The equation 8x − 2y = 25 represents a linear function. Which equation represents the same function?


A. The number of minutes m to cook c cups of rice

B. The volume V of a cube with side length s

C. The distance walked after m minutes at r feet per minute

D. The cost C for t tickets to a museum


HELP OR DIE

As per the given dimension, we need 343 small cubes to completely cover the larger cube.

To determine how many small cubes are needed to cover the larger cube, we need to think about how many of the smaller cubes can fit inside the larger cube.

We can start by looking at the dimensions of the larger cube. Each side is 7cm long, so the volume of the cube can be calculated by multiplying the length, width, and height:

7cm x 7cm x 7cm = 343 cubic centimeters

Now let's consider the dimensions of the smaller cubes. Each cube is 1cm x 1cm x 1cm, so the volume of each cube is:

1cm x 1cm x 1cm = 1 cubic centimeter

To determine how many of these smaller cubes are needed to cover the larger cube, we need to divide the volume of the larger cube by the volume of each small cube:

343 cubic centimeters ÷ 1 cubic centimeter = 343

So we need 343 small cubes to completely cover the larger cube.

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Complete Question:

How many cubes of dimensions 1cm*1cm*1cm are required to cover a cube of dimensions 7cm*7cm*7cm?

1 marble is equal in weight to 84 jujubes.

Let's start by writing down the given information in equations:

5m = 3s (where m represents one marble and s represents one toy soldier)

7s = 5c (where c represents one plush chipmunk)

3c = 14j (where j represents one jujube)

We want to find out how many jujubes equal the weight of one marble, so we need to eliminate all the other variables except for j and m. We can do this by using substitution and algebraic manipulation.

First, we can solve the second equation for s in terms of c:

7s = 5c

s = (5/7)c

Then, we can substitute this expression for s in the first equation:

5m = 3s

5m = 3(5/7)c

m = (3/7)c

Next, we can solve the third equation for c in terms of j:

3c = 14j

c = (14/3)j

Now we can substitute this expression for c in the previous equation:

m = (3/7)c

m = (3/7)(14/3)j

m = 2j

So we have found that one marble is equal in weight to 2 jujubes. But the question asks for the weight of one marble in terms of jujubes, not in terms of jujubes and toy soldiers and plush chipmunks. We can use the other equations to eliminate the other variables:

5m = 3s

5m = 3(5/7)c

5m = (15/7)c

m = (3/7)c

7s = 5c

7s = 5(14/3)j

s = (10/3)j

Putting this all together:

m = (3/7)c

m = (3/7)(7s/5)

m = (3/5)s

m = (3/5)(10/3)j

m = 2j

So we have found that one marble is equal in weight to 2 jujubes. Finally, we can use the third equation to find how many jujubes are equal in weight to 1 marble:

3c = 14j

c = (14/3)j

m = (3/7)c

m = (3/7)(14/3)j

m = 2j

1 marble = 2 jujubes

1 jujube = 1/2 marble

1 marble = 2 jujubes = 2(84) = 168 jujubes

Therefore, one marble is equal in weight to 84 jujubes.

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The closest option to this value is d) 10.75, but none of the options is an exact match.

To find the interquartile range (IQR) of the data, we need to first find the first quartile (Q1) and the third quartile (Q3).

To do this, we can arrange the data in order from smallest to largest:

98, 100, 104, 105, 106, 110, 112, 115, 116, 118, 120, 122, 126, 130, 136, 140

The median of the data is the average of the two middle values, which are 112 and 115. So, the median is (112 + 115) / 2 = 113.5.

To find Q1, we need to find the median of the data values below the median. These are:

98, 100, 104, 105, 106, 110, 112, 115

The median of these values is (106 + 110) / 2 = 108.

To find Q3, we need to find the median of the data values above the median. These are:

116, 118, 120, 122, 126, 130, 136, 140

The median of these values is (122 + 126) / 2 = 124.

Now we can calculate the interquartile range (IQR) as the difference between Q3 and Q1:

IQR = Q3 - Q1 = 124 - 108 = 16.

Therefore, the interquartile range of the data is 16, or in decimals 16.00.

The closest option to this value is d) 10.75, but none of the options is an exact match.

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Answer:

1/169

Step-by-step explanation:

Answer:

1

Step-by-step explanation:

part a.

the percentage of eggs between 42 and 45mm is 48.48%

part b.

