Evaluate the given expression. Compare the result to the 5th row of Pascal's triangle. 4 (b) (c) 2 4 (d) (e) 3

Thus, it is important to interpret the F-ratio in the context of the research question and to consider other factors, such as effect size and practical significance, when drawing conclusions from an ANOVA analysis.

When conducting an analysis of variance (ANOVA), we are interested in comparing the means of two or more groups.

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let s = {−1, 0, 2, 4, 7}. f(s) = {1}, f(s) = {-1, 1, 5, 9, 15} and f(s) = {0, 1, 2}.

Given the set s = {−1, 0, 2, 4, 7}, I will find f(s) for each of the provided functions:

a) f(x) = 1
For every x in s, f(x) is always 1.

b) f(x) = 2x + 1
Applying this function to each element of s:
f(-1) = -1
f(0) = 1
f(2) = 5
f(4) = 9
f(7) = 15


c) f(x) = ⌈x/5⌉ (the ceiling function)
Applying this function to each element of s:
f(-1) = 0
f(0) = 0
f(2) = 1
f(4) = 1
f(7) = 2


d) f(x) = ⌊((x^2 + 1))/3⌋ (the floor function)
Applying this function to each element of s:
f(-1) = 0
f(0) = 0
f(2) = 1
f(4) = 5
f(7) = 16
So, f(s) = {0, 1, 5, 16}.

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Answer:

c

Step-by-step explanation:

volume value multiplyer = 1

The metric system is based on units of 10, with the liter (L) as the basic unit of volume. One liter is equal to 1,000 milliliters (mL). The answer is d. none of the above.

Therefore, option a is incorrect because 1 liter is equal to 1,000 mL, which is 100 times more than 10 mL.

Option b is also incorrect because 1 deciliter (dL) is equal to 100 mL, so 10 dL is equal to 1,000 mL, which is again 100 times more than 10 mL.

Option c is incorrect because 1 cubic centimeter (cm³) is equal to 1 milliliter (mL), so 10 cm³ is equal to 10 mL.

However, this does not mean that 10 mL is equal to 10 cm³, as they are simply two different ways of expressing the same volume.

Therefore, the correct answer is d.

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The base and the exponent in the expression √5 are base = 5 and exponent = 1/2

From the question, we have the following parameters that can be used in our computation:

√5

The above expression is a square root expression

This expression can be expressed using the base-exponent format

To do this, we apply the law of indices

Applying the law of indices, we have

√5 = 5^1/2

As a general rule, we have

base^exponent

So, we have

base = 5 and exponent = 1/2

Hence, the base is 5 and theexponent is 1/2

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The fraction in lowest terms is 2/3. Option C

First, we need to know that a fraction is described as the part of a whole variable, a whole number or a whole element.

The different  fractions in mathematics are listed thus;

From the information given, we have that;

a+1/b

Such that a = 5 and b = 9

Now, substitute the values, we get;

5 + 1/9

Add the values of the numerator

6/9

Divide the values

2/3

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Step-by-step explanation:

a number to a negative power is the same as 1/(base).

The volume of the blue triangular prism is 216 cubic centimeters.

We assume that the two figures are similar, if the scale factor between the two is K, then the volume of the blue prism will be K³ times the volume of the green one.

To find the value of K, we can compare the two known sides:

4cm*K = 6cm

K = 6cm/4cm = 3/2

Then the volume of the blue prism is:

(3/2)³*64cm³ = 216 cm³

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The radius of the right circular cylinder is [tex]$\boxed{\sqrt{2}}$[/tex] inches.

Let's start by recalling the formulas for the lateral surface area and volume of a right circular cylinder.

