Find the circumference of a circle with diameter, d = 9.06m. Give your answer rounded to 2 DP

Given:

Four students took a math test on decimals before being tutored. After attending tutoring sessions for a week, they took another test to see if they could improve their scores.

Required:

Here we have to find who's score was lower after tutoring.

Explanation:

Now by considering the red and blue bar in the given figure, we can say that

Kim score was lower after tutoring.

Final Answer:

Kim score was lower after tutoring.

The dependent variable of interest in examining whether the degree of parental involvement differs based on students' grade in school is the degree of parental involvement.

In this scenario, the independent variable is the students' grade in school (i.e., first, second, third, etc.). The dependent variable is the variable that is expected to be influenced by the independent variable. In this case, the dependent variable of interest is the degree of parental involvement.

The degree of parental involvement refers to the level or extent to which parents are engaged in their child's education and school-related activities. It can encompass various aspects, such as attending parent-teacher meetings, volunteering at school, helping with homework, and participating in school events.

By examining whether the degree of parental involvement differs based on students' grade in school, researchers or educators aim to explore any potential patterns or differences in parental involvement as children progress through different grade levels. This investigation can provide insights into how parental involvement may vary across grade levels and inform strategies to enhance parental engagement in education at specific stages of a child's schooling. Thus, the dependent variable of interest is the degree of parental involvement.

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Answer:

9 students played basketball

Step-by-step explanation:

Number of students that played tag:

36*5/12=

180/12 = 15 students that played tag

Number of students that played kickball:

36*1/3=

36/3 = 12 students that played kickball

Number of students that played basketball:

First, find the total number of students that DIDN'T play basketball:

15 students played basketball and 12 played kickball, so:

15 + 12 = 27 students didn't play basketball

36 - 27 = 9 students played basketball

Answer:

9 students played basketball

Step-by-step explanation:

36 students in the class

5/12 played tag (of the 36)

1/3 played kickball

Find a common denominator and add the fractions

5/12 + 1/3 = 5/12 + 4/12 = 9/12

9/12 simplifies to 3/4

3/4 played kickball and tag, the rest played basketball

Subtract kickball and tag students from 1, to find the fraction of basketball players

1 - 3/4 = 1/4 of the students played basketball

The total is 36 students, 1/4 played basketball

1/4 of 36 = 36/4 = 9

9 students played basketball

Let me know if you have any questions!

The interest estimated by compounding methods are-

(a) Annual: n = 1; CI = $860

(b) Semiannual: n = 2; CI = $862.

(c) Monthly: n = 12; CI = $865

(d) Daily: n = 365; CI = $855

The formula for computing the compound interest is -

CI = A - P

A = P(1 + r/100nt)∧rt

In which,

(a) Annual: n = 1

A = 10,400(1 + 8/500)∧5

A = 10,400 x 1.08

A = 11,260

CI = 11,260 - 10,400

CI = $860

(b) Semiannual: n = 2

A = 10,400(1 + 8/1000)∧10

A = 10,400 x 1.089

A = 11,262

CI = $862.

(c) Monthly: n = 12

A = 10,400(1 + 8/6000)∧60

A = 10,400 x 1.083

A = 11,265

CI = $865

(d) Daily: n = 365

A = 10,400(1 + 8/182500)∧1825

A = 10,400 x 1.0832

A = 11,255

CI = $855

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Answer:

All the boxes in the store will weigh b. 4,250 lbs.

Step-by-step explanation:

Well, we know that the storage facility can hold a maximum of 50 large boxes, each weighing about 85 pounds.

85 x 50 = 4,250

Therefore, all the boxes in the store will weigh b. 4,250 lbs.

Hope this helps! :D

Answer:

Step-by-step explanation:

As in Binomial theorem :-

\( = {(4t - 3)}^{5} \)

\(= {(4t)}^{5} - 5 {(4t)}^{4} (3)+ 10 {(4t)}^{3} {(3)}^{2} - 10 {(4t)}^{2} {(3)}^{3} + 5(4t) {(3)}^{4} - {(3)}^{5} \)

\(= {1024t}^{5} - 5 ({256t}^{4} )(3)+ 10 ({64t}^{3} ) (9 ) - 10 ({16t}^{2} )(27) + 5(4t) (81) - 243\)

\(= {1024t}^{5} - 3840{t}^{4} + 5760{t}^{3} - 4320{t}^{2} + 1620t- 243\)

Hence, the required answer is :-

\({1024t}^{5} - 3840{t}^{4} + 5760{t}^{3} - 4320{t}^{2} + 1620t- 243\)

Answer: Danita gets 2×900 = 1800 and Juanita gets 2700

Step-by-step explanation: add ratios together. You get 2+3 = 5

Divide 4500 by 5 you get 900. Multiplying this by 2 and 3

Answer:

Brainliest???