The median width is approximately (42+45)/2 = 43.5mm.

The median length is approximately (56+59)/2 = 57.5mm.

part c.

The width of grade A chicken eggs has a range of about 24mm.

part d.

I think its impossible to determine because we don't have the value for the standard deviation.

The second option should be correct.

A histogram is described as an approximate representation of the distribution of numerical data.

part a.

From the histogram, we see that the frequency for the bin that ranges from 42 to 45mm is 4 and we have  a total of 33 eggwe use this values and calculate the percentage of eggs between 42 and 45mm is 48.48%.

part b.

we have an  estimation  that the median of the width is 48mm and the median of the length is around 60mm.

part c.

Also from the histogram, we notice  that the smallest value is around 36mm and the largest value is around 66mm, hence the width of grade A chicken eggs has a range of about 24mm.

In a histogram, the range is the width that the bars cover along the x-axis and these are approximate values because histograms display bin values rather than raw data values.

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The cost for one donut is $0.32, so the cost for one dozen donuts is:

12 donuts x $0.32/donut = $3.84

The cost for the cardboard box is $0.18 per square foot of cardboard, and there are 144 square inches in 1 square foot, so the cost per square inch of cardboard is:

$0.18 / 144 sq in = $0.00125/sq in

If t represents the total surface area of the box in square inches, then the cost of the box is:

t x $0.00125/sq in

To convert square inches to square feet, we divide by 144:

t/144 square feet x $0.18/square foot = t x $0.00125/sq in

Thus, the expression to model the total cost for one dozen donuts where t represents the total surface area of the box in square feet is:

$3.84 + (t/144) x $0.18

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The function that could be described is f(x) = 10cos(2πx/3), where the amplitude is 10, the period is 3, and the maximum value is 20.

In a cosine function, the amplitude represents the vertical distance from the midline to the maximum or minimum value. Here, the maximum value is 20, which means the amplitude is half of that, i.e., 10. The period of the function is the distance it takes for one complete cycle, and in this case, it is 3 units.

By using the formula f(x) = A*cos(2πx/P), where A is the amplitude and P is the period, we can determine that the given function matches the described characteristics.

The function f(x) = 10cos(2πx/3) has a maximum value of 20 and a minimum value of 0, and it completes one cycle over the interval of the period, which is 3 units.

In conclusion, the function f(x) = 10cos(2πx/3) satisfies all the given conditions and represents the described function.

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The function has a local maximum at x = -6.

To determine the intervals on which the function f(x) = 4x³ - 12x² - 576x + 6 is increasing or decreasing, we first find its derivative, f'(x), and then analyze its critical points.

f'(x) = 12x² - 24x - 576

Now, set f'(x) = 0 and solve for x:

12x² - 24x - 576 = 0

Divide by 12:
x² - 2x - 48 = 0

Factor:
(x - 8)(x + 6) = 0

So, the critical points are x = 8 and x = -6.

Analyze the intervals:
f'(-7) > 0, so increasing on (-∞, -6)
f'(0) < 0, so decreasing on (-6, 8)
f'(9) > 0, so increasing on (8, ∞)

The function f(x) is decreasing on the interval (-6, 8). It is increasing on the interval (-∞, -6) and the interval (8, ∞). The function has a local maximum at x = -6.

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Answer: Let's use the Pythagorean theorem to solve this problem.

Let x be the common factor of the ratio 3:4, so the sides of the rectangle are 3x and 4x.

The Pythagorean theorem states that for any right triangle, the sum of the squares of the two shorter sides is equal to the square of the length of the hypotenuse (the longest side).

So, for the rectangle with sides 3x and 4x, we have:

(3x)^2 + (4x)^2 = (diagonal)^2

9x^2 + 16x^2 = 100

25x^2 = 100

x^2 = 4

Taking the square root of both sides, we get:

x = 2

Therefore, the sides of the rectangle are:

3x = 3(2) = 6 cm

4x = 4(2) = 8 cm

So, the length and width of the rectangle are 6 cm and 8 cm, respectively.

The total surface area of the foam pit can be calculated by finding the area of each face and adding them together.

Since the pit is a rectangular prism, it has six faces: the top, bottom, front, back, left, and right. The area of each face can be calculated using the formula for the area of a rectangle, which is length times width.

