show that the properties of a probability distribution for a discrete random variable are satisfied.

The answer will of given mathematical logic will be T F T T F respectivelly.

A negation, conjunction, and disjunction are the fundamental mathematical logics. The symbols for negation, conjunction, and disjunction in mathematical logic are "," "," and "v," respectively.

Logical proofs frequently employ mathematical logic. Proofs are legitimate arguments that establish the veracity of mathematical assertions. A series of statements make up an argument. The conclusion is the last assertion, and the premises are all the statements that came before it (or hypothesis).

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The complete truth table is shown in the below diagram.

≈ q → ≈ p: True  False  True  True

The contrapositive of a conditional statement is a new conditional statement that is formed by negating both the hypothesis (the "if" part) and the conclusion (the "then" part) of the original statement, and switching their positions. The truth table for the contrapositive of a conditional statement has the same number of rows as the truth table for the original statement.

For example, if the original statement is "If it is raining, then the ground is wet", then the contrapositive would be "If the ground is not wet, then it is not raining."

According to the given table the contrapositive of a conditional statement q and p is defines as;

True

False

True

True

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Accοrding tο the data, the answers tο Questiοns 1 and 2 are: Harris shοuld anticipate spinning a sum οr less 10 apprοximately 367 times and a tοtal that is 10 οr larger apprοximately 133 times.

Yοur example is an equatiοn since an equatiοn that's any statement with an equal's sign. The usage οf equatiοns in mathematical expressiοns is widespread because mathematicians adοre equal signs. An equatiοn is a cοllectiοn οf guidelines fοr prοducing a specific οutcοme.

Part A: Tο calculate the likelihοοd that Harris will spinning a sum οr less 10, multiply the οverall number οf spins by the chance οf οbtaining a sum οr less 10. The οutcοmes οf the first spinner's spin are 1, 2, and 3, while the results οf the secοnd spinner's spin are 4, 5, 6, 7, and 8. Hence, the amοunts οr less 10 are:

1 + 4 = 5

1 + 5 = 6

1 + 6 = 7

1 + 7 = 8

1 + 8 = 9

2 + 4 = 6

2 + 5 = 7

2 + 6 = 8

2 + 7 = 9

3 + 4 = 7

3 + 5 = 8

3 + 6 = 9

There are 11 amοunts that cοuld be less than ten. The number οf successful results divided by the entire number οf pοssibilities, which is 11/15, represents the likelihοοd οf receiving a payοut οf less than 10 in a single spin. Harris will therefοre spin a tοtal less than 10 times, and the equatiοn tο estimate this is:

11/15 = x/500

After finding x, we οbtain:

x = (11/15) x 500

x = 366.67, which rοunds up tο 367

Sο, Harris shοuld expect tο spin a sum less than 10 abοut 367 times.

Part B: Tο determine hοw frequently Harris shοuld anticipate spinning a sum οf ten οr mοre, we can deduct the times that he shοuld anticipate spinning a sum lοwer than ten frοm the οverall number οf spins:

500 - 367 = 133

Therefοre, Harris shοuld expect tο spin a sum that is 10 οr greater abοut 133 times.

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a. Histogram shows tip amounts ranging between $6 and $25, skewed to the right with a longer tail of higher tips.

b. Increasing the $8 tip to $18 would increase the mean since total tip amount increases by $10 spread out over 60 customers. Median won't be affected since changing one value does not alter the middle value.

a. The histogram shows that Elizabeth received a range of tip amounts, with the majority of tips falling between $6 and $25. The distribution is skewed to the right, with a longer tail of higher tip amounts.

b. i. The mean would increase because the total tip amount would increase by $10, and this increase would be spread out over the 60 customers.

ii. The median would not be affected because it is the middle value when the data is ordered, and changing one value does not change the middle value.

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The likelihood that one or more of the dice will land on an even number is 1296.

The likelihood of an event is quantified by its probability, which is a number. It is stated as a number between 0 and 1, or in percentage form, as a range between 0% and 100%. The likelihood of an event increasing with probability of occurrence.

Four 6-sided dice are rolled what is the probability that at least two dice show least 2 die the same.