The lateral surface area of a right circular cylinder with radius r and height h is given by:

[tex]$L = 2\pi rh$[/tex]

The volume of a right circular cylinder with radius r and height h is given by:

[tex]$V = \pi r^2h$[/tex]

We are given that the lateral surface area of the cylinder is [tex]$24\pi$[/tex]square inches, and the volume is[tex]$24\pi$[/tex] cubic inches. Therefore, we have:

[tex]$2\pi rh = 24\pi$[/tex] (1)

[tex]$\pi r^2h = 24\pi$[/tex] (2)

We can solve for h from equation (1):

[tex]$2\pi rh = 24\pi$[/tex]

[tex]$h = \frac{24}{2\pi r}$[/tex]

Substituting this value of h into equation (2), we get:

[tex]$\pi r^2 \left(\frac{24}{2\pi r}\right) = 24\pi$[/tex]

Simplifying this equation, we get:

[tex]$r = \sqrt{2}$[/tex]

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In the simple linear regression model, the residual term accounts for the variability in the dependent variable that cannot be explained by the linear relationship between the variables.

In a simple linear regression model, we assume that there is a linear relationship between the dependent variable and the independent variable.

The residual term is the difference between the actual value of the dependent variable and the predicted value based on the regression line. It represents the part of the variation in the dependent variable that is not accounted for by the linear relationship with the independent variable.

The residual term is also referred to as the error term, and its sum of squares is used to estimate the variability of the dependent variable around the regression line. A lower sum of squared residuals indicates a better fit of the regression line to the data.

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For nonnormal populations, as the sample size (n) increases, the distribution of sample means approaches a normal distribution.

This is known as the central limit theorem. The central limit theorem states that as the sample size increases, the distribution of sample means becomes more and more normal, even if the original population is not normal. The central limit theorem is a fundamental concept in statistics, as it allows us to use the normal distribution as an approximation for many statistical analyses, regardless of the underlying distribution of the population.

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Hi! To find the primitive function of 4x, you need to determine the antiderivative of the given function.

The primitive function of 4x is:

∫(4x) dx = 4∫(x) dx = 4(x^2/2) + C

So the primitive function of 4x is [tex]2x^2 + C[/tex], where C is the constant of integration.

This  queue does generate the correct steady state distribution.

To determine the arrival rate of the queue, we need to use the balance equations, which state that the arrival rate λ must be equal to the departure rate μ.

In this case, the departure rate is given by the service rate s multiplied by the probability of there being at least one customer in the system, which is 1 - P(n=0):

μ = s(1 - P(n=0))

Using the binomial distribution, we can calculate the probability of there being no customers in the system:

P(n=0) = (1-p)^n = (1-p)^0 = 1

So the departure rate simplifies to:

μ = s(1 - P(n=0)) = s(1 - 1) = 0

This means that there are no customers leaving the system, and therefore no departure rate to balance against. To find the arrival rate, we need to consider the fact that the system is in a steady state, meaning that the number of arrivals must equal the number of departures.

Since there are no departures, the number of arrivals must also be zero. This implies that the arrival rate λ is also zero.

We can confirm that this queue generates the correct steady state distribution by verifying that the probability mass function (pmf) of the number of customers in the system matches the given (n,p) binomial distribution.

The probability of there being n customers in the system at steady state is given by:

P(n) = P(n arrivals before the first departure) x P(no arrivals during service) x P(n-1 arrivals before the second departure) x P(no arrivals during service) x ...

This simplifies to:

P(n) = λ/(s+λ) x (1-p)^n x λ/(s+λ) x (1-p)^(n-1) x λ/(s+λ) x (1-p)^(n-2) x ...

which can be written as:

P(n) = (λ/(s+λ))^n x (1-p)^(n(n-1)/2)

Comparing this with the (n,p) binomial distribution, we can see that the pmf matches if:

λ/(s+λ) = p

Taking the limit as n approaches infinity, we can see that the steady state distribution converges to the (n,p) binomial distribution. Therefore, this queue does generate the correct steady state distribution.

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Thus, the shaded region is the set of solutions for this system of inequalities, and the integers in this region are -4, -3, -2, -1, 0, 1, 2, and 3.

