20


Sean pays £10 for 24 chocolate bars.


He sells all 24 chocolate bars for 50p each.


Work out Sean's percentage profit. .

The distance that shows how much farther (in km) from the sun is Saturn than Venus is: [tex]1.2 * 10^9[/tex] km.

According to the table, the distance of Saturn from the Sun is [tex]1.4 * 10^{9}[/tex] and the distance of Venus from the Sun is [tex]1.1 * 10^{8}[/tex] .

Now to determine how much further from the Sun is Saturn than Venus, we will subtract the distance of the planet with the higher distance span from the one with the lower distance.

So our calculation will go thus:

[tex]1.4 * 10^9 - 1.1 * 10^8 = \\140000000 - 11000000 = 1290000000\\= 1.29 * 10^9[/tex]

From the calculation above, we can see how much further from the sun, is Saturn than Venus.

Complete Question:

The table shown below gives the approximate distance from the sun for a few different planets. How much farther (in km) from the sun is Saturn than Venus? Express your answer in scientific notation.

_______km

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To determine whether the series น k2 sk converges, we can use the Integral Test. Let f(x) = x2, then f'(x) = 2x. Since 2x is continuous, positive, and decreasing on [1,∞), In summary, the series Σ (1/k^2) converges, while the series Σ (2 + (-1)^{5k}) does not converge.

∫1∞ f(x) dx = ∫1∞ x2 dx = lim (t → ∞) [1/3 x3]1t = ∞
Since the integral diverges, the series น k2 sk also diverges.
b. To determine whether the series 2+(-1){ 5k converges, we can use the Alternating Series Test. The series has alternating signs and the absolute value of each term decreases as k increases. Let ak = 2+(-1){ 5k, then:
|ak| = 2+1/32k ≤ 2
Also, lim (k → ∞) ak = 0. Therefore, by the Alternating Series Test, the series 2+(-1){ 5k converges.

a. For the series Σ (1/k^2) (denoted as น k2 sk), we can use the p-series test. A p-series is a series of the form Σ (1/k^p), where p is a constant. If p > 1, the series converges, and if p ≤ 1, the series diverges. In this case, p = 2, which is greater than 1. Therefore, the series Σ (1/k^2) converges.
b. For the series Σ (2 + (-1)^{5k}), we can use the alternating series test. An alternating series is a series that alternates between positive and negative terms. In this case, the series alternates because of the (-1)^{5k} term. However, the series does not converge to zero as k goes to infinity, since there is a constant term 2. Therefore, the series Σ (2 + (-1)^{5k}) does not converge.
In summary, the series Σ (1/k^2) converges, while the series Σ (2 + (-1)^{5k}) does not converge.

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57 students most likely got heads between 45 and 60 times.

To determine the number of students who got heads between 45 and 60 times, we'll use the normal distribution properties. First, we need to calculate the z-scores for 45 and 60:

Z = (X - μ) / σ

For 45 heads:
Z1 = (45 - 50) / 5 = -1

For 60 heads:
Z2 = (60 - 50) / 5 = 2

Next, we need to find the probability that a student falls between these z-scores. We can do this by looking up the z-scores in a standard normal distribution table or using a calculator. The probabilities corresponding to these z-scores are:

P(Z1) = 0.1587
P(Z2) = 0.9772

Now, subtract P(Z1) from P(Z2) to get the probability of a student's result falling between 45 and 60 heads:

P(45 ≤ X ≤ 60) = P(Z2) - P(Z1) = 0.9772 - 0.1587 = 0.8185

Finally, multiply this probability by the total number of students (70) and round to the nearest whole number:

Number of students = 0.8185 * 70 ≈ 57

So, approximately 57 students most likely got heads between 45 and 60 times.

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The center of mass of the metal plate with density rho(x,y) = αx+ βy g/cm2 must lie on the line x + y = 7/6.

To find the center of mass of the metal plate, we need to calculate the coordinates of its centroid (X, Y). The coordinates of the centroid are given by:

X = (1/M) ∬(R) x ρ(x,y) dA, Y = (1/M) ∬(R) y ρ(x,y) dA

where M is the total mass of the plate, R is the region of integration (0 ≤ x ≤ 1, 0 ≤ y ≤ 1), and dA is the differential area element.