For 2 of the same: 5×5×642) =900

For 3 of the same: 5×643) =120

For 4 of the same: 644) =6

Combined: 900+120+6=1026

Total possibilities: 64=1296

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The probability that at least one of the dice lands on an even number is 15/16 or approximately 0.938.

What is probability?

Probability is a measure of the likelihood of an event occurring. It is a number between 0 and 1, where 0 means the event is impossible and 1 means the event is certain to happen. The probability of an event can be calculated by dividing the number of favorable outcomes by the total number of possible outcomes.

We can solve this problem by finding the probability that all four dice land on odd numbers and then subtracting this probability from 1 to get the probability that at least one of the dice lands on an even number.

The probability that one dice lands on an odd number is 3/6 = 1/2, and the probability that all four dice land on odd numbers is:

(1/2) × (1/2) × (1/2) × (1/2) = 1/16

Therefore, the probability that at least one of the dice lands on an even number is:

1 - 1/16 = 15/16

So the probability that at least one of the dice lands on an even number is 15/16 or approximately 0.938.

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The workload of the resource is 20.5 units.

To calculate the workload of a resource that serves different flow unit types, one must know the amount of flow units, the processing time for each flow unit, and the number of resources available. This is best calculated using Little's Law, which states that the average number of flow units in a system is equal to the average rate of flow units multiplied by the average time they spend in the system.

For example, if a resource is serving 3 flow unit types, A, B and C, with 10, 8 and 5 units respectively, and a processing time of 2 minutes, 1 minute and 3 minutes respectively, with 2 resources available, the workload can be calculated as follows:

Workload = (10*2 + 8*1 + 5*3) / 2

       = 41 / 2

       = 20.5 units

Therefore, the workload of the resource is 20.5 units.

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Complete question

What are the flow unit types that the resource is serving?

Answer:

Step-by-step explanation:

D

In response to the stated question, we may state that As a result, the 66th derivative of f(x) = 4 sin(x) is 4 sin(x) (x).

In mathematics, the derivative of a function with real variables measures how sensitively the function's value varies in reaction to changes in its parameters. Derivatives are the fundamental tools of calculus. Differentiation (the rate of change of a function with respect to a variable in mathematics) (in mathematics, the rate of change of a function with respect to a variable). The use of derivatives is essential in the solution of calculus and differential equation problems. The definition of "derivative" or "taking a derivative" in calculus is finding the "slope" of a certain function. Because it is frequently the slope of a straight line, it should be enclosed in quotation marks. Derivatives are rate of change metrics that apply to almost any function.

Using the chain rule and the derivative of the sine function repeatedly yields the 66th derivative of the function [tex]f(x) = 4 sin (x).[/tex]

The derivative of sin(x) is cos(x), and the derivative of cos(x) is -sin(x), and this pattern repeats itself every two derivatives.

As a result, the first derivative of f(x) is:

[tex]f'(x) = 4 cos (x)[/tex]

The second derivative is as follows:

[tex]f"(x) = -4 sin (x)[/tex]

The third derivative is as follows:

[tex]f"'(x) = -4 cos (x)[/tex]

The fourth derivative is as follows:

[tex]f""(x) = 4 sin (x)[/tex]

And so forth.

[tex]f^{(66)(x)} = 4 sin (x)[/tex]

Because the pattern repeats every four derivatives, the 66th derivative is the same as the second, sixth, tenth, fourteenth, and so on.

As a result, the 66th derivative of f(x) = 4 sin(x) is 4 sin(x) (x).

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The congruence x + 13 ≡ 55 (mod 34) simplifies to x ≡ 12 (mod 34). There are three solutions for x less than 100 that satisfy this congruence.

The given congruence is x + 13 ≡ 55 (mod 34). Simplifying this, we get x ≡ 12 (mod 34).

We need to find the number of solutions for x that are less than 100 and satisfy this congruence.

The general solution for the congruence x ≡ 12 (mod 34) is x = 12 + 34k, where k is an integer.

The solutions that are less than 100 are obtained when k = 0, 1, or 2.

Thus, the number of solutions is 3.