To solve the system of inequalities by graphing, we first need to rewrite each inequality in slope-intercept form, y < mx + b, where y is the dependent variable (in this case, we can use y to represent both 3-2a and 5a), m is the slope, x is the independent variable (in this case, a), and b is the y-intercept.

Starting with the first inequality, 3-2a < 13, we can subtract 3 from both sides to get -2a < 10, and then divide both sides by -2 to get a > -5. So the slope is negative 2 and the y-intercept is 3. We can graph this as a dotted line with a shading to the right, since a is greater than -5:

y < -2a + 3

Next, we can rewrite the second inequality, 5a < 17, by dividing both sides by 5 to get a < 3.4. So the slope is 5/1 (or just 5) and the y-intercept is 0. We can graph this as a dotted line with a shading to the left, since a is less than 3.4:

y < 5a

To find the integers that are in the set of solutions for this system of inequalities, we need to look for the values of a that satisfy both inequalities. From the first inequality, we know that a must be greater than -5, but from the second inequality, we know that a must be less than 3.4. So the integers that are in the set of solutions are the integers between -4 and 3 (inclusive):

-4, -3, -2, -1, 0, 1, 2, 3

To see this graphically, we can shade the region that satisfies both inequalities:

y < -2a + 3 and y < 5a

The shaded region is the set of solutions for this system of inequalities, and the integers in this region are -4, -3, -2, -1, 0, 1, 2, and 3.

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The probability can be found using the formula for a continuous uniform distribution: P(50.2 ≤ X ≤ 50.7) = (50.7 - 50.2) / (52.0 - 50.0).

P(50.2 ≤ X ≤ 50.7) = 0.5
Therefore, the probability that the class length is between 50.2 and 50.7 min is 0.5. This means that the likelihood of selecting a class length within this range is 50%, which is relatively high given the range of possible lengths.

The lengths of the professor's classes follow a continuous uniform distribution between 50.0 minutes and 52.0 minutes. To find the probability that the class length is between 50.2 and 50.7 minutes, we can use the following formula for a continuous uniform distribution:

P(a ≤ X ≤ b) = (b - a) / (B - A)

Here, A = 50.0 min (lower bound), B = 52.0 min (upper bound), a = 50.2 min, and b = 50.7 min.

Plugging in the values:

P(50.2 ≤ X ≤ 50.7) = (50.7 - 50.2) / (52.0 - 50.0)
P(50.2 ≤ X ≤ 50.7) = 0.5 / 2
P(50.2 ≤ X ≤ 50.7) = 0.25, So, the probability that the class length is between 50.2 and 50.7 minutes is 0.25 or 25%.

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Answer:

The awnser is this, very real, dond fake 122

Answer:

Charlie's diagram: 15 inches by 6 inches

Zach's diagram: 60 inches by 24 inches

The total number of outcomes where all three dice show different numbers is 120.

To determine the number of outcomes in which all three dice show different numbers, you can use the multiplication principle. First, choose a number on the red die (6 options). Then, choose a different number on the blue die (5 options, since it cannot be the same as the red die). Finally, choose a different number on the green die (4 options, since it cannot be the same as the red or blue die). Multiply the options together: 6 x 5 x 4 = 120. Therefore, there are 120 possible outcomes in which the three dice all show different numbers.

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This is true, we can verify that the equation n(A∪B)=n(A)+n(B) holds for the given disjoint sets A={a,e,i,o,u} and B={g,h,k,l,m}.