Step-by-step explanation:

Reflection over the y axis

Answer:

\(\boxed {\boxed {\sf 53}}\)

Step-by-step explanation:

This is a right triangle, which we know because of the little square in the corner.

We can use the Pythagorean Theorem:

\(a^2+b^2=c^2\)

where a and b are the legs and c is the hypotenuse.

In this triangle, 28 and 45 are the legs, because they make up the right angle. The unknown side is the hypotenuse, because it is opposite the right angle.

\(a=28 \\b=45 \\\)

\((28)^2+(45)^2=c^2\)

Solve the exponents.

\(784+(45)^2=c^2\)

\(784+2025=c^2\)

Add the two numbers.

\(2809=c^2\)

Solve for c by isolating it on one side of the equation. c is being squared. The Inverse of a square is the square root. Take the square root of both sides of the equation.

\(\sqrt{2809} =\sqrt{c^2}\)

\(\sqrt{2809} = c\)

\(53=c\)

The third side and hypotenuse of the triangle is 53

\(\pink{\sf Third \: side \: of \: the \: triangle = 53}\)

As, the given triangle is a right angled triangle,

Hence, We can use the Pythagoras' Theorem,

\(\star\:{\boxed{\sf{\pink {H^{2} = B^{2} + P^{2}}}}}\)

Here,

In given triangle,

Now, by Pythagoras' theorem,

\(\star\:{\boxed{\sf{\pink {H^{2} = B^{2} + P^{2}}}}}\)

\( \sf : \implies H^{2} = (45)^{2} + (28)^{2}\)

\( \sf : \implies H^{2} = 45 \times 45 + 28 \times 28\)

\( \sf : \implies H^{2} = 2025 + 784\)

\( \sf : \implies H^{2} = 2809\)

By squaring both sides :

\( \sf \sqrt{H^{2}} = \sqrt{2809}\)

\( \sf : \implies H^{2} = \sqrt{2809}\)

\( \sf : \implies H^{2} = \sqrt{(53)^{2}}\)

\( \sf : \implies H^{2} = 53 \)

\(\pink{\sf \therefore \: Third \: side \: of \: the \: triangle \: is \: 53}\)

━━━━━━━━━━━━━━━━━━━━━

Answer:

8 cubes

If we want to cover the base of the prism having dimensions 2 cm × 1 cm with cm length of cube, then there will be (4 × 2) = 8 cubes that can occupy the base

Step-by-step explanation:

The total surface area of the triangular prism is 1008cm² with dimensions 13cm, 5cm, 12cm, 9cm, 15cm, and 20cm.

To find the total surface area of a triangular prism with dimensions 13cm, 5cm, 12cm, 9cm, 15cm, and 20cm, we first need to identify the different faces of the prism. The triangular prism has two triangular faces and three rectangular faces.

The area of one triangular face can be calculated as

(1/2) x base x height, where the base is 12cm and the height is 9cm. Therefore, the area of one triangular face is

(1/2) x 12cm x (9+5)cm = 84cm²

Since there are two triangular faces, the total area of the triangular faces is 2 x 84cm² = 168cm².

The area of one rectangular face can be calculated as length x width, where the (length, width) is (13cm, 20cm), (14cm, 20cm) and (15cm, 20cm) = 15 × 20 + 13 × 20 + 14 × 20

= 840cm²

Thus, the total surface area of the triangular prism is the sum of the areas of the two triangular faces and three rectangular faces, which is

168cm² + 840cm² = 1008cm²

Therefore, the total surface area of the triangular prism is 1008cm².

