served.

In probability terms, the student's wait time is a random variable with two possible outcomes - wait time less than 5 minutes, or wait time greater than or equal to 5 minutes. Since there is an equal chance of either outcome occurring, the probability of the wait time being greater than or equal to 5 minutes is 50%.

This is also known as the Law of Large Numbers.

To further illustrate this concept, imagine that the student flips a fair coin. The two possible outcomes of the coin toss are heads or tails. Since each outcome has an equal chance of occurring, the probability of either heads or tails is 50%.

In this case, the probability of the student's wait time being at least 5 minutes is the same as the probability of the coin toss being heads or tails, hence making the probability 50% or 0.5.

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Answer:

Sure, here are the numbers arranged from greatest to least

Step-by-step explanation:

Order from greatest to least

1.3, 4.5%, 1/3, -4/5, -0.04, -21.5%

Answer:

From the greatest to least

Step-by-step explanation:

1.3, 4.5%, 1/3, -4/5, -0.04, -21.5%

You can identify which number is greater than other by using a number line.

What is a number line?

A number line is what a math student can use to find the answer to addition and subtraction questions. A straight line, theoretically extending to infinity in both positive and negative directions from zero, that shows the relative order of the real numbers.

Hope this helps :)

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