To calculate the total surface area of a rectangular prism with an open top, we need to add the areas of all six faces together.

The area of each face can be calculated using the formula for the area of a rectangle (length times width).

The top of the foam pit is open, so we don't need to include it in our calculation.

After finding the area of each face, we simply add them all together to get the total surface area.

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The position of particle in a straight line with v = t(t^2 + 1)³ + 3t is (1/8)t⁸ + (3/6)t⁶ + (3/4)t⁴ + 2t² C.

To find an expression for the position s after a time t, we need to integrate the velocity function v with respect to time t.

Using the power rule of integration and the constant of integration C, we have:

s = ∫v dt = ∫[t(t² + 1)³ + 3t] dt

after expanding t(t² + 1)³ using binomial theorem we have-

(t^2 + 1)³ = t⁶ + 3t⁴ + 3t² + 1

Substituting this into the integral, we get:

s = ∫[t(t⁶ + 3t⁴ + 3t^2 + 1) + 3t] dt

s = ∫[t^7 + 3t⁵ + 3t³ + t + 3t] dt

s = ∫t^7 dt + 3∫t⁵ dt + 3∫t³ dt + ∫4t dt

s = (1/8)t⁸ + (3/6)t⁶ + (3/4)t⁴ + 2t² + C

Therefore, the expression for the position s after a time t is:

S = (1/8)t⁸ + (3/6)t⁶ + (3/4)t⁴ + 2t² + C, where C is the constant of integration.

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Answer: 1,004.8 or 320[tex]\pi[/tex]

Step-by-step explanation:

[tex]\frac{1}{3} \pi 8^{2} 15=1,004.8[/tex]

The domain of the function is 0 ≤ b ≤ 6, and the range of the function is 30 ≤ c ≤ 174.

The function that models the total weight of a crate with b boxes inside is given as:

c = 24b + 30

We know that each crate can hold a maximum of 6 boxes. Therefore, the number of boxes inside the crate can only take values from 0 to 6.

Domain:

The number of boxes b can take values from 0 to 6. Therefore, the domain of the function is:

0 ≤ b ≤ 6

Range:

To find the range of the function, we need to consider the maximum and minimum values that c can take when

0 ≤ b ≤ 6.

When b = 0, the crate is empty, and the total weight of the crate is:

c = 24(0) + 30 = 30.

When b = 6, the crate is full with 6 boxes, and the total weight of the crate is:

c = 24(6) + 30 = 174.

Therefore, the range of the function is:

30 ≤ c ≤ 174

We can then say the domain of the function is 0 ≤ b ≤ 6, and the range of the function is 30 ≤ c ≤ 174.

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The completed two column table in the question showing that the measure of the angle m∠GKB = 120° can be presented as follows;

Statement [tex]{}[/tex]                                                Reason

[tex]\overline{AB}[/tex] || [tex]\overline{EF}[/tex]         [tex]{}[/tex]                                          Given

m∠ELJ = 120°

m∠ELJ + m∠ELK = 180°     [tex]{}[/tex]                        Linear pair Postulate

m∠BKL + m∠GKB = 180°   [tex]{}[/tex]                        Linear pair Postulate

m∠ELJ + m∠ELK = mBKL + m∠GKB  [tex]{}[/tex]      Transitive property

∠ELK ≅ ∠BKL                         [tex]{}[/tex]                   1. Alternate Interior Angles

m∠ELK = m∠BKL [tex]{}[/tex]                                      2. Definition of congruent angles

m∠ELJ + m∠ELK = m∠ELK + m∠GKB[tex]{}[/tex]      Substitution property

m∠ELJ = m∠GKB[tex]{}[/tex]                                      Subtraction property

m∠GKB = m∠ELJ [tex]{}[/tex]                                     Symmetric property

m∠GKB = 120° [tex]{}[/tex]                                          Substitution

An angle is the figure formed at the point of intersection of two rays that have the same starting point. The parts of an angle includes; The vertex, which is the point of intersection of the rays, and the sides or arms of the angle, which are the two rays forming the angle.

The details of the the statements that completes the above table used to prove the measure of the angle m∠GKB = 120° are as follows;

Alternate interior angles theorem

The alternate interior angles theorem states that the alternate interior angle formed by the two parallel lines and their shared transversal are congruent.

Definition of congruent angles

Congruent angles are angles that have the same measure.