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Answer:

Using the cosines law

Step-by-step explanation:

So, we need to find the length of the other side, we will use the version of the Cosine Rule where a2 is the subject of the formula, which is given by, a2=b2+c2−2bccosA, Side a is the one we are trying to find. Sides band care the other two sides, and angle Ais the angle opposite side a. c2=a2+b2−2abcosC

Answer:

n = 2

Step-by-step explanation:

Given

\(\frac{1}{2}\)(n - 4) - 3 = 3 - (2n + 3)

Multiply through by 2 to clear the fraction

n - 4 - 6 = 6 - 2(2n + 3) ← distribute

n - 10 = 6 - 4n - 6 ( add 4n to both sides )

5n - 10 = 0 ( add 10 to both sides )

5n = 10 ( divide both sides by 5 )

n = 2

According to this property, when both sides of an equation are multiplied by the same real number, both sides of the equation always remain the same. The formula for this property can be expressed in real numbers a, b and c. If a × c = b × c.

Property of Equality:

The equivalence property describes the relationship between two equal quantities. If you apply a math operation to one side of the equation, you must also apply it to the other side of the equation to maintain balance.

That is, a property that does not change the truth value of an equation or does not affect the equivalence of two or more quantities is called an equality property. These equality properties help solve various algebraic equations and define equivalence or equilibrium relationships.

Multiplicative Property of Equality:

According to this property, when both sides of an equation are multiplied by the same real number, both sides of the equation always remain the same.

The formula for this property can be expressed as for the real numbers a, b, and c.

If a × b, then, a × c = b × c.

In algebra, the multiplicative property of an equation helps extract unknown terms from an equation. Because multiplication and division are opposites of each other.

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Assuming these triangles are similar, first find the standard ratio between them by dividing the corresponding sides of triangle VUT by RSQ:

\(R=\frac{36}{54} =\frac{2}{3}\)

Now we know that if we multiply any side of triangle RSQ by the ratio (2/3), we can find the corresponding side of triangle VUT. Using this logic, multiply side RQ by our common ratio to find what UT should equal:

\(24*\frac{2}{3} =UT\)

\(16=UT\)

Finally, set up an equation for this side:

\(x+5=16\)

Solve for x:

\(x=11\)

a. If every student who attends shares the facility cost equally, the function model will be 1200/N

domain : 1 ≤ N ≤ (312 - additional adult chaperones)           N ∈ Z

1200/N ≤ 7.50

1200 / 7.50 ≤ N

N ≥ 1200/7.5

N ≥ 160

More than 160 students must attend to make the cost per student no more than $ 7.50

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Using the given information we found that the equation of the parabola is:

y = (-4/9)*(x - 3)^2 + 5

And its graph is below.

For a parabola with vertex (h, k), the equation is:

y = a*(x - h)^2 + k

Here the vertex is (3, 5), so the equation is:

y = a*(x - 3)^2 + 5

And the y-intercept is y = 1, this means that:

1 = a*(0 - 3)^2 + 5

1 = a*9 + 5

1 - 5 = a*9

-4/9 = a

So the parabola is:

y = (-4/9)*(x - 3)^2 + 5

And its graph is below.

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Answer:

34.166

Step-by-step explanation:

Mean=(Sum of all the numbers)/12=410/12=34.166

\(\huge\textsf{Hey there!}\)

\(\bullet \large\textsf{ Finding the mean you have to add up all of your numbers \& divide it}\\\large\textsf{by the total numbers in your data plot}\)

\(\mathsf{\dfrac{20 + 5 + 45 + 90 + 60 + 45 + 30 + 10 + 20+ 45 + 15 + 25}{12}}\)

\(\mathsf{20 + 5 + 45 + 90 + 60 + 45 + 30 + 10 + 20+ 45 + 15 + 25}\)

\(\mathsf{= 25 + 45 + 90 + 60 + 45 + 30 + 10 + 20 + 45 + 15 + 25}\)

\(\mathsf{= 70 + 90 + 60 + 45 + 30 + 10 + 20 + 45 + 15 + 25}\)

\(\mathsf{= 160 + 60 + 45 + 30 + 10 + 20 + 45 + 15 + 25}\)

\(\mathsf{= 220 + 45 + 30 + 10 + 20 + 45 + 15 + 25}\)

\(\mathsf{= 265 + 30 + 10 + 20 + 45 + 15 + 25}\)

\(\mathsf{= 295 + 10 + 20 + 45 + 15 + 25}\)