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Answer:

150 cm2

Step-by-step explanation:

Given side of cube's area = 25. Since Side's a square,

Edge^2 = 5^2 = 5 cm

Total surface area: 6*a² = 6*5*5 = 150 cm2

The cardinality of set from the given vein diagram is found as 2.

Think about set A. The set A is said to be finite and so its cardinality is same to the amount of elements n if it includes precisely n items, where n  ≥  0. |A| stands for the cardinality of such a set A.

It turns out that there are two kinds of infinite sets that we need to determine between since one form is much "bigger" than the other. Particularly, one type is referred to as countable and the other as uncountable.

From the given figure

Set A = {8 , 8, 3, 6}

Compliment of Set B (elements not present in set B):

Set [tex]B^{c}[/tex] = {8, 8, 6(pink), 3(white)}

Thus,

(A∩ [tex]B^{c}[/tex] ) = {8, 8} (present in both set)

n (A∩ [tex]B^{c}[/tex] ) = 2 (cardinal number)

Thus, the cardinality of the set from the given vein diagram is found as 2.

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Answer:

Step-by-step explanation:

To solve this problem, we can use the formula for the Pythagorean theorem, which states that for any right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

In this case, we are given the length of two sides of the triangle (the legs) and we need to find the length of the hypotenuse.

Let's label the sides of the triangle:

The shorter leg is the vertical side opposite the angle marked 55 degrees, so let's call it "a".

The longer leg is the horizontal side adjacent to the angle marked 55 degrees, so let's call it "b".

The hypotenuse is the side opposite the right angle, so let's call it "c".

Using trigonometry, we can determine the value of "a" and "b":

a = b * tan(55°) (since tangent = opposite/adjacent, we solve for opposite which is "a" in this case)

a = 100 * tan(55°) = 100 * 1.428 = 142.8

b = 100

Now, we can use the Pythagorean theorem to find the length of the hypotenuse:

c^2 = a^2 + b^2

c^2 = 142.8^2 + 100^2

c^2 = 20484.84 + 10000

c^2 = 30484.84

c = sqrt(30484.84)

c ≈ 174.6

Therefore, the length of the hypotenuse is approximately 174.6 units (the units are not given in the problem, but we can assume they are consistent with the units used for the given values of "a" and "b").

The problem does not specify the orientation or scale of the graph, but we can assume that it is a right triangle with the angle marked 55 degrees in the upper left corner.

The vertical side (the shorter leg) of the triangle should be labeled with a length of approximately 142.8 units (assuming the units used for the problem are consistent with the values given for "a" and "b"). The horizontal side (the longer leg) should be labeled with a length of 100 units.

The hypotenuse (the side opposite the right angle) should be drawn as a diagonal line connecting the endpoints of the vertical and horizontal sides. The hypotenuse should be labeled with a length of approximately 174.6 units.

The angle marked 55 degrees should be labeled as such, and the other two angles of the triangle (the right angle and the angle opposite the longer leg) should be labeled accordingly.

In linear equation, 11.85 pounds is the weight of the wire.

What is  linear equation?

A linear equation is a first-order (linear) term plus a constant in the algebraic form y=mx+b, where m is the slope and b is the y-intercept. The variables in the previous sentence, y and x, are referred to as a "linear equation with two variables" at times.

Total weight of pool having 16 wires     =13.6 pounds

Weight of the pool                                 =1.75

Therefore the weight of the wire alone = 13.6 - 1.75

                                                             =  11.85 pounds

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The value of b is 73° as opposite angles of congruent sides are equal in an isosceles triangle.

An angle is created by cοmbining twο rays (half-lines) that have a cοmmοn terminal. The angle's vertex is the latter, while the rays are alternately referred tο as the angle's legs and its arms.

An angle within a shape that has the shape's base as οne οf its sides is knοwn as the base angle οf a shape in geοmetry. Cοnsider the triangle in the image as an example. We can οbserve that the triangle's base side is made up οf an angle B side and an angle C side. As a result, the triangle's base angles are angles B and C.

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The volume of the new rectangular prism is 160 in³ after it has been dilated with a scale factor of 2.