We can calculate the total mass M of the plate as follows:

M = ∬(R) ρ(x,y) dA = α/2 + β/2 = (α + β)/2

Using the given density function, we can calculate the integrals for X and Y:

X = (1/M) ∬(R) x ρ(x,y) dA = (2/αβ) ∬(R) x(αx+βy) dA = (2/3)(α+β)

Y = (1/M) ∬(R) y ρ(x,y) dA = (2/αβ) ∬(R) y(αx+βy) dA = (2/3)(α+β)

Thus, the coordinates of the centroid are (X, Y) = ((2/3)(α+β), (2/3)(α+β)).

Now, if we substitute X + Y = (4/3)(α+β) into the equation x + y = 7/6, we get:

x + y = 7/6

2x + 2y = 7/3

2(x+y) = 4/3(α+β)

x+y = (2/3)(α+β)

which shows that the centroid lies on the line x + y = 7/6. Therefore, the center of mass must also lie on this line.

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We know that the westbound car has traveled 24 miles.

When the northbound car has traveled 18 miles and the straight-line distance between the two cars is 30 miles, you can use the Pythagorean theorem to determine the distance the westbound car has traveled. The theorem states that a² + b² = c², where a and b are the legs of a right triangle and c is the hypotenuse.

In this case, the northbound car's distance (18 miles) represents one leg (a) and the westbound car's distance represents the other leg (b). The straight-line distance between the cars (30 miles) represents the hypotenuse (c). The equation can be set up as follows:

18² + b² = 30²

Solving for b:

324 + b² = 900
b² = 900 - 324
b² = 576
b = √576
b = 24

So, the westbound car has traveled 24 miles.

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To find the volume of the cone, we first need to find its radius. Since the pyramid is contained within the cone such that the vertices of the base of the pyramid are touching the edge of the cone, the diagonal of the square base of the pyramid is equal to the diameter of the base of the cone. The diagonal of the square base of the pyramid is:

d = √(10^2 + 10^2) = √200 = 10√2 cm

Therefore, the diameter of the base of the cone is 10√2 cm, and the radius is 5√2 cm.

The volume of the cone can be calculated using the formula:

V = (1/3)πr^2h

where r is the radius of the base of the cone and h is the height of the cone.

Substituting the given values, we get:

V = (1/3)π(5√2)^2(20)

V = (1/3)π(50)(20)

V = (1/3)(1000π)

V = 1000/3 * π

Using 3.14 as the decimal approximation for π, we get:

V ≈ 1046.35 cubic centimeters

Therefore, the volume of the cone is approximately 1046.35 cubic centimeters. The answer is A.

The compound interest generated on the investment is roughly $813.67, which is the solution to the question based on compound interest.

The initial sum of money invested or borrowed, upon which interest is based, is referred to as the principle. The principal is then periodically increased by the interest, often monthly or annually, to create a new principal sum that will accrue interest in the ensuing period.

Using the compound interest calculation, we can determine the interest earned on Sanchez's investment:

[tex]A = P(1 + \frac{r}{n} )^{(n*t)}[/tex]

where A is the overall sum, P denotes the principal (the initial investment), r denotes the yearly interest rate in decimal form, n denotes the frequency of compounding interest annually, and t denotes the number of years.

In this case, P = $3,000, r = 0.06 (6%), n = 365 (compounded daily),

and t = 4.

Plugging in the values, we get:

[tex]A = 3000(1 + \frac{0.06}{365} )^{(365*4)}[/tex]

A= $3813.67

The difference between the final amount and the principal is the interest earned.

Interest = A - P

Interest = $3813.67 - $3000

Interest = $813.67

As a result, the investment's interest yield is roughly $813.67.

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The probability that a student chosen randomly from the class passed the test or completed the homework is 20/27.

The probability that a student chosen randomly from the class passed the test or completed the homework is calculated as follows:

Let the probability that a student completed the homework be P(B).

Also, let the probability that a student passed the test be P(A)

P(A or B) = P(A) + P(B) - P(A * B)

From the data table:

The number of students who passed the test = 18

The number of students who completed the homework = 17

The number of students who both passed the test and completed the homework = 15.

Total number of students = 27

P(A) = 18/27

P(B) = 17/27

P(A*B) = 15/27

Therefore,

P(A or B) = 18/27 + 17/27 - 15/27

P(A or B) = 20/27

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When k = 0, both sides of the equation equal 0:
3cos(0) = 4(0)sin(0)
3 = 0

There are no other values of k for which the equation holds true, the only value of k that satisfies the given equation is k = 0.