Since A and B are disjoint sets, meaning they have no common elements, we can say that A∩B=∅. Therefore, the equation n(A∪B)=n(A)+n(B) becomes:
n({a,e,i,o,u,g,h,k,l,m}) = n({a,e,i,o,u}) + n({g,h,k,l,m})
Counting the elements, we see that n({a,e,i,o,u,g,h,k,l,m})=10, n({a,e,i,o,u})=5, and n({g,h,k,l,m})=5.
Substituting these values back into the equation, we get:
10 = 5 + 5
Hi! To verify the equation n(A∪B) = n(A) + n(B) for the given disjoint sets A = {a, e, i, o, u} and B = {g, h, k, l, m}, we first need to find the union of sets A and B.
Since A and B are disjoint (meaning they have no elements in common), the union of A and B, denoted as A∪B, simply combines the elements of both sets. So, A∪B = {a, e, i, o, u, g, h, k, l, m}.
Now, let's find the number of elements (n) in each set:
n(A) = 5 (as there are 5 elements in set A)
n(B) = 5 (as there are 5 elements in set B)
n(A∪B) = 10 (as there are 10 elements in the union of A and B)
Now, we can verify the equation:
n(A∪B) = n(A) + n(B)
10 = 5 + 5

The equation holds true for the given disjoint sets A and B.

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To find the point with the largest z-coordinate that lies on the intersection of the plane x + y + 0 and the sphere x^2 + y^2 + z^2 = 9, we can start by finding the intersection curve of the plane and the sphere.

Substituting y = -x into the equation of the sphere, we get.

x^2 + (-x)^2 + z^2 = 9

2x^2 + z^2 = 9

z^2 = 9 - 2x^2

Substituting y = -x into the equation of the plane, we get:

x + (-x) + 0 = 0

x = 0

So the intersection curve is given by the parametric equations:

x = 0

y = -x = 0

z = ±√(9 - 2x^2)

Since we want the point with the largest z-coordinate, we need to find the point on the curve where z is maximized. Since z^2 is a decreasing function on the interval [0, √(9/2)], we know that z is maximized at x = 0 or x = ±√(9/2). We can evaluate z at these three points:

(0, 0, 3)

(√(9/2), 0, √(9/2 - 9/2)) = (√(9/2), 0, 0)

(-√(9/2), 0, √(9/2 - 9/2)) = (-√(9/2), 0, 0)

Therefore, the point with the largest z-coordinate that lies on the intersection of the plane x + y + 0 and the sphere x^2 + y^2 + z^2 = 9 is (0, 0, 3).

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All of the examples listed are unit rates. A unit rate is a ratio between two different units where one of the terms has a value of 1.

In all of the options given, there is a value of 1 associated with one of the units. For example, in option A, the unit rate is 1.5 cups of flour per 1 cup of milk. Similarly, in option B, the unit rate is 1.33 sandwiches per 1 person. In option C, the unit rate is 1 stamp per 1 person. In option D, the unit rate is 11.4 cars per 1 hour. In option E, the unit rate is 6 coins per 1 player (since 24 divided by 4 is 6). Finally, in option F, the unit rate is 30 miles per 1 hour.

Here are the examples that are unit rates from the given options:

C. 5 stamps per person
D. 11.4 cars per hour
F. 30 miles per hour

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The probability that all 4 players have an ace is approximately 0.00000369 or about 0.00037%.

To find the probability that all 4 players have an ace, we can use the principle of multiplication, which states that the probability of two independent events occurring together is the product of their individual probabilities.

There are 4 aces in the deck of 52 cards. The probability that the first player gets an ace is 4/52. After the first ace has been dealt, there are 51 cards remaining, including 3 aces. Therefore, the probability that the second player gets an ace is 3/51. Similarly, the probability that the third player gets an ace is 2/50, and the probability that the fourth player gets an ace is 1/49.

Using the principle of multiplication, we can multiply these probabilities to find the probability that all four players get an ace:

P(all 4 players have an ace) = (4/52) x (3/51) x (2/50) x (1/49) = 0.00000369

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The mean estimation of cholesterol change  for the population of aquarobic program participants using 95% confidence falls between 8.17 and 11.83 mg/dL.

To estimate the mean cholesterol change using 95% confidence, we need to use a confidence interval. The formula for a confidence interval is:

Mean cholesterol change ± (t-value * standard error)

We can use a t-distribution with n-1 degrees of freedom, where n is the number of participants in the aquarobic program. We can assume that the sample is randomly selected and independent, and that the population of cholesterol changes follows a normal distribution.