In this case, the scale factor is 2, which means that the dimensions of the original figure will be multiplied by 2 to get the dimensions of the new figure.

Volume of rectangular prism = length x width x height

20 = l x w x h

Next, we need to find the new dimensions of the rectangular prism after it has been dilated by a scale factor of 2. We can do this by multiplying each dimension of the original rectangular prism by 2.

New length = 2 x l

New width = 2 x w

New height = 2 x h

Now we can find the volume of the new rectangular prism by using the same formula as before, but with the new dimensions:

Volume of new rectangular prism = (2 x l) x (2 x w) x (2 x h)

Simplifying this expression, we get:

Volume of new rectangular prism = 8 x (l x w x h)

We know that l x w x h is equal to the volume of the original rectangular prism, which is 20 in³. So we can substitute this value into the expression to get:

Volume of new rectangular prism = 8 x 20 in³

Volume of new rectangular prism = 160 in³

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The probability of a randomly selected, normally distributed value lying between 20 below the mean and 20 above the mean is 0.9545 (rounded to the nearest hundredth).

A number between zero and one represents the probability that an occurrence will take place. An event is a predefined set of random variable outcomes. Only one mutually exclusive event can occur at a time. Exhaustive events encompass or include all possible outcomes.

We can calculate the probabilities of a randomly chosen value falling between different z-scores using the provided standard normal distribution table.

a) The probability of a value being 0 below or above the mean is the same as the probability of a value being -1 to 1 standard deviations from the mean. According to the standard normal distribution table, the probability of a z-score between -1 and 1 is 0.6827. As a result, the probability of a randomly chosen, normally distributed value falling between 0 below and 0 above the mean is 0.6827. (rounded to the nearest hundredth).

b) The probability of a value falling between 20 and 20 standard deviations from the mean is the same as the probability of a value falling between -20/10 and 20/10 standard deviations from the mean (since the standard deviation is 10). According to the standard normal distribution table, the probability of a z-score between -2 and 2 is 0.9545. As a result, the probability of a randomly chosen, normally distributed value falling between 20 below and 20 above the mean is 0.9545. (rounded to the nearest hundredth).

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The homotopy classes of paths from x to x in a space X refer to a set of equivalence classes of continuous paths that start and end at the same point, x, in the space X, where equivalence is defined in terms of homotopy.

In other words, for any two paths, there exists a continuous transformation (called a homotopy) between them such that the endpoints remain fixed. Two paths are said to be homotopic if they can be continuously deformed into each other while keeping their endpoints fixed. The set of all paths that are homotopic to each other forms an equivalence class.

The homotopy classes of paths from x to x are important in algebraic topology, as they provide a way to study the topological structure of a space by analyzing the properties of the paths within it. They can also be used to define higher algebraic structures such as the fundamental group and higher homotopy groups.

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In linear equation, 65.625% of the pupils travel by bus.

What is  linear equation?

A linear equation is a first-order (linear) term plus a constant in the algebraic form y=mx+b, where m is the slope and b is the y-intercept. The variables in the previous sentence, y and x, are referred to as a "linear equation with two variables" at times.

A) 1800 * 0.55 * 0.3 = 297 Girls.

B)  1800 * 0.45 * 0.7 = 567 boys

C)  Girl

      297/864 * 100%  = 34.375%

  boy -

       567 ÷ (297 + 567 ) * 100%  = 65.625%

         864 = 297 + 567

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Answer: (f⋅g)(1) = 10.

Step-by-step explanation:

To find (f⋅g)(x), we need to multiply the two functions f(x) and g(x) together. This can be done by multiplying each term of f(x) by each term of g(x), and then combining like terms. We get:

(f⋅g)(x) = f(x) * g(x)

= (-7x+3) * (3x^2 - 4x - 1)

= -21x^3 + 28x^2 + x - 3x^2 + 4x + 1

= -21x^3 + 25x^2 + 5x + 1

To find (f⋅g)(1), we can substitute x=1 into the expression for (f⋅g)(x):

(f⋅g)(1) = -21(1)^3 + 25(1)^2 + 5(1) + 1

= -21 + 25 + 5 + 1

= 10

Therefore, (f⋅g)(1) = 10.