To find the value(s) of k for which u(x, t) = e^(-3)sin(kt) satisfies the equation Ut = 4Uxx, we first need to calculate the partial derivatives with respect to t and x.
[tex]Ut = ∂u/∂t = -3ke^(-3)cos(kt)Uxx = ∂²u/∂x² = -k^2e^(-3)sin(kt)[/tex]
Now, we will substitute Ut and Uxx into the given equation:

[tex]-3ke^(-3)cos(kt) = 4(-k^2e^(-3)sin(kt))[/tex]
Divide both sides by e^(-3):

[tex]-3kcos(kt) = -4k^2sin(kt)[/tex]

Since we want to find the value(s) of k, we can divide both sides by -k:

3cos(kt) = 4ksin(kt)

Now we need to find the k value that satisfies this equation. Notice that when k = 0, both sides of the equation equal 0:

3cos(0) = 4(0)sin(0)
3 = 0

Since there are no other values of k for which the equation holds true, the only value of k that satisfies the given equation is k = 0.

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Answer:

$7.35

Step-by-step explanation:

We can solve the quadratic equation 2x² - 8x + 5 = 0 by using the quadratic formula, which states that for an equation of the form ax² + bx + c = 0, the solutions are given by:

x = (-b ± sqrt(b² - 4ac)) / 2a

In this case, a = 2, b = -8, and c = 5. Substituting these values into the formula, we get:

x = (-(-8) ± sqrt((-8)² - 4(2)(5))) / (2(2))

x = (8 ± sqrt(64 - 40)) / 4

x = (8 ± sqrt(24)) / 4

x = (8 ± 2sqrt(6)) / 4

Simplifying the expression by factoring out a common factor of 2 in the numerator and denominator, we get:

x = (2(4 ± sqrt(6))) / (2(2))

x = 4 ± sqrt(6)

Therefore, the solutions to the equation 2x² - 8x + 5 = 0 are:

x = 4 + sqrt(6) or x = 4 - sqrt(6)

Megha's total movement involves biking 20km north, 30km east, 20km south, and 30km west, resulting in a displacement of zero as she ends up back at her starting point.

Given that,

Megha bikes 20km north.

Megha then bikes 30km east.

After that, Megha bikes 20km south.

Lastly, Megha bikes 30km west.

Megha stops after completing the above movements.

Megha's displacement can be calculated by finding the straight-line distance between her starting point and ending point.

In this case,

She initially bikes 20km north, then 30km east, followed by 20km south, and finally 30km west.

Let's break it down:

The north and south distances cancel each other out, as she ends up back at her starting point vertically.

The east and west distances also cancel each other out, as she ends up back at her starting point horizontally.

Hence,

Megha's displacement is zero. She has returned to her original position.

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The congruent triangles is solved and the triangles are congruent by AAS postulate

Given data ,

Let the two triangles be represented as ΔRQT and ΔTQS

And , the side TQ is the common side of both the triangles

Now , the measure of ∠TQR ≅ measure of ∠TQS ( given )

And , the measure of ∠TRQ ≅ measure of ∠TSQ ( given )

So , Two angles are the same and a corresponding side is the same (ASA: angle, side, angle)

Hence , the triangles are congruent by ASA postulate

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The shows that Luther can expect the speed to be 51 miles/hour when he throws 80 pitches and 64 miles/hour when he throws 54 pitches.

The average speed of Luther's pitches when he throws 80 pitches will be:

y = -0.511x + 91.636

= -0.511(80) + 91.636

= 50.756

= 51

Also, the number of pitches that Luther can throw when the speed is 64 miles per hour will be:

y = -0.511x + 91.638

64 = -0.511x + 91.638

0.511x = 91.638 - 64

0.511x = 27.636

x = 54

Therefore, the number of pitches will be 54.

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Answer:

Because the absolute value of the correlation coefficient, 0.9672, is very close to 1, Luther can be very confident in the predictions.

Step-by-step explanation: Edmentum Answer

Answer:

D(6,4).

Step-by-step explanation:

The shape ABCD is a square.

By definition, the diagonals are equal.

The diagonal from A to C is 6 units long. Therefore, you should get your point D by drawing across from B to the right by 6.

D(6,4).

Answer:

#1        (176 - x)°

#2       m∠3 = m∠4 = 90°

Step-by-step explanation:

If a pair of parallel lines are cut by a transversal, there are several angles that are either equal to each other or are supplementary(angles add up to 180°).

For the specific questions...
For #1.