To find the t-value, we need to use a t-table or calculator with the appropriate degrees of freedom and confidence level. For 95% confidence and n=sample size, the t-value is:

t-value = 2.306

To calculate the standard error, we can use the formula:

standard error = standard deviation / sqrt(n)

If the standard deviation is not given, we can use the sample standard deviation instead. We can assume that the sample standard deviation is a good estimate of the population standard deviation.

Once we have the standard error, we can substitute it into the confidence interval formula along with the t-value and the mean cholesterol change. This will give us the 95% confidence interval for the mean cholesterol change.

For example, if the mean cholesterol change is 10 mg/dL and the standard deviation is 3 mg/dL, and there were 20 participants in the aquarobic program, then the 95% confidence interval would be:

10 ± (2.306 * (3 / sqrt(20)))
10 ± 1.83

The confidence interval would be (8.17, 11.83). This means that we can be 95% confident that the true mean cholesterol change for the population of aquarobic program participants falls between 8.17 and 11.83 mg/dL.

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This means there is approximately a 3.33% chance that the student will be able to solve all 6 problems on the exam.

We have a student who can solve 7 out of the 10 problems. The instructor will select 6 questions at random for the exam. We want to find the probability that the student can solve all 6 problems on the exam.

To determine this probability, we will use the concept of combinations. A combination is a selection of items from a larger set, where the order of the items does not matter. In this case, we will calculate the combinations of problems the student can solve and the total possible combinations of problems on the exam.

The student can solve 7 problems, so there are C(7,6) combinations of problems she can solve, where C(n,k) represents the number of combinations of n items taken k at a time. There are a total of 10 problems, so there are C(10,6) possible combinations of problems that could appear on the exam.

The probability that the student can solve all 6 problems on the exam is given by the ratio of the combinations of solvable problems to the total possible combinations of problems on the exam:

Probability = C(7,6) / C(10,6)

Using the formula for combinations, we find:

C(7,6) = 7! / (6!(7-6)!) = 7
C(10,6) = 10! / (6!(10-6)!) = 210

So, the probability that the student can solve all 6 problems on the exam is:

Probability = 7 / 210 = 1/30 ≈ 0.0333

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Source of Variation Sum of Squares Degrees of Freedom Mean Squares ​F-Test Statistic Treatment 400 2 200 1.333 Error 5699 22 259.045 Total 6099 24.

To complete the ANOVA table, we'll calculate the missing values using the given information.
1. Calculate the Total Sum of Squares:
Total Sum of Squares = Treatment Sum of Squares + Error Sum of Squares
Total Sum of Squares = 400 + 5699 = 6099
2. Calculate the Total Degrees of Freedom:
Total Degrees of Freedom = Treatment Degrees of Freedom + Error Degrees of Freedom
Total Degrees of Freedom = 2 + 22 = 24
3. Calculate the Mean Squares for Treatment and Error:
Mean Squares for Treatment = Treatment Sum of Squares / Treatment Degrees of Freedom
Mean Squares for Treatment = 400 / 2 = 200
Mean Squares for Error = Error Sum of Squares / Error Degrees of Freedom
Mean Squares for Error = 5699 / 22 = 259
4. Calculate the F-Test Statistic:
F-Test Statistic = Mean Squares for Treatment / Mean Squares for Error
F-Test Statistic = 200 / 259 = 0.772
Now, we can fill in the ANOVA table:
Source of Variation | Sum of Squares | Degrees of Freedom | Mean Squares | F-Test Statistic
Treatment          |      400       |          2         |     200      |      0.772
Error              |     5699       |         22         |     259      |
Total              |     6099       |         24         |              |

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The probability that a randomly chosen bit string of length ten and beginning with a 1 has exactly two 1's is 0.28125.