The shear force and bending moment diagrams for the given beam will have multiple segments of different shapes and slopes, reflecting the variation of loads along the length of the beam.

To construct the shear and bending diagrams for the given beam, we need to analyze the beam for the different sections where the load is applied. We can break down the beam into five sections:

Leftmost section (0 ≤ x ≤ L/4)

Second section (L/4 < x ≤ L/2)

Third section (L/2 < x ≤ 5L/12)

Fourth section (5L/12 < x ≤ 7L/12)

Rightmost section (7L/12 < x ≤ L)

We can use the equations for shear and bending moments to create the plots:

For section 1: 0 ≤ x ≤ L/4

The shear force diagram will be constant since there is no load applied in this section. The bending moment diagram will be a sloping line, which will be zero at x = 0 and will increase linearly with x as we move toward the right end of the section.

For section 2: L/4 < x ≤ L/2

The shear force diagram will start from the value of P at x = L/4 and remain constant up to x = L/2. The bending moment diagram will be a parabolic curve, which will be zero at x = L/4 and L/2 and will reach a maximum value at the midpoint of the section.

For section 3: L/2 < x ≤ 5L/12

The shear force diagram will start from the value of P at x = L/4 and remain constant up to x = L/2. At x = 5L/12, a load of P/3 is added, causing the shear force to increase suddenly. The bending moment diagram will be a cubic curve, which will be zero at x = L/4 and L/2, and will have a local minimum at x = 5L/12.

For section 4: 5L/12 < x ≤ 7L/12

The shear force diagram will start from the value of P + P/3 at x = 5L/12 and remain constant up to x = 7L/12. The bending moment diagram will be a cubic curve, which will be zero at x = L/4 and L/2, and will have a local maximum at x = 7L/12.

For section 5: 7L/12 < x ≤ L

The shear force diagram will start from the value of P + P/3 at x = 5L/12 and remain constant up to x = 7L/12. At x = L/3, a load of P/3 is added, causing the shear force to decrease suddenly. The bending moment diagram will be a parabolic curve, which will be zero at x = L/4 and L/2 and will reach a minimum value at the midpoint of the section.

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The inequality is X > 0 and a word sentence represent the graph is  X  the graph of a number line with an open circle on zero and an arrow pointing to the right.

The inequality X > 0 represents the graph of a number line with an open circle on zero to left and an arrow pointing to the right. This means that any value of X that is greater than zero is a valid solution for the inequality.

In other words, X can be any positive number, such as 1, 2, 3, and so on. However, X cannot be zero or any negative number, as those values do not satisfy the inequality. Therefore, the word sentence that represents this inequality is "X is greater than zero."

This means that X must be a positive number, and it can be any value that is greater than zero.

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For the computation of confidence interval for the difference in two population proportions following are negative,

p₁(cap) - p₂(cap)

Lower bound of the confidence interval

Upper bound of the confidence interval

For the computation of confidence interval,

The difference in two population proportions,

p₁ - p₂, can be negative or positive.

This implies,

The sample estimate of the difference in proportions,

p₁(cap) - p₂(cap), can also be negative or positive.

The standard error and critical value are always positive values and cannot be negative.

The lower and upper bounds of the confidence interval can be negative or positive.

Depending on the sample estimate and the margin of error.

So, both the lower and upper bounds can be negative.

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The above question is incomplete, the complete question is:

You are computing a confidence interval for the difference in 2 population proportions. which of the following could be negative?

Select all.

a. p₁

b. p₁(cap) - p₂(cap)

c. Standard error

d. Critical value

e. Lower bound of the confidence interval

f. Upper bound of the confidence interval

Answer:

I don't knowI am sorry I will let someone else answer

Step-by-step explanation:

there is an 85% chance that the two cards drawn at random from an ordinary deck of 52 cards will not be sixes.

The probability of drawing a card from an ordinary deck without replacement can be determined using the concept of conditional probability. Conditional probability is the probability of an event occurring, assuming that another event has already occurred.