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The amount that Benjamin must save every month to pay off the discounted premium is $ 40. 80

The total premium for the year would be $ 637. 20

The amount that Benjamin's discounted premium would come to for the year is:

= 1, 080 x ( 1 - 66 %)

= $ 367. 20

The amount he would need to save every month on deployment is :

= 367. 20 / 9

= $ 40. 80

His total premium would be :

= 367. 20 + ( 1, 080 / 12 x 3 months when he comes back )

= $ 637. 20

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The total weight of the cards inside the box is approximately 217.6 oz.

Each deck of cards weighs 3.4 oz, and there are 64 decks of cards in the box. Therefore, the total weight of the cards inside the box is 3.4 oz/deck x 64 decks = 217.6 oz. As the answer needs to be rounded to the nearest ounce, we round 217.6 to the nearest ounce, which gives us 218 oz.

However, the question asks for the weight of the cards, which is only accurate to one decimal place. Therefore, we round 217.6 to one decimal place, which gives us 217.6 oz. Hence, the total weight of the cards inside the box is approximately 217.6 oz.

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The number of days 3 3/4 bags of tortillas will last is 5 days.

To solve this problem, we need to use the concept of fractions. We know that Lucy's burrito stand uses 3/4 of a bag of tortillas every day. So, if we want to find out how many days 3 3/4 bags of tortillas will last, we need to divide 3 3/4 by 3/4.

To do this, we can convert 3 3/4 to an improper fraction, which is 15/4. Then, we can divide 15/4 by 3/4 using the following steps:

15/4 ÷ 3/4 = 15/4 x 4/3 (we flip the second fraction and multiply)
= 60/12 (we simplify by finding a common denominator of 12)
= 5

Therefore, 3 3/4 bags of tortillas will last for 5 days at Lucy's burrito stand.

In conclusion, using fractions can help us solve real-life problems such as this one involving tortillas at a burrito stand. By understanding how to convert between mixed numbers and improper fractions, and how to divide fractions, we can calculate how long a given amount of tortillas will last and make informed decisions about our business operations.

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Therefore, the probability that Jack wins the race and Martina finishes last is 1/6 or approximately 0.167.

There are 3 people racing, so there are 3! = 6 possible ways the race can end (assuming no ties).

These are:

Jack, Martina, Napier

Jack, Napier, Martina

Martina, Jack, Napier

Martina, Napier, Jack

Napier, Jack, Martina

Napier, Martina, Jack

Of these 6 outcomes, there is only 1 where Jack wins the race and Martina finishes last: outcome 2.

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Answer:

(5,725)1.045^5

Step-by-step explanation:

(5,725)1.045^5

5,725 is the original amt of $

1.045 is the % of interest

5 is the # of years

Solve this and round the nearest

cent.

Therefore, it is possible for Diego to draw D 19 times, O 10 times, and G 1 time when picking 30 letters from the bag with replacement, although it is highly unlikely.

Probability is a measure of the likelihood or chance of an event occurring. It is expressed as a number between 0 and 1, with 0 indicating that the event is impossible, and 1 indicating that the event is certain to occur.

Here,

Yes, it is possible for Diego to draw D 19 times, O 10 times, and G 1 time when picking 30 letters from the bag with replacement.

The probability of drawing the letter D on one pick is 1/3, since there is 1 D out of 3 letters in the bag. Similarly, the probability of drawing the letter O on one pick is also 1/3, and the probability of drawing the letter G on one pick is 1/3.

Since Diego replaces each letter he picks, the probability of drawing D 19 times in a row is (1/3)¹⁹, the probability of drawing O 10 times in a row is (1/3)¹⁰, and the probability of drawing G 1 time is 1/3.

The probability of all these events happening in this order is the product of their individual probabilities, which is:

(1/3)¹⁹ * (1/3)¹⁰ * 1/3 = (1/3)³⁰

This probability is very small, but it is still greater than zero.

