A popcorn container is in the shape of a square pyramid the container meausers 9 inches high and each side measures 6 inches how many cubic inches does the popcorn container hold?


NEED HELP SO LOST AND CONFUSED PLS HELPP IM NOT A VERY SMART PERSON SO HELPPPP IM LOSIN ME MIND

The number of balls, given the ratio of white balls to yellow balls that could not be in the bag is C. 42 balls.

The number of balls that could not be in the bag, given the ratio of white balls to yellow balls, is the number that would not give a whole number when multiplied by the ratio of white balls to yellow balls.

26 balls :

= 3 / ( 10 + 3 white balls ) x 26

= 3 / 13 x 26

= 6 balls

39 balls :

= 3 / 13 x 39

= 9 balls

42 balls :

= 3 / 13 x 42

= 9.69 balls

There therefore cannot be 42 balls.

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Options for this question include:

The area of the sector is ≈134 sq units and the length of the arc is ≈16.75

The sector of a circle produced by a portion of the circumference (arc) and the circle's radii at both ends of the arc. A sector of a circle can be compared to a piece of pizza or a pie in shape.

The length of an arc is the distance between two point in the section of a curve.

According to the given information:

Given that radius= 16 in, ∅ = 60°

The formula for finding the area of sector= (∅/360°)[tex]\pi r^{2}[/tex]

Substituting values in the formula we get

= (60°/360°)*3.14*[tex]16^{2}[/tex]

=133.9733

≈134 sq units.

The formula for Finding Arc Length =  (∅/360°)[tex]2\pi r[/tex]

On substituting the values

= (60°/360°)*2*3.14*16

=16.74666

≈16.75 in

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Angle 1 = Angle 2 = Angle 3 = Angle 4, we can infer.

A right angle is an angle with a measure that is exactly equal to 90 degrees (or /2).

We can observe that a transversal divides two parallel lines in the given diagram. Let's give the angles names:

The angle created by the junction of line m and line p is Angle 1, which is located in the top left corner.

The angle created by the intersection of line n and line p is Angle 2, which is located in the top right corner.

Line n and line q intersect at angle 3, which is located in the lower right corner.

The angle created by the junction of line m and line q is angle 4, which is located in the lower left corner.

As they are on the same side of the transversal and in corresponding locations with respect to the parallel lines, we can see that angles 1 and 2 are corresponding angles. When lines are cut by a transversal, corresponding angles are congruent, hence we know that:

1 + 2 = an angle.

Due to their placement on opposing sides of the transversal and in-between the parallel lines, angles 2 and 3 can also be seen to represent alternate internal angles. When lines are cut by a transversal, alternate inner angles are congruent, hence we know that:

2 + 3 = a right angle

As they are on the same side of the transversal and in corresponding locations with respect to the parallel lines, we can see that angles 3 and 4 are corresponding angles. When lines are cut by a transversal, corresponding angles are congruent, hence we know that:

Angle 3 = angle 4 =...

So, Angle 1 = Angle 2 = Angle 3 = Angle 4 follows.

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The range of the given graph will be [-1, 2.9].

Think about the function y = f. (x).

The range of the function is the range of all the y values, from least to maximum. Substitute all possible values of x into the provided expression of y to see whether it is positive, negative, or equal to other values.

A bar chart with bars for each dimension (category) that range between a start value and an end value is called a range bar chart. To track a project schedule, for instance, Range Bar charts can be used to show the start and finish of activities or tasks along a time axis, much like a Gantt chart.

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We can write the following system of equations:

x + y = 28 (equation 1)

50x + 80y = 1850 (equation 2)

An equation is a statement that two expressions are equal. It consists of two expressions, separated by an equals sign. The expression on the left-hand side of the equals sign is equal to the expression on the right-hand side. Equations can be used to represent relationships between variables, to solve problems, and to describe mathematical models. They are an important tool in algebra, calculus, and other areas of mathematics.

A linear equation is an algebraic equation of the first degree, which means that the variables in the equation have an exponent of one and are not multiplied or divided by any other variables. Linear equations can have one or more variables and are typically represented graphically by a straight line.