In order to calculate the probability that the two cards drawn are not sixes, we can use the formula:

P(A and B) = P(A) x P(B|A)

Where A and B represent two independent events, P(A) is the probability of event A occurring, and P(B|A) is the conditional probability of event B occurring given that event A has already occurred.

The probability of drawing the first card that is not a six is:

P(A) = 48/52 = 0.9231

The probability of drawing the second card that is not a six, given that the first card drawn was not a six, is:

P(B|A) = 47/51 = 0.9216

Therefore, the probability of drawing two cards at random from an ordinary deck of 52 cards and having neither of them be a six is:

P(A and B) = P(A) x P(B|A) = 0.9231 x 0.9216 = 0.8503 or approximately 85%.

This means that there is an 85% chance that the two cards drawn at random from an ordinary deck of 52 cards will not be sixes.

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Answer: $27.98

Step-by-step explanation:

22.00 × .2= 4.40

22 + 4.40 = 26.40

26.40 × .06 = 1.584

26.40 + 1.584 = 27.984

Round to the nearest hundred so the total paid by the customer would be 27.98

The graph that shows the electricity usage on a record-breaking summer day is Sacramento, California is a function.

The domain is 24 hours of a day.

The number of megawatts used at 8 am is 1, 200 megawatts.

The time with the most electricity used was 4 pm to 6 pm and least used was 4 am.

f ( 12 ) would be 1, 900 megawatts.

Usage is increasing from 4 am to 5 pm and decreasing from 5 pm to 4 am.

The graph is a function because each point on the graph represents a distinct megawatt usage. The domain would be 24 hours of a day as this graph of electricity usage shows the usage per day.

The megawatts used at 8 am is:

= 1, 300 - ( 200 / 2 )

= 1, 200 megawatts

From 4 am to 5 pm, we see that electricity usage is increasing as people are getting ready for work and going to work, but from 5 pm to 4 am, electricity usage decreases.

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Answer:

To write an exponential decay function for this situation, we can use the formula:

P(t) = P₀e^(rt)

where:

P(t) = the population at time t

P₀ = the initial population

r = the annual rate of decrease (as a decimal)

t = time in years

We are given P₀ = 3,846 and r = -0.0027 (since the population is decreasing).

To approximate the population in 2022, we need to find t, the number of years from 1996 to 2022. That is:

t = 2022 - 1996 = 26 years

Now we can plug in the values we have:

P(t) = 3,846 e^(-0.0027t)

To find P(2022), we plug in t = 26:

P(26) = 3,846 e^(-0.0027(26))

≈ 3,200.62

Therefore, we can approximate the population of the city in 2022 to be about 3,201 people.

Answer:

3,101

Step-by-step explanation:

Please hit brainliest if this helped!

To write an exponential decay function for the population of the city, we can use the formula:

P(t) = P₀e^(-rt)

where P(t) is the population at time t, P₀ is the initial population, r is the decay rate, and e is the base of the natural logarithm.

In this problem, P₀ = 3,846 and r = 0.0027 (0.27% expressed as a decimal). We want to find the population in 2022, which is 26 years after 1996.To use the formula, we need to convert 26 years to the same time units as the decay rate. Since the decay rate is per year, we can use 26 years directly. Therefore, the exponential decay function for the population is:

P(t) = 3,846e^(-0.0027t)

To find the population in 2022 (t = 26), we substitute t = 26 into the function:

P(26) = 3,846e^(-0.0027*26) ≈ 3,101

Therefore, the population in 2022 is approximately 3,101.

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Answer:

If the time is 3:45 how many minutes is it slow or fast

Answer:

Math Quotient Verification

G For each ordered pair, determine whether it is a solution to the system of equations. 9x+2y=-5 2x-3y=-8 (x, y) (1, -7) (0, -4) (5,6) (-1,2) Is it a solution? Yes No X 5

To check if an ordered pair is a solution to a system of equations, we substitute the values of x and y into both equations and see if both equations are satisfied.

Let's check each ordered pair one by one:

(1, -7):

9x + 2y = -5 becomes 9(1) + 2(-7) = -5, which is false.

2x - 3y = -8 becomes 2(1) - 3(-7) = -8, which is true.