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a) There are 1540 ways that the teacher can select the three students.

b) The probability that all three students selected are girls is approximately 0.176 or 17.6%.

c) The probability that at least one boy is selected is approximately 0.824 or 82.4%.

a)

To find the number of ways the teacher can select three students out of 22 students (7 boys and 15 girls), we can use the combination formula. The number of ways to select r items from a set of n items is given by:

nCr = n! / (r! * (n-r)!)

where n! represents the factorial of n (i.e., n! = n x (n-1) x (n-2) x ... x 3 x 2 x 1), and r! represents the factorial of r. Applying this formula, we get:

22C3 = 22! / (3! * (22-3)!) = 22! / (3! * 19!) = (22 x 21 x 20) / (3 x 2 x 1) = 1540

Therefore, there are 1540 ways that the teacher can select the three students.

b)

To find the probability that all three students selected are girls, we can use the formula for the probability of an event occurring. Since there are 15 girls and 7 boys, the probability of selecting a girl is 15/22 for the first selection, 14/21 for the second selection (since there are now 14 girls left out of 21 remaining students), and 13/20 for the third selection. Applying the formula, we get:

P(all three are girls) = (15/22) x (14/21) x (13/20) ≈ 0.176

Therefore, the probability that all three students selected are girls is approximately 0.176 or 17.6%.

c)

To find the probability that at least one boy is selected, we can use the complement rule. The complement of selecting at least one boy is selecting all three girls, which we calculated in part (b) to be approximately 0.176. Therefore, the probability of selecting at least one boy is:

P(at least one boy) = 1 - P(all three are girls) ≈ 1 - 0.176 ≈ 0.824

Therefore, the probability that at least one boy is selected is approximately 0.824 or 82.4%.

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The value of y that satisfies the equation [tex]\int x^5 dx = \int x^y dx[/tex] is y = -1.

We know that the indefinite integral of x^5 dx is (1/6) x^6 + C, where C is

the constant of integration. Therefore:

[tex]\int x^5 dx = (1/6) x^6 + C[/tex]

We want to find y such that [tex]\int x^5 dx = \int x^y dx[/tex]. Using the power rule of integration, the indefinite integral of [tex]x^y[/tex] dx is [tex](1/(y+1)) x^{(y+1)} + C[/tex], where C is the constant of integration. Therefore:

[tex]\int x^y dx = (1/(y+1)) x^{(y+1)} + C[/tex]

For these two integrals to be equal, we need:

[tex](1/6) x^6 + C = (1/(y+1)) x^{(y+1) } + C[/tex]

Subtracting C from both sides, we get:

[tex](1/6) x^6 = (1/(y+1)) x^{(y+1)}[/tex]

Multiplying both sides by (y+1), we get:

[tex](1/6) x^6 (y+1) = x^{(y+1)}[/tex]

Now, we can equate the powers of x on both sides:

[tex]x^6 (y+1) = x^{(y+1)}[/tex]

Using the fact that[tex]x^a \times x^b = x^{(a+b)}[/tex], we can simplify the left-hand side:

[tex]x^(6(y+1)) = x^{(y+1)}[/tex]

Now, we can equate the exponents on both sides:

6(y+1) = y+1

Simplifying, we get:

6y + 6 = y + 1

5y = -5

y = -1

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The original price of the television was $240.

Let's denote the original price of the television by "x".

From the first sentence, the website applies a 4.5% processing fee and charges $6.00 for shipping. Therefore, the cost of the television with these fees is:

x + 0.045x + 6.00 = 1.045x + 6.00

From the second sentence, the customer uses a coupon that takes 15% off of the final price. Therefore, the price after the discount is:

0.85(1.045x + 6.00) = 0.88825x + 5.10

The problem states that the customer paid $218.28 for the order. Therefore, we can set up the following equation:

0.88825x + 5.10 = 218.28

Solving for x, we get:

0.88825x = 213.18

x = 240

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We can decompose the given rational function as follows:

f(x) = (16x^3 + 12x^2 + 10x + 2) / [(x^4 – 4x^2)(x^2 + x + 1)^2(x^2 – 3x + 2)(x^4 + 3x^2 + 2)]

To find the partial fraction decomposition, we first factor the denominator completely:

x^4 – 4x^2 = x^2(x^2 – 4) = x^2(x – 2)(x + 2)

x^2 + x + 1 = (x + 1/2)^2 + 3/4

x^2 – 3x + 2 = (x – 1)(x – 2)

x^4 + 3x^2 + 2 = (x^2 + 1)(x^2 + 2)

Substituting these factorizations into the denominator, we get:

f(x) = (16x^3 + 12x^2 + 10x + 2) / [x^2(x – 2)(x + 2)(x + 1/2)^2(3/4)^2(x – 1)(x – 2)(x^2 + 1)(x^2 + 2)]

We can now write the partial fraction decomposition as:

f(x) = A/x + Bx + C/(x – 2) + D/(x + 2) + E/(x + 1/2) + F/(x + 1/2)^2 + G/(x – 1) + H/(x^2 + 1) + I/(x^2 + 2)

where A, B, C, D, E, F, G, H, and I are constants to be determined.