In the given question,

Let x be the number of small boxes shipped and y be the number of large boxes shipped.

From the problem, we know that each small box of paper weighs 50 pounds and each large box of paper weighs 80 pounds. We also know that a total of 28 boxes of paper were shipped weighing 1850 pounds altogether.

Using this information, we can write the following system of equations:

x + y = 28 (equation 1)

50x + 80y = 1850 (equation 2)

In equation 1, x + y = 28 represents the total number of boxes shipped, which is the sum of the number of small boxes shipped (x) and the number of large boxes shipped (y).

In equation 2, 50x + 80y = 1850 represents the total weight of the boxes shipped, which is the sum of the weight of each small box (50x) and the weight of each large box (80y).

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a) The system of inequalities that can be used to represent the situation is x + y ≤ 15 and 4x + 8y ≥ 80

b) One combination of hours that will allow Edith to earn at least $80 per week while working no more than 15 hours is 10 hours babysitting and 5 hours as a library assistant.

To start, we know that Edith can work no more than 15 hours per week. Therefore, we can write the following inequality:

x + y ≤ 15

Next, we know that Edith earns $4 per hour babysitting and $8 per hour as a library assistant. We want her to earn at least $80 per week, so we can write the following inequality:

4x + 8y ≥ 80

This inequality states that the total amount Edith earns from both jobs (4x + 8y) must be greater than or equal to $80.

Now that we have two inequalities, we have a system of inequalities that can be used to represent the situation:

x + y ≤ 15

4x + 8y ≥ 80

To determine one combination of hours that will allow Edith to earn at least $80 per week while working no more than 15 hours, we can solve this system of inequalities.

Let's solve the first inequality for y:

y ≤ 15 - x

Now we can substitute this expression for y into the second inequality:

4x + 8(15 - x) ≥ 80

Simplifying and solving for x, we get:

-4x + 120 ≥ 80

-4x ≥ -40

x ≤ 10

To find the number of hours she should work as a library assistant, we can substitute x = 10 into our expression for y:

y ≤ 15 - x

y ≤ 15 - 10

y ≤ 5

So Edith should work no more than 5 hours as a library assistant per week.

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Answer:

Let x be the price of one adult ticket, and y be the price of one student ticket.

From the information given in the problem, we can set up the following system of equations:

15x + 5y = 265 (equation 1)

15x + 16y = 419 (equation 2)

We can solve for x by subtracting equation 1 from equation 2:

15x + 16y - (15x + 5y) = 419 - 265

11y = 154

y = 14

Now we can substitute y = 14 into either equation 1 or equation 2 to solve for x. Let's use equation 1:

15x + 5(14) = 265

15x + 70 = 265

15x = 195

x = 13

Therefore, one adult ticket costs $13, and one student ticket costs $14.

Answer:

x = 15

Step-by-step explanation:

Isolate the variable, x. Note the equal sign, what you do to one side, you do to the other.

Do the opposite of PEMDAS.

PEMDAS is the order of operations, and stands for:

Parenthesis

Exponents (& Roots)

Multiplications

Divisions

Additions

Subtractions

~

First, add 10 to both sides of the equation:

[tex]4x - 10 = 50\\4x - 10 (+10) = 50 (+10)\\4x = 50 + 10\\4x = 60[/tex]

Next, divide 4 from both sides of the equation:

[tex]4x = 60\\\frac{4x}{4} = \frac{60}{4} \\x = \frac{60}{4} = 15\\x = 15[/tex]

~

x = 15 is your answer.

~

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Answer:

Step-by-step explanation:

[tex]\tt 4x -10 = 50[/tex]

Add 10 to both sides:-

Divide both sides by 4:-

___________________

Hope this helps!

well, you'd have the 2000 plus the 7% :|

[tex]\begin{array}{|c|ll} \cline{1-1} \textit{\textit{\LARGE a}\% of \textit{\LARGE b}}\\ \cline{1-1} \\ \left( \cfrac{\textit{\LARGE a}}{100} \right)\cdot \textit{\LARGE b} \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{7\% of 2000}}{\left( \cfrac{7}{100} \right)2000}\implies 140~\hfill \underset{ new~balance }{\stackrel{ 2000~~ + ~~140 }{\text{\LARGE 2140}}}[/tex]

If you are rolling two dice and one coin. the probability of rolling a three on the first dice, a four on the second dice, and flipping a heads on the coin is B. 1/72.

Probability of rolling a three on the first dice is 1/6, the probability of rolling a four on the second dice is 1/6, and the probability of flipping a heads on the coin is 1/2.

To find the probability of all three events happening together, we need to multiply the probabilities of each individual event:

(1/6) x (1/6) x (1/2) = 1/72

Therefore, the probability is B. 1/72.

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The exponential function between (-1, 2/5) and (3, 250) is as follows:

[tex]f(x) = 2 * 5^x[/tex]

By combining the fourth roots from both sides, we arrive at:

b = 5

When we use the expression we discovered for a and this value of b, we get:

a = (2/5) * 5 = 2

As a result, the exponential function between (-1, 2/5) and (3, 250) is as follows:

[tex]f(x) = 2 * 5^x[/tex]

A relation between a collection of inputs and outputs is known as a function. A function is, to put it simply, a relationship between inputs in which each input is connected to precisely one output. Each function has a range, codomain, and domain. The usual way to refer to a function is as f(x), where x is the input. A function is typically represented as y = f. (x).

In mathematics, a function is a unique arrangement of the inputs (also referred to as the domain) and their outputs (sometimes referred to as the codomain), where each input has exactly one output and the output can be linked to its input.

from the question:

This is the shape of the exponential function:

f(x) = a *[tex]b^x[/tex]

where a represents the starting point and b represents the exponential function's base.

We must solve the system of equations to determine the values of a and b that meet the requirements:

a * [tex]b^(-1)[/tex] = 2/5 (equation 1) 

a *[tex]b^3[/tex]= 250 (equation 2) 

We can solve for an in equation 1 by multiplying both sides by b:

a = (2/5) * b

Substituting this expression into equation 2, we get:

(2/5) * b *[tex]b^3[/tex] = 250

Simplifying, we get:

[tex]b^4 = 3125[/tex]

By combining the fourth roots from both sides, we arrive at:

b = 5

When we use the expression we discovered for a and this value of b, we get:

a = (2/5) * 5 = 2

As a result, the exponential function between (-1, 2/5) and (3, 250) is as follows:

[tex]f(x) = 2 * 5^x[/tex]

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As a result, when x = -5a /, the subject of the formula a/b = 2x/(x + 5) is x. (a - 2b).

For instance, C is the subject of the formula C = 2r, which calculates the circumference of a circle. To modify the subject of a formula is to rewrite it such that the quantities are still related in the right way. M/2 equals D if M = 2D.

In order to eliminate the fraction and make x the subject of the formula a/b = 2x/(x + 5), we can cross-multiply:

a(x + 5) = 2bx

Next, we can distribute the a:

ax + 5a = 2bx

We can now separate the x terms from the constant terms on each side:

ax - 2bx = -5a

x(a - 2b) = -5a

Finally, we can solve for x by dividing both sides by (a - 2b):

x = -5a / (a - 2b).

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a) Compare the populations using measures of center and variation. Mean and MAD of vollyball game attendence are 86, 19.6 respectively. Mean and MAD of basketball game attendence are 185, 17.7 respectively.

b) Difference in the measures of center as a -4.6 to 5.1 multiple of the measure of variation. The difference in the means is about -4.6 to 5.1 times the MAD.

We have two tables consists attendances at volleyball games and basketball games at a school during the year. See the above figure carefully. Now, Compare the populations using measures of center and variation : As we know mean of data is calculated by dividing the sum of data values to the number of data values. Here, Sum of vollyball attendence values, ∑xᵢ = 1720

Number of days = 20

So, mean of vollyball game attendence

= 1720/20 = 86

Sum of basketball ball attendence values, ∑yᵢ = 3700

Number of days = 20

So, mean of basketball ball game attendence = 3700/20 = 185

Now, measure absolute deviations, MAD

MAD of vollyball game attendence is

= [∑( mean of vollyball game attendence - attendance values (i))]/number of days

= 392/20 = 19.6

MAD of basketball ball game attendence is = [∑( mean of basketball ball game attendence - attendance values (i))]/number of days

= 354/20 = 17.7

b) Now, we check the relation between difference in the measures of center as a multiple of the measure of variation.