\(\mathsf{= 305 + 20 + 45 + 15 + 25}\)

\(\mathsf{= 325 + 45 + 15 + 25}\)

\(\mathsf{= 370 + 15 + 25}\)

\(\mathsf{= 385 + 25}\)

\(\mathsf{= \bf 410}\)

\(\mathsf{= \dfrac{410}{12}}\)

\(\mathsf{\dfrac{410}{12} \approx \bf 34.1667}\)

\(\boxed{\boxed{\huge\textsf{Thus, your ANSWER is: \boxed{\mathsf{ \bf \underline{{\star 34.1667\star}}}}}}}}\huge\checkmark\)

\(\huge\textsf{Good luck on your assignment \& enjoy your day!}\)

~\(\frak{Amphitrite1040:)}\)

Prove \((pv q) ^ (qv ^ (rv\) is true if all p, q, r same truth value, false otherwise. Compare true/false for each clause with un/negated variable.

Step 1: We should prove that \((p v q) ^ (qv ^ (rv\) is true if and only ifp, q, and r have the same truth value.

Step 2: Suppose p, q, and rare all true. Then, (p v -) (qv (rv -p) is true as each clause has an unnegated variable.

Step 3: Suppose p, q, and rare all false. Again, (p v g) A (qvA (rv -p) is true as each clause has a negated variable.

Step 4: Now, suppose one of the variables p, q, or r is true and the other two are false. Then, the clause that contains the negation of the true variable will be false.

Step 5: Similarly, when one of the variables is false and the other two are true, the clause that contains the negation of the true variable will be true.

Step 6: Hence,\((p v q) ^ (q ν ^ (r v p)\) is true when R q, and r have the same truth value and it is false otherwise.

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Answer:

f(x) = 2(x+4)(x+4)(x+4)(x-10)

Step-by-step explanation:

Here, we want to write a polynomial function with leading coefficient of 2, root - 4 with multiplicity 3 ( meaning it appears 3 times);

and root 10 with multiplicity 1

If x = -4

Then in linear form, we have x + 4

if x = 10, in linear form, we have x -10

So the polynomial is;

f(x) = 2(x+4)(x+4)(x+4)(x-10)

Step-by-step explanation:

As I do not have the work shown to calculate the number of ways to choose the passengers, I cannot identify the specific error made. However, I can provide some insight on how to approach this type of problem correctly.

When choosing 2 people from a group of 10, we can use combinations, which are denoted as "n choose k" and represented by the formula:

n choose k = n! / (k! * (n-k)!)

where n is the total number of items in the group and k is the number of items to be chosen.

In this case, we want to choose one passenger for first class and one passenger for coach. Therefore, we can separate the group of 10 into two subgroups: one subgroup with 1 person (for first class) and another subgroup with 9 people (for coach).

Using combinations, we can calculate the number of ways to choose 1 person from the subgroup of 1 and 1 person from the subgroup of 9:

(1 choose 1) * (9 choose 1) = 1 * 9 = 9

Therefore, there are 9 ways to choose the passengers to fill the seats.

Answer:

5 2/3 or 5.6

Step-by-step explanation:

The diagonal of a room measuring 5 ft and 12 ft along the walls should be 14.6 ft.

The diagonal of the room can be calculated using the Pythagorean theorem, which states that the square of the hypotenuse (the longest side of the triangle) is equal to the sum of the squares of the other two sides. In this case, the hypotenuse would be the diagonal, the other two sides would be 5 ft and 12 ft.

Thus, the equation would be: a² + b² = c², with a = 5 ft, b = 12 ft, c = 14.6 ft.

The calculation for this equation is as follows:

5² + 12² = c²; 25 + 144 = c²; 169 = c²; √169 = c; c = 14.6 ft.

This means that the diagonal of a room measuring 5 ft and 12 ft along the walls is 14.6 ft.

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the required sample size is approximately 178. Answer (c) is correct.To determine the required sample size, we can use the formula:

n = (z * σ / E)^2

where:
- n is the sample size
- z is the z-score for the desired level of confidence (99% corresponds to z = 2.58)
- σ is the standard deviation of the population
- E is the maximum error we want to allow in our estimate of the population mean

Plugging in the values given in the problem, we get:

n = (2.58 * 14 / 4)^2 ≈ 178

Therefore, the required sample size is approximately 178. Answer (c) is correct.