Note that the term E/(x + 1/2) has a repeated linear factor (x + 1/2)^2, so we need to include a second term F/(x + 1/2)^2 in the decomposition.

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The equation of the sinusoidal function is y = 5sin(x).

To solve this, we need to find the equation of the sinusoidal function that has a maximum point at (0,5) and a minimum point at (2π,-5).

First, we know that the function is a sine function because it has a maximum at (0,5) and a minimum at (2π,-5).

Second, we can find the amplitude of the function by taking half the difference between the maximum and minimum values. In this case, the amplitude is (5-(-5))/2 = 5.

Third, we can find the vertical shift of the function by taking the average of the maximum and minimum values. In this case, the vertical shift is (5+(-5))/2 = 0.

Finally, we can find the period of the function by using the formula T=2π/b, where b is the coefficient of x in the equation of the function. In this case, we know that the function completes one cycle from x=0 to x=2π, so the period is 2π.

Putting it all together, the equation of the function is y = 5sin(x)

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Answer:

48.91

Step-by-step explanation:

r=cos^-1(.92)

r=23.07

cos(23.07)=45/y

y=45/cos(23.07)

48.91

The rate of increase, we can see that the function is an exponential growth model with a base of 1.04, which means that the population increases by 4% each year.

There seems to be an error in the problem statement. If the function P(x) = 88,200(1.04)^x models the population after the year 2002, then it doesn't make sense to ask for the population in the year 2000, which is two years before 2002.

Assuming that the function is correctly stated and represents the population after 2002, we can find the population after a certain number of years by plugging that number into the function. For example, to find the population after 5 years (in 2007), we would use:

P(5) = 88,200(1.04)^5 = 105,159.43

This means that the population of the city in 2007 would be approximately 105,159 people.

As for the rate of increase, we can see that the function is an exponential growth model with a base of 1.04, which means that the population increases by 4% each year.

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The formula that models the ratio y is:

y = 6 / (6 + x)

Let y be the ratio of the number of gallons of acid in the container compared to the total volume of liquid in the container, and let x be the number of gallons of water added to the container.

Initially, the container has 6 gallons of acid and 0 gallons of water, for a total volume of 6 gallons. When x gallons of water is added, the total volume of liquid becomes 6 + x gallons, and the amount of acid remains at 6 gallons.

Therefore, the formula that models the ratio y is:

y = 6 / (6 + x)

This formula gives the ratio of the number of gallons of acid in the container compared to the total volume of liquid in the container when x gallons of water is added.

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a. The tube dilution in tubes 1, 3, and 5:
- Tube 1: 1:5 (1 ml serum + 4 ml saline)
- Tube 3: 1:125 (1:5 dilution from Tube 1 x 1:5 dilution from Tube 2 x 1:5 dilution from Tube 3)
- Tube 5: 1:3125 (1:125 dilution from Tube 3 x 1:5 dilution from Tube 4 x 1:5 dilution from Tube 5)

b. The solution dilution in tubes 1, 2, and 7:
- Tube 1: 1:5
- Tube 2: 1:25 (1:5 dilution from Tube 1 x 1:5 dilution from Tube 2)
- Tube 7: 1:78125 (1:3125 dilution from Tube 5 x 1:5 dilutions for Tubes 6 and 7)

c. The total volume and solution dilution in tube 10 before transfer:
- Total volume: 3 ml (2 ml saline + 1 ml transferred from Tube 9)
- Solution dilution: 1:1953125 (1:78125 dilution from Tube 7 x 1:5 dilutions for Tubes 8, 9, and 10)

d. The amount or volume of serum in tube 6 before transfer and after transfer:
- Before transfer: 0.00064 ml (2 ml x 1:3125 dilution from Tube 5)
- After transfer: 0.00032 ml (1 ml x 1:3125 dilution from Tube 5, as half the volume was transferred to Tube 7)

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