There are 2¹⁰ possible bit strings of length ten. Among them, the number of bit strings that begin with a 1 is 2⁹ (since there are 2 choices for each of the remaining 9 bits).

Now, we need to count the number of bit strings of length ten and beginning with a 1 that have exactly two 1's. There are 9 choices for the position of the second 1 (since the first bit is already a 1), and then 8 choices for the position of the third 1 (since the first two bits are already fixed), and 8 choices for each of the remaining 8 bits.

So, the total number of bit strings of length ten and beginning with a 1 that have exactly two 1's is 982⁸.

Therefore, the probability that a randomly chosen bit string of length ten and beginning with a 1 has exactly two 1's is (982⁸)/(2⁹) = 0.28125.

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The calculated dimensions of the paper are 6 inches by 10 inches

From the question, we have the following parameters that can be used in our computation:

Let the dimensions be

L = Length

W = Width

So, we have

L * W = 60

2 * (L + W) = 32

This gives

L * W = 60

(L + W) = 16

Let L = 10

So, we have

W = 6

Testing these values, we have

6 * 10 = 60 -- true

2 * (6 + 10) = 32 -- true

Hence, the dimensions of the paper are 6 inches by 10 inches

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When Solver is unable to find a solution to a problem, it can be an indication of various issues. One possible cause is that the problem may be too complex for Solver to solve within the given constraints.

In such cases, it may be necessary to adjust the problem parameters or seek alternative solutions. Another possible cause of Solver's inability to find a solution could be due to incorrect input data. This can lead to inconsistent or contradictory constraints, making it impossible for Solver to arrive at a feasible solution. Lastly, Solver may fail to find a solution due to numerical errors or limitations in the algorithm used. These issues can arise when dealing with large datasets or highly non-linear problems. In any case, when Solver is unable to find a solution, it is important to carefully examine the problem and its parameters, and consider alternative approaches. Sometimes, a small adjustment to the input data or constraints can make all the difference in arriving at a successful solution.  In such cases, Solver may struggle to find an accurate or optimal solution. Ill-conditioned problems typically involve numerical instability or poor scaling, while infeasible problems occur when the given constraints cannot be satisfied simultaneously. It is crucial to analyze and refine the problem to enable Solver to find a viable solution effectively. When there is a problem with Solver being unable to find a solution, it often indicates the presence of a(n) ill-conditioned or infeasible problem.

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When we conclude that the results we have gathered from our sample are probably also found in the population from which the sample was drawn, we say that the results are statistically significant.

In statistical analysis, we use hypothesis testing to determine whether the results of a sample are likely to be representative of the population as a whole. If the results are statistically significant, we can infer that there is a low probability that the observed differences between the sample and the population occurred by chance alone.

This allows us to generalize the findings from the sample to the larger population with a reasonable degree of confidence.

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The expected value per share, taking into account the possibility of financial distress, is 7.92.

If the probability of financial distress is 10%, we need to adjust our valuation to account for the possibility of that scenario occurring.

Let's denote the probability of no financial distress as p and the probability of financial distress as (1-p). Then, the expected value of equity per share can be calculated as:

Expected value = p × 9.50 + (1-p) × 2.0

We know that (1-p) = 0.10, so we can substitute that in:

Expected value = p × 9.50 + 0.10  2.0

Solving for p, we get:

p = (Expected value - 0.10 × 2.0) / 9.50

Substituting the given values, we get:

p = (9.50 - 0.10 × 2.0) / 9.50 = 0.789

This means that the probability of no financial distress is 0.789, and the probability of financial distress is 0.211.

Now, we can calculate the value per share under the scenario of financial distress:

Value per share (distress) = 2.0

And the value per share under the scenario of no financial distress:

Value per share (no distress) = 9.50

So, the expected value per share is:

Expected value per share = p × Value per share (no distress) + (1-p) × Value per share (distress)

Expected value per share = 0.789 × 9.50 + 0.211 × 2.0 = 7.50 + 0.42 = 7.92

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