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Answer:

Its A i took this question

Step-by-step explanation:

The work required to pump all the water over the edge of the tank is approximately 271,433.64 foot-pounds.

To calculate the work required to pump all the water over the edge of the tank, we need to consider the weight of the water in the tank and the height it is lifted.

First, let's find the volume of the water in the tank. The volume of a cone is given by the formula V = (1/3)πr²h, where r is the radius and h is the height. Plugging in the values, we have:

V = (1/3)π(3²)(12)

= (1/3)π(9)(12)

= 36π

Next, we need to find the weight of the water. The weight of an object is given by the formula W = mg, where m is the mass and g is the acceleration due to gravity. The mass of the water can be found by multiplying its volume by the density of water, which is approximately 62.4 pounds per cubic foot:

m = (36π)(62.4)

≈ 22619.47 pounds

Now, we can calculate the work done by multiplying the weight of the water by the height it is lifted. In this case, the height is 12 feet:

W = (22619.47)(12)

≈ 271433.64 foot-pounds

Therefore, the work required to pump all the water over the edge of the tank is approximately 271,433.64 foot-pounds.

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The work required to pump water out of an inverted conical tank involves calculating the pressure-volume work at infinitesimally small volumes within the tank and integrating this over the entire volume of the tank. This provides an interesting application of integral calculus in Physics.

The question requires the concept of work in Physics applied to a fluid, in this case, water lying within an inverted conical tank. Work is done when force is applied over a distance, as stated by work = force x distance. In the fluid analogy, the 'force' link is the pressure exerted on the water and the distance is the change in volume of the fluid. Therefore, work done (W) = Pressure x Change in Volume (ΔV).

In this scenario, you are required to pump out water from an inverted conical tank, hence, the work you do is against the gravitational force pulling the water downwards. To calculate the total work done, you have to consider the work done at each infinitesimally small (hence, constant pressure) strip of volume and integrate over the entire volume of the tank.

The detail of calculation would require the knowledge of integral calculus and the formula for volume of a cone. I recommend considering this as an interesting application of integrals in Physics. Also remember that the volume of a cone = 1/3πr²h, where 'r' is the radius of base and 'h' is the height of cone.

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Answer:

\(\frac{400}{3} \pi\)

Step-by-step explanation:

Formula for Cone: π\(r^{2}\frac{h}{3}\)

Since we have all the components, we can find the volume of the cone.

R = 10

H = 4

π\(10^{2}\frac{4}{3}\)

10×10 = 100

100π\(}\frac{4}{3}\)

\(}\frac{4}{3}\)×100

4       100          400

---  ×   -----    =    ------

3          1              3

Answer: \(\frac{400}{3} \pi\)

Hope this helped.

Find the volume of a cone with radius 10 feet and height of 4 feet.

Given Data :

Radius = 10 feet

Height = 4 feet

Formulae :

\( \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \star \: \blue{ \underline{ \overline{ \green{ \boxed{ \frak{{ \sf V}olume_{(Cone)} = \pi {r}^{2} \frac{h}{3}}}}}}}\)

Putting the values we get

\( \frak{ Volume_{(Cone)} = 3.14 × (10)² × \frac{4}{3} }\)

\( \frak{Volume_{(Cone)} = 3.14 × 100 × \frac{4}{3} }\)

\( \frak{Volume_{(Cone)} = 314 \times \frac{4}{3} }\)

\( \frak{Volume_{(Cone)} = 418.67 \: ft³ }\)

Henceforth, the volume is 418.67 ft³

The area of a desert in Africa, which is 4 times the area of a desert in Asia (m square miles), can be expressed as 4m square miles.

To find the area of the desert in Africa, we start with the given information that the area of the desert in Asia is m square miles. Since the area of the desert in Africa is 4 times that of Asia, we multiply the area of the desert in Asia (m) by 4. This gives us the algebraic expression 4m, which represents the area of the desert in Africa.

By multiplying the value of m by 4, we are scaling up the area by a factor of 4. This accounts for the statement that the desert in Africa is 4 times the size of the desert in Asia. Therefore, if we know the area of the desert in Asia represented by m, we can express the area of the desert in Africa as 4m, reflecting its fourfold increase in size compared to Asia.

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