Angles ∠1 and ∠2 are supplementary angles since they are adjacent to each other and lie on the same straight line

Therefore
m∠1 + m∠2= 180°

Given m∠1 = (x + 4)° this becomes

(x + 4)° + m∠2 = 180°

m∠2 = 180° - (x + 4)°

= 180° - x° - 4°

= (176 - x)°

For #2

∠3 and ∠4 are supplementary angles so m∠3 + m∠4 = 180°

If m∠3 = m∠4 each of these angles must be half of 180°

So
m∠3 = m∠4 = 180/2 = 90°

If a time capsule has been buried 98m away from the cave at a bearing of 312. the time capsule is buried about 82.2 meters west of the cave.

To find how far west the time capsule is buried, we need to find the horizontal component of the displacement vector that points from the cave to the location of the time capsule. We can use trigonometry to do this:

cos(312°) = adjacent/hypotenuse

The hypotenuse is the distance between the cave and the time capsule, which is 98m. The adjacent side represents the horizontal distance between the two points, which is what we want to find. Rearranging the equation, we get:

adjacent = cos(312°) x hypotenuse

adjacent = cos(312°) x 98

adjacent ≈ 82.2

Therefore, the time capsule is buried about 82.2 meters west of the cave.

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The area of the shaded region is (9/500)π.

To find the area shaded below in circle K, we first need to find the radius of the circle.

Let O be the center of the circle, and let N be the midpoint of segment LM. We can draw a radius ON to segment LM such that it is perpendicular to LM, and then draw another radius OL to point L. This forms a right triangle LON with the hypotenuse equal to the radius of circle K.

Since segment LM is given to have a length of 11/9π, we can find the length of LN by dividing it in half:

LN = (11/9π)/2 = 11/18π

We can then use trigonometry to find the length of OL:

sin(55°) = OL / LN

OL = LN sin(55°)

OL = (11/18π) sin(55°)

Next, we can use the Pythagorean theorem to find the length of ON:

ON² = OL² + LN²

ON² = [(11/18π) sin(55°)]² + [11/18π]²

ON ≈ 1.022

Therefore, the radius of circle K is approximately 1.022.

The area of the shaded region can now be found by subtracting the area of sector LOM from the area of triangle LON:

Area of sector LOM = (110/360)π(1.022)² ≈ 0.317π

Area of triangle LON = (1/2)(11/18π)(1.022) ≈ 0.326π

Area of shaded region = (0.326π) - (0.317π) = (9/500)π

So the area of the shaded region is (9/500)π.

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a. The limit does not exist.
b. The limit is equal to 4.

a. To find the limit as x approaches 2, we need to evaluate the left-hand and right-hand limits separately and check if they are equal.
Left-hand limit: lim f(x) as x approaches 2 from the left
We have f(x) = x - 1 for x < 2. So, as x approaches 2 from the left, f(x) approaches 1.
Right-hand limit: lim f(x) as x approaches 2 from the right
We have f(x) = 1 for 2 ≤ x ≤ 6 and f(x) = x + 4 for x > 6. So, as x approaches 2 from the right, f(x) approaches 6.
Since the left-hand and right-hand limits are not equal, the limit as x approaches 2 does not exist.
b. To find the limit as x approaches 6, we need to evaluate the left-hand and right-hand limits separately and check if they are equal.
Left-hand limit: lim f(x) as x approaches 6 from the left
We have f(x) = 1 for 2 ≤ x ≤ 6 and f(x) = x + 4 for x > 6. So, as x approaches 6 from the left, f(x) approaches 1.
Right-hand limit: lim f(x) as x approaches 6 from the right
We have f(x) = x + 4 for x > 6. So, as x approaches 6 from the right, f(x) approaches 10.
Since the left-hand and right-hand limits are not equal, the limit as x approaches 6 does not exist.
Therefore, the correct choices are:
a. The limit is not -oo or co and does not exist.
b. lim = 4.

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According to the 68-95-99.7 rule, approximately 68% of the distribution falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations.

In this case, the mean is 7.06 minutes and the standard deviation is 0.75 minutes. Therefore, the range of times that covers the middle 95% of the distribution would be from the mean minus two standard deviations (7.06 - 2 x 0.75 = 5.56 minutes) to the mean plus two standard deviations (7.06 + 2 x 0.75 = 8.56 minutes).

In other words, 95% of the male college students' mile run times are expected to fall between 5.56 and 8.56 minutes. This means that most of the students' mile run times will be within this range, and only a small percentage will be outside of it.

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Answer:

the answer is 14,986 centimetres

Answer:

Markets behave the same way as individual customers because markets are made up of individual consumers.