Therefore, (1, -7) is not a solution to the system of equations.

(0, -4):

9x + 2y = -5 becomes 9(0) + 2(-4) = -8, which is false.

2x - 3y = -8 becomes 2(0) - 3(-4) = 12, which is false.

Therefore, (0, -4) is not a solution to the system of equations.

(5, 6):

9x + 2y = -5 becomes 9(5) + 2(6) = 41, which is false.

2x - 3y = -8 becomes 2(5) - 3(6) = -8, which is true.

Therefore, (5, 6) is not a solution to the system of equations.

(-1, 2):

9x + 2y = -5 becomes 9(-1) + 2(2) = -11, which is false.

2x - 3y = -8 becomes 2(-1) - 3(2) = -8, which is true.

Therefore, (-1, 2) is not a solution to the system of equations.

Therefore, the answer is "No" for all the ordered pairs given in the problem.

(please mark my answer as brainliest)

Answer:

g(y) = √(3/2 y)

Step-by-step explanation:

To find the inverse of a function, we need to solve for x in terms of y and interchange x and y. That is, we need to write the given function f(x) = 2/3x^2 in the form y = 2/3x^2 and then solve for x in terms of y.y = 2/3x^2

Multiplying both sides by 3/2, we get:

3/2 y = x^2

Taking the square root of both sides, we get:x = ± √(3/2 y)

Note that we have two possible values of x for each value of y, because the square root can be either positive or negative. However, for a function to have an inverse, it must pass the horizontal line test, which means that each value of y can only correspond to one value of x.Therefore, we need to restrict the domain of the original function to ensure that it is one-to-one. The simplest way to do this is to take the range of the function and use it as the domain of the inverse function.The range of f(x) = 2/3x^2 is all non-negative real numbers, or [0, ∞). Therefore, we can define the inverse function g(y) as:

g(y) = ± √(3/2 y)

where we choose the positive square root to ensure that the function is one-to-one.Thus, the inverse of the function f(x) = 2/3x^2 is:

g(y) = √(3/2 y)

with domain [0, ∞).

Let me know if this helped by hitting brainliest! If you have a question, please comment below and I'll get back to you ASAP!

(a) The interval (-∞, ∞).

(b) The interval (-∞, ∞).

(c) The interval (-∞, ∞).

(d) The interval (-π/2, π/2) \ {0}.

(a) The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is the interval where the coefficient function, 3t, is continuous and bounded. Since 3t is a continuous and bounded function for all t in the interval (-∞, ∞), the given initial value problem is certain to have a unique twice-differentiable solution for all t in (-∞, ∞).

(b) The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is the interval where the coefficient functions, t(t - 4), 3t, and 4, are continuous and bounded. Since t(t - 4), 3t, and 4 are continuous and bounded functions for all t in the interval (-∞, ∞), the given initial value problem is certain to have a unique twice-differentiable solution for all t in (-∞, ∞).

(c) The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is the interval where the coefficient functions, cost and In|t|, are continuous and bounded. Since cost and In|t| are continuous and bounded functions for all t in the interval (-∞, ∞), the given initial value problem is certain to have a unique twice-differentiable solution for all t in (-∞, ∞).

(d) The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is the interval where the coefficient functions, x - 2, 1, and (x - 2)tanx, are continuous and bounded. Since x - 2, 1, and (x - 2)tanx are continuous and bounded functions for all x in the interval (-π/2, π/2) \ {0} , the given initial value problem is certain to have a unique twice-differentiable solution for all x in (-π/2, π/2) \ {0}.

Learn more about twice-differentiable solution here

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The given question is incomplete, the complete question is:

determine the longest interval in which the given initial value problem is certain to have a unique twice- differentiable solution. Do not attempt to find the solution. (a) ty" + 3y = 1, y(1) = 1, y'(1) = 2   (b)   t(t – 4)y" + 3ty' + 4y = 2, y(3) = 0, y'(3) = -1 (c)   y" + (cost)y' + 3( In |t|) y = 0, y(2) = 3, y'(2) = 1 (d) (x - 2)y"+y' +(x - 2)(tan x) y = 0, y(3) = 1, y'(3) = 2