A : vollyball game attendence

B : basketball game attendence

difference in the measures of center

= Mean (A) - Mean(B) or Mean(B) - Mean (A)

measure of variation, MAD(A) , MAD(B)

[Mean (A) - Mean (B)]/MAD(A)

= ( 86 - 185)/19.6

= -4.643 ~ -4.6

[Mean (B) - Mean (A)]/MAD(B)

= ( 185 - 86)/17.7

= 91/17.7 = 5.1412 ~ 5.1

Hence, difference in the means is about -4.6 to

5.1 times the MAD.

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Complete question :

The above tables show the attendances at volleyball games and basketball games at a school during the year. Volleyball Game Attendance Basketball Game Attendance 112 95 84106 62 68 75 88 93 127 98 117 60176 141 152 181 198 21 49 5485 74 88 132 5 202 190 173 155 169 188 195 4 179 163 186 184207 219 228

a) Compare the populations using measures of center and variation.

b) Express the difference in the measures of center as a multiple of the measure of variation. Round your answers to the nearest tenth. The difference in the means is about (blank) to (blank) times the MAD.

The value of the calculated perimeter, in inches, of the floor in the scale drawing is 50 inches

The dimensions of the rectangular floor is calculated as

Dimension = 28 feet in length and 22 feet in width

The length of the scale is given as

Scale length = 14 inches

When the above is compared, we can see that

The length is divided by 2 to get the scale length

So, we have

Scale width = 22 inches/2

Evaluate

Scale width = 11 inches

The perimeter is then calculated as

Perimeter = 2 * (Length + width)

So, we have

Perimeter = 2 * (14 + 11)

Evaluate

Perimeter = 50 inches

Hence, the perimeter is 50 inches

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We can see here that the best description of the meaning of the y-intercept of the regression line, as given by the equation ý = 27.273x + 18.182, is: C. "When the time to play is 0 hours, the predicted rating is 18.182. This interpretation is not meaningful since a game cannot have a playtime of 0 hours."

The y-intercept is a term used in the context of a linear equation or a graph to describe the point at which the graph intersects or crosses the y-axis. It represents the value of the dependent variable (y) when the independent variable (x) is equal to zero.

The y-intercept is the value of ŷ (the predicted rating) when x (the time to play) is zero. However, in this context, it does not make sense for a game to have a playtime of zero hours. Therefore, the interpretation of the y-intercept in this case is not meaningful or applicable to the real-world scenario of board games.

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so yeah the answers

Answer:

6*3

n=26

For each of the following situations the critical value are:

a) H0: μ=115 vs. HA: μ ≠115 at α= 0.05; n= 41,  the critical values are ±1.96.
b) H0: p=0.14 vs. HA: p>0.14 at α= 0.10,  the critical value is 1.28.
c) H0: p= 0.6 vs. HA: p≠ at α=0.01 - z = 2.326, the critical values are ±2.58.
d) H0: p=0.8 vs. HA: p<0.8 at α=0.01; n=500, the critical value is -2.33.
e) H0​: p=0.9 vs. HA: p< at α=0.01, the critical value is -2.33.

a) Since n=41 is greater than 30, we can use a z-test. The test is two-tailed because the alternative hypothesis is μ≠115. Using a significance level of 0.05, the critical values are ±1.96. Therefore, the rejection region is z < -1.96 or z > 1.96.

b) Since we are testing a proportion, we can use a z-test for proportions. The test is one-tailed because the alternative hypothesis is p > 0.14. Using a significance level of 0.10, the critical value is 1.28. Therefore, the rejection region is z > 1.28.

c) Since we are testing a proportion, we can use a z-test for proportions. The test is two-tailed because the alternative hypothesis is p≠0.6. Using a significance level of 0.01, the critical values are ±2.58. Therefore, the rejection region is z < -2.58 or z > 2.58.

d) Since n=500 is greater than 30, we can use a z-test for proportions. The test is one-tailed because the alternative hypothesis is p < 0.8. Using a significance level of 0.01, the critical value is -2.33. Therefore, the rejection region is z < -2.33.

e) Since we are testing a proportion, we can use a z-test for proportions. The test is one-tailed because the alternative hypothesis is p < 0.9. Using a significance level of 0.01, the critical value is -2.33. Therefore, the rejection region is z < -2.33.

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The event C is: C = { HHT, HTH, THH, HTT, THT, TTH }
P(C) = 1/2

(a) P(notB) = 1/2
(b) P((notA) or C) = 5/8
(c) P(A or B or C) = 7/8

Solution:
A fair coin is tossed three times, and the events A, B, and C are defined as follows: A: { At least one head is observed } B: { At least two heads are observed } C: { The number of heads observed is odd }

The sample space of the given event is as follows:
S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Probability of the occurrence of A:
If at least one head is observed, then it can be possible in the following ways:
(i) HHT, HTH, THH, or
(ii) HHT, HTH, THH, or HTT, THT, TTH, or
(iii) HHT, HTH, THH, or HTT, THT, TTH, or HHH
Therefore, the event A is: A = { HHT, HTH, THH, HTT, THT, TTH, HHH }
P(A) = 7/8
Probability of the occurrence of B:
If at least two heads are observed, then it can be possible in the following ways:
(i) HHT, HTH, THH, or
(ii) HTT, THT, TTH, or
(iii) HHH
Therefore, the event B is: B = { HHT, HTH, THH, HTT, THT, TTH, HHH }
P(B) = 7/8
Probability of the occurrence of C:
If the number of heads observed is odd, then it can be possible in the following ways:
(i) HHT, HTH, THH, or
(ii) HTT, THT, TTH
Therefore, the event C is: C = { HHT, HTH, THH, HTT, THT, TTH }
P(C) = 1/2

Probability of the occurrence of notB:
Not B means "at least one head" or "no head." So, the event notB is not the event B, which means the event (notB) is the event in which we do not get at least two heads.
The (notB) event is: (notB) = { TTH, THT, HTT, TTT }
P(notB) = 4/8 = 1/2

Probability of the occurrence of (notA) or C:
The event (notA) means that we don't get at least one head, which means we get all tails. The (notA) event is: (notA) = { TTT }
P(notA) = 1/8
Now, we will find P((notA) or C).
For this, we can take the union of (notA) and C.
(notA) ∪ C = { HHT, HTH, THH, HTT, THT, TTH, TTT }
P((notA) or C) = 7/8

Probability of the occurrence of A or B or C:
The probability of the occurrence of A or B or C can be found as follows:
P(A or B or C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)
Here, A ∩ B = { HHT, HTH, THH, HHH }, A ∩ C = { HHT, HTH, THH }, B ∩ C = { HHT, HTH, THH }, and A ∩ B ∩ C = { HHT, HTH, THH }.
Substituting these values, we get:
P(A or B or C) = 7/8 + 7/8 + 1/2 - 3/8 - 3/8 - 1/8 + 1/8
P(A or B or C) = 7/8

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If a student is equally likely to select pizza, nachos, or chicken for lunch, the probability that the student does not select chicken is 2/3.

Since the student has three equally likely options to choose from, the probability of selecting each option is 1/3. Therefore, the probability of the student selecting chicken is 1/3.

To calculate the probability that the student does not select chicken, we need to subtract the probability of selecting chicken from 1. This is because the sum of all possible outcomes must equal 1, and if we know the probability of one outcome, we can find the probability of its complement (not selecting chicken) by subtracting it from 1.

Therefore, the probability that the student does not select chicken is:

1 - 1/3 = 2/3

So, the correct answer is 2/3.

In summary, when the probability of each possible outcome is equal, we can find the probability of an event (in this case, not selecting chicken) by subtracting the probability of that outcome from 1.

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As a result, the answer to the following question, We may use the SAS criteria to argue that the two triangles are congruent if we know that KM is congruent to SU.

A triangle is a polygon because it contains four or more parts. It features a simple rectangular shape. A triangle ABC is a rectangle with the edges A, B, and C. When the sides are not collinear, Euclidean geometry produces a single plane and cube. If a triangle contains three components and three angles, it is a polygon. The corners are the points where the three edges of a triangle meet. The sides of a triangle sum up to 180 degrees.  

Option b: KM is congruent to SU is the additional information required by SAS to demonstrate Triangle KLM is congruent to Triangle UTS.

To demonstrate that two triangles are congruent using the SAS (Side-Angle-Side) congruence criterion, we must demonstrate:

Two comparable side pairs are congruent.

The angle included (the angle formed by these two sets of sides) is congruent.

In this situation, we are told that LM is congruent to ST, but we must still demonstrate that another pair of sides is congruent. We may use the SAS criteria to argue that the two triangles are congruent if we know that KM is congruent to SU.

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Answer: 0.72 pts.

Step-by-step explanation:

Answer:f(1) = f(3) = f(9) = f(27) = f(81).

Step-by-step explanation:To show that f(1) = f(3) = f(9) = f(27) = f(81), we can use the given equation and substitute x with different values.

Let x = 1:

f(3x) = f(3) + f(x)

f(3(1)) = f(3) + f(1)

f(3) = f(3) + f(1) (since f(3x) = f(3) + f(x) when x = 1)

f(1) = 0

Now, let x = 3:

f(3x) = f(3) + f(x)

f(3(3)) = f(3) + f(3)

f(9) = 2f(3)

Since we already know that f(3) = f(3) + f(1), we can substitute it into the equation above:

f(9) = 2f(3) = 2(f(3) + f(1)) = 2f(3) + 2f(1)

f(9) - 2f(3) = 2f(1)

Now let x = 9:

f(3x) = f(3) + f(x)

f(3(9)) = f(3) + f(9)

f(27) = f(3) + f(9)

We can substitute the expression for f(9) we just derived:

f(27) = f(3) + (f(9) - 2f(3))

f(27) = f(9) - f(3)

f(27) = 2f(3) - f(3)

f(27) = f(3)

Finally, let x = 27:

f(3x) = f(3) + f(x)

f(3(27)) = f(3) + f(27)

f(81) = f(3) + f(27)

We can substitute the expression for f(27) we just derived:

f(81) = f(3) + f(27)

f(81) = f(3) + f(3)

f(81) = 2f(3)

Since we know that f(27) = f(3), we can substitute it into the equation above:

f(81) = 2f(3) = 2(f(27)) = 2(f(9) - f(3)) = 2f(9) - 4f(3)

We also know that f(9) = 2f(3) + 2f(1) from our earlier work, so we can substitute it into the equation above:

f(81) = 2f(9) - 4f(3) = 2(2f(3) + 2f(1)) - 4f(3) = 4f(3) + 4f(1) - 4f(3) = 4f(1)

Therefore, we have shown that f(1) = f(3) = f(9) = f(27) = f(81).

b.  [tex]\dfrac{x^\frac{1}{2} }{y^{-1}}[/tex] is an equivalent expression of y√x.

d. [tex]\rm (xy^2)^{\tfrac{1}{2} }[/tex] is an equivalent expression of y√x.

Equivalent expressiοns are expressiοns that wοrk the same even thοugh they lοοk different. If twο algebraic expressiοns are equivalent, then the twο expressiοns have the same value when we plug in the same value(s) fοr the variable(s).

Given expression y√x

a. Is not equivalent as power is rational and exponent are non equal,

b. [tex]\dfrac{x^\frac{1}{2} }{y^{-1}}[/tex]

=  [tex]\dfrac{\sqrt{x} }{\frac{1}{y}}[/tex]

= y√x

Thus, b.  [tex]\dfrac{x^\frac{1}{2} }{y^{-1}}[/tex] is an equivalent expression of y√x.

c. Is not a n equivalent expression at has a different variable.

d. [tex]\rm (xy^2)^{\tfrac{1}{2} }[/tex]

[tex]\rm\sqrt{ (xy^2)} }[/tex]

[tex]\rm y \sqrt{ (x)} }[/tex]

Thus,  d. [tex]\rm (xy^2)^{\tfrac{1}{2} }[/tex] is an equivalent expression of y√x.

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Result obtained is $460

Answer:Let X denote the number of tests that need to be performed until a satisfactory result is obtained. Then X has a geometric distribution with p = 0.25. Therefore,X ~ Geo(0.25) E[X] = 1/p = 1/0.25 = 4 tests.For the first 5 tests, the cost is $100 each. So the cost of running the first 5 tests is 5 × $100 = $500. Thereafter, the cost of each test is $40. The expected cost of running the tests until a satisfactory result is obtained is given by:E(Cost) = $500 + E(X – 5) × $40Expected Cost = $500 + (E(X) – 5) × $40= $500 + (4 – 5) × $40= $460Therefore, the expected cost of running the tests until a satisfactory result is obtained is $460.

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The inverse of the function is t = (r - 1.13) / 0.83.

What is function ?

In mathematics, a function is a relation between a set of inputs (the domain) and a set of possible outputs (the codomain) with the property that each input is related to exactly one output. A function can be thought of as a rule or a machine that takes an input value and produces a corresponding output value.

The given equation is: r = 0.83t + 1.13

To find the inverse of the function, we need to solve the equation for t.

r = 0.83t + 1.13

Subtract 1.13 from both sides:

r - 1.13 = 0.83t

Divide both sides by 0.83:

(r - 1.13)/0.83 = t

So the inverse function is:

t = (r - 1.13)/0.83

Or we can write it as:

[tex]f^{(-1)(r)} = (r - 1.13)/0.83[/tex], where [tex]f^{(-1)}[/tex]  represents the inverse function.

Therefore, the inverse of the function is t = (r - 1.13) / 0.83.

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Answer:I could possibly be wrong but I say no. From what I remember, with these things you can put the points wherever as long as it follows the rules you were given for each triangle.

Step-by-step explanation:

I actually drew it out, and to me it appears as it’s not, yet I could be wrong and I’m sorry if I am! Angles aren’t my best, but I’m really good at math! If you have other problems I can go over them

From the right triangle figure the sides are solved to be

The side b and side d is worked using SOH CAH TOA

Sin = opposite / hypotenuse - SOH

Cos = adjacent / hypotenuse - CAH

Tan = opposite / adjacent - TOA

The figure describes a right triangle, and side b is calculated using sin, SOH

Sin 45 = opposite / hypotenuse

Sin 45 = b / 8

b = 8 * sin 45

b = 4√2

Solving for side d

sin 60 = d / 7

d = 7 * sin 60

d = 7√3 / 2

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Ratio analysis is the best way to ensure reliable financial information, as it allows investors to evaluate the financial strength of a company and the reliability of its financial information by calculating ratios such as the current ratio, quick ratio, and debt-to-equity ratio.

Reliable financial information is information that is accurate and up-to-date. It is the foundation of any investor’s decision-making process, as it allows them to make informed decisions. One of the best ways to ensure reliable financial information is to perform a ratio analysis. Ratio analysis is the process of evaluating financial information by calculating ratios such as the current ratio, quick ratio, and debt-to-equity ratio.

The current ratio is calculated by dividing a company’s current assets by its current liabilities. This ratio measures a company’s ability to pay its short-term obligations. The quick ratio is calculated by dividing a company’s current assets minus inventory by its current liabilities. This ratio measures a company’s ability to pay its short-term obligations without having to sell its inventory. The debt-to-equity ratio is calculated by dividing a company’s total liabilities by its  total number. This ratio measures how much debt a company has relative to its equity.

By looking at these ratios, investors can get an idea of the financial strength of a company and the reliability of its financial information. The higher the ratios, the more reliable the financial information is likely to be

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Complete question:

How can investors ensure reliable financial information?

Shapes 1 and 2  that can be classified as rectangles.

Rectangles are geometric shapes with four sides and four angles. They are characterized by having opposite sides that are parallel and of equal length, and all four angles are right angles (90 degrees).

Rectangles are also known as quadrilaterals, which means they have four sides.

The second shape can be classified as a rectangle as it has four sides of equal length, two pairs of opposite sides that are parallel, and four right angles.

The first shape can also be classified as a rectangle as it has two pairs of parallel lines of equal length, with four right angles.

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