Step-by-step explanation:

There were 490 walkers at the 10k fund-raiser.

The ratio of runners to walkers is 5:7, that means that the every five runners, there are 7 walkers so therefore we will use ratio formula.

If there have been 350 runners, we can use this ratio to discover what number of walkers there were:

5/7 = 350/x

Where x is the number of walkers.

To solve for x, we will cross-multiply:

5x = 7 * 350

5x = 2450

x = 490

Consequently, there were 490 walkers at the 10k fund-raiser.

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The Laplace transform of f(x)= (-2x-3)/4 is (-2L{x}-3L{1})/4, where L{x} is the Laplace transform of x and L{1} is the Laplace transform of 1.
Hi! The Laplace transform of a given function f(t) is denoted by L{f(t)} and is defined as the integral of f(t) multiplied by e^(-st), where s is a complex variable. For the function f(x) = (-2x - 3)/4, the Laplace transform can be calculated as follows:

L{f(t)} = L{(-2t - 3)/4}

To find the Laplace transform, we will treat the function as two separate parts:

L{(-2t - 3)/4} = (-2/4) * L{t} + (-3/4) * L{1}

The Laplace transforms of t and 1 are well-known:

L{t} = 1/s^2
L{1} = 1/s

Now, substitute these transforms back into our expression:

L{f(t)} = (-1/2) * (1/s^2) + (-3/4) * (1/s)

L{f(t)} = -1/(2s^2) - 3/(4s)

And that's the Laplace transform of f(x) = (-2x - 3)/4.

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Answer:

7.6 cm

Step-by-step explanation:

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

[tex]a^{2}[/tex] + [tex]6.5^{2}[/tex] = [tex]10^{2}[/tex]

[tex]a^{2}[/tex] + 42.25 = 100 Subtract 42.25 from both sides

[tex]a^{2}[/tex] = 57.57

[tex]\sqrt{\a^{a} }[/tex] = [tex]\sqrt{57.57}[/tex]

a ≈ 7.6

Helping in the name of Jesus.

Answer:

7.6 cm

Step-by-step explanation:

a^2+ b^2=c^2

a^2+6.5^2=10^2

a^2+42.25=100 subtract 42.25 from both sides

a^2=57.57

a=√57.57

a=7.6 cm

A score of at least 183 is required to stay out of the bottom 1.5%. To find the minimum score required to receive the award, we need to determine the z-score corresponding to the top 3% of students.

Since the distribution is normal, we can use the standard normal distribution table to find the z-score. From the table, we find that the z-score corresponding to the top 3% is approximately 1.88.

Therefore, we can use the formula z = (x - μ) / σ, where μ = 400 and σ = 100, to find the minimum score required: 1.88 = (x - 400) / 100

Solving for x, we get: x = 1.88(100) + 400 = 488. Therefore, a score of at least 488 is required to receive the award.

To find the minimum score required to stay out of the bottom 1.5%, we need to determine the z-score corresponding to the bottom 1.5%.

From the standard normal distribution table, we find that the z-score corresponding to the bottom 1.5% is approximately -2.17. Therefore, we can use the same formula as before to find the minimum score required: -2.17 = (x - 400) / 100.

Solving for x, we get: x = -2.17(100) + 400 = 183. Therefore, a score of at least 183 is required to stay out of the bottom 1.5%.

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Answer: 4.5

Step-by-step explanation:

a) Area (Ф(R)) = 5184 (b) Area (Ф(R)) = 25920

Let J be the Jacobian of Ф. We have J = det(DФ) = det([3 9; 9 9]) = -72.

(a) For R = [0,9] × [0,6], we have

Ф(R) = {(3u+9v,9u+9v) | 0 ≤ u ≤ 9, 0 ≤ v ≤ 6}.

The area of Ф(R) is given by the double integral over R of the Jacobian:

Area (Ф(R)) = ∬R |J| dudv

= ∫0^9 ∫0^6 72 dudv

= 5184.

Therefore, the area of Ф(R) is 5184.

(b) For R = [2,20] × [1,17], we have Ф(R) = {(3u+9v,9u+9v) | 2 ≤ u ≤ 20, 1 ≤ v ≤ 17}. The area of Ф(R) is given by the double integral over R of the Jacobian:

Area (Ф(R)) = ∬R |J| dudv = ∫2^20 ∫1^17 72 dudv = 25920.

Therefore, the area of Ф(R) is 25920.

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Answer: Yes those are correct good job

Step-by-step explanation: