In an office, each worker eats 2 biscuits per day. There are 20 workers in the office
every day of the week. Biscuits come in packets of 14 and cost £1.30 per packet.
a) How many biscuits are eaten in the office per week?
b) How much does the office spend on biscuits per week?
******
£...

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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Option D is the graph that represent the variation of the gravitational energy (E) with the distance (r) from the centre of a spherical shell of mass M radius R

When R is less than r, the graph we would have  is a straight line. When R = r, it has constant value and finally if R is more than r it is a hyperbola.

The intensity that is inside the spherical shell would be given as 0. Then we would have this represented as

I = 0

upto r = R

I ∝ I/r²

r > R

Hence D

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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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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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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 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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Let's label the regions in the Venn diagram as follows:

F: students who do only field events

T: students who do only track events

S: students who do only swimming events

FT: students who do both field and track events

ST: students who do both swimming and track events

FST: students who do all three events

From the information given in the problem, we can fill in some of the values in the Venn diagram:

F + FT + 15 = 33 (33 students do field events, and 15 do field events only)

T + FT + 14 = 40 (40 students do track events, and 14 do both field and track events)

S + ST + 10 = 23 (23 students do swimming, and 10 do both swimming and track events)

F + T + S + 2FT + ST + FST = 68 (there are 68 students in total)

We can simplify these equations to:

F + FT = 18

T + FT = 26

S + ST = 13

F + T + S + 2FT + ST + FST = 68

To solve for the remaining unknowns, we need to use some algebra. We can start by solving for FT:

F + FT = 18

T + FT = 26

Adding the two equations, we get:

F + T + 2FT = 44

Rearranging, we get:

FT = (44 - F - T) / 2

Now we can substitute this expression for FT into the equation for the total number of students:

F + T + S + 2FT + ST + FST = 68

F + T + S + 2((44 - F - T) / 2) + ST + FST = 68

F + T + S + 44 - F - T + ST + FST = 68

Simplifying, we get:

S + ST + FST = 26

Now we have two equations involving S, ST, and FST:

S + ST = 13

S + ST + FST = 26

We can solve for S and ST by subtracting the first equation from the second:

S + ST + FST = 26

(S + ST) + FST = 26

Substituting S + ST = 13:

13 + FST = 26

FST = 13

So there are 13 students who do all three events. Now we can use the equation S + ST = 13 to solve for S:

S + ST = 13

ST = 10 (given in the problem)

S = 13 - 10 = 3

So there are 3 students who do swimming only. Similarly, we can use the equation T + FT = 26 to solve for T:

T + FT = 26

FT = (44 - F - T) / 2

Substituting F + FT = 18:

T + (18 - F) / 2 = 26

Multiplying both sides by 2:

2T + 18 - F = 52

The solutions can be written as x = 8 + 10i and x = 8 - 10i

The graph of a quadratic equation is a parabola, which is a symmetrical U-shaped curve. The direction and shape of the parabola depend on the value of a. If a is positive, the parabola opens upward, and if a is negative, the parabola opens downward.

Quadratic equations are used in many areas of mathematics and science, including physics, engineering, and finance. They are also used in real-life situations to model and solve problems related to motion, distance, time, and other quantities.

To solve this quadratic equation, we can rearrange it to be in standard form:

x² - 16x + 164 = 0

We can then use the quadratic formula to find the solutions:

x = (-b ± √(b² - 4ac)) / 2a

In this case, a = 1, b = -16, and c = 164. Substituting these values into the formula, we get:

x = (16 ± √((-16)² - 4(1)(164))) / 2(1)

Simplifying, we get:

x = (16 ± √(256 - 656)) / 2

x = (16 ± √(-400)) / 2

We have a negative number inside the square root, which means that the solutions are complex. To express them in the form x = a + bi and x = a - bi, we can simplify the square root by factoring out -1:

x = (16 ± √(400) i) / 2

x = (16 ± 20i) / 2

x = 8 ± 10i

Therefore, the solutions to the quadratic equation x² - 16x + 164 = 0 are:

x = 8 + 10i and x = 8 - 10i

So, a = 8 and b = 10. Therefore, the solutions can be written as:

x = 8 + 10i and x = 8 - 10i

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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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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.

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.

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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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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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]

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?

Answer:

$1920

Step-by-step explanation:

If Tammy's water bill is deducting $160 every other month for 2 years, (or $160 a month for 1 year since we skip a month every two months), then you would multiply $160 by 12 months to get $1920.

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!

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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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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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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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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Recursive fοrmula οf a1 = 3, d = -8 is -29 = -21 + (-8)

Any term οf a series can be defined by its preceding term in a recursive fοrmula (s).

The recursive fοrmula fοr an arithmetic sequence with first term a1=3 and cοmmοn difference d=-8 is:

an = an-1 + d, where n >= 2

Expanding the fοrmula fοr the first few terms, we get:

a2 = a1 + d = 3 - 8 = -5

a3 = a2 + d = -5 - 8 = -13

a4 = a3 + d = -13 - 8 = -21

a5 = a4 + d = -21 - 8 = -29

Nοw,

Recursive fοrmula οf a1 = 3, d = -8 is -29 = -21 + (-8)

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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).

Answer: 42$

Step-by-step explanation:

egiryofhvjg

so yeah the answers

Answer:

6*3

n=26

Answer:

21,265.65 in³

Step-by-step explanation:

Formula: Volume (V) = πr²h

V = πr²h

Slotting in values,

V = 3.14 x 1.5² x 3.01

V = 3.14 x 2.25 x 3.01

V = 21.26565 in³

But that is for 1 cupcake

To fill a thousand of them, you will need to multiply your answer by the number of cupcakes in total.

Therefore,

Answer = V x 1000

Answer = 21.26565 x 1000

Answer = 21,265.65 in³

The answer is already rounded to the nearest hundredth,

So they would need 21,265.65 cubic inches (in³) to fill 1000 cupcakes.

Please mark brainliest.

Thanks.

Another representation of the given algebraic expression 52s + 30b as required is; 2 (26s + 15b).

As evident in the task content; it follows that the equivalent expression to 52s + 30b is to be determined.

Given; 52s + 30b

Since 2 is a common factor to both terms of the equation; we have that;

2 (26s + 15b)

Ultimately, the equivalent expression is; 2 (26s + 15b).

Complete question:

What is 52s+30b in a factorized form.

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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 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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