Use Routh's stability criterion to determine how many roots with positive real parts the following equations have: a. s4+8s3+32s2+80s+100=0 b. s5+10s4+30s3+80s2+344s+480=0 c. s4+2s3+7s2−2s+8=0 d. s3+s2+20s+78=0

Answer:

In China, $0.45

Hope it helps!

The spring can be stretched up to 8 feet beyond its natural length with a force not exceeding 24 pounds.

Hooke's Law states that the force required to maintain a spring stretched x units beyond its natural length (F) is proportional to x. Mathematically, this can be expressed as:

F ∝ x     (Equation 1)

The work done on a spring can be calculated using the formula:

Work = (1/2) k x^2     (Equation 2)

where k is the spring constant.

Given that the work required to stretch the spring from 2 feet beyond its natural length to 4 feet beyond its natural length is 18 ft-lb, we can write the following equation using Equation 2:

(1/2) k (4^2 - 2^2) = 18

Simplifying the equation:

k (16 - 4) = 36

12k = 36

k = 36/12

k = 3 ft-lb/ft^2

Now, we can use Equation 1 and the given force limit of 24 pounds to determine the maximum stretch beyond the natural length (x_max). We know that the force (F) is proportional to x:

F = kx

Substituting the values:

24 = 3x_max

Solving for x_max:

x_max = 24/3

x_max = 8 feet

Therefore, the spring can be stretched up to 8 feet beyond its natural length with a force not exceeding 24 pounds.

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The particular solution of the differential equation (1) is given by

yp = (x raised to power of 2 - x)e raised to power x

The general solution of the differential equation (1) is given by

y = c1x + c2e raised to power of x + (x raised to power of 2 - x)e^x

where c1 and c2 are arbitrary constants.

The complementary equation of the differential equation (1) is given by

(x−1)y′′−xy′+y=0

The general solution of the complementary equation is given by

y = c1x + c2e^x

where c1 and c2 are arbitrary constants.

To find a particular solution of the differential equation (1), we can use the method of variation of parameters. In this method, we assume that the particular solution is of the form

yp = u(x)x + v(x)e^x

where u(x) and v(x) are functions to be determined.

Substituting this expression into the differential equation (1), we get

(x−1)u′′(x)x + (x−1)u′(x)e^x - xu′(x)x - xu′(x)e^x + u(x)x + v(x)e^x = (x−1)^2e^x

Simplifying this equation, we get

(x−1)u′′ + (x−1)u′ - xu′ + u + v = (x−1)^2e^x

Matching the coefficients of the different powers of x on both sides of the equation, we get the following system of equations:

u′′ = 2e^x

u′ = x - 2

u = x^2 - x

v = 0

Solving this system of equations, we get

u(x) = x^2 - x

v(x) = 0

Substituting these expressions into the expression for yp, we get the following particular solution:

yp = (x^2 - x)e^x

The general solution of the differential equation (1) is given by the sum of the general solution of the complementary equation and the particular solution, which is given by

y = c1x + c2e^x + (x^2 - x)e^x

where c1 and c2 are arbitrary constants.

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In this case, Christopher divides the athletes into groups of varsity and junior varsity, which creates the strata. The type of sampling used in this scenario is stratified sampling.

Stratified sampling is a sampling method where the population is divided into homogeneous subgroups or strata, and individuals are randomly selected from each stratum in proportion to their representation in the population. In this case, Christopher divides the athletes into groups of varsity and junior varsity, which creates the strata.

By randomly selecting a proportionate number of individuals from each group, Christopher ensures that both varsity and junior varsity athletes are represented in the sample, maintaining the proportional representation of each group in the population. This method allows for more accurate and representative results by capturing the characteristics of both groups separately.


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The converse of the following statement: If x < 0, then x5 < 0 is If x5 < 0, then x < 0. The answer is option D.

The converse "If x5 < 0, then x < 0" is true. Conditional statements are made up of two parts: a hypothesis and a conclusion. If the hypothesis is valid, the conclusion is also true, according to conditional statements. The inverse, converse, and contrapositive are three variations of a conditional statement that have different implications. The converse of a conditional statement is produced by exchanging the hypothesis and the conclusion. A converse is valid if and only if the original conditional is valid and the hypothesis and conclusion are switched. The hypothesis "x < 0" and the conclusion "x5 < 0" are the two parts of the conditional statement "If x < 0, then x5 < 0."

Therefore, the converse of this statement is "If x5 < 0, then x < 0." This converse is correct since it is always valid. If x5 is less than zero, x must be less than zero because a negative number to an odd power is still negative.

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Answer: $826.82 to $873.18

Step-by-step explanation:

The probability of getting less than or equal to 17 headaches is approximately 0.0281.The drug is effective in the given situation as the percentage of headaches is less than 8% of the treated subjects

We have 270 trials with a probability of success 8%. Here, n = 270, p = 0.08, and q = 1 - p = 0.92. We need to find the probability of getting less than or equal to 17 headaches.The mean of the normal distribution is given as μ = np = 270 × 0.08 = 21.6.The variance is given by the formula σ² = npq.

Therefore, σ = sqrt(npq) = sqrt(270 × 0.08 × 0.92) = 2.4095.To standardize the normal distribution, we need to find the z-score. The formula for z-score is given by z = (x - μ) / σWhere x = 17Plug in the values, we get z = (17 - 21.6) / 2.4095 = -1.9122.We need to find P(z < -1.9122)Using a standard normal table, we find P(z < -1.9122) = 0.02813

Therefore, the probability of getting less than or equal to 17 headaches is approximately 0.0281.The drug is effective in the given situation as the percentage of headaches is less than 8% of the treated subjects

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The inverse of an invertible symmetric matrix A, denoted as A^(-1), is also a symmetric matrix.

The inverse of an invertible symmetric matrix A, denoted as A^(-1), is also a symmetric matrix.

To prove this, let's start with the given information: A is an nxn invertible symmetric matrix, meaning A^T = A. We want to show that A^(-1) is also symetric, i.e., (A^(-1))^T = A^(-1).

Since A is an invertible matrix, it has a unique inverse A^(-1). We can use the properties of transpose and matrix inversion to demonstrate that (A^(-1))^T = A^(-1).

Taking the transpose of both sides of the equation A^T = A, we have (A^(-1))^T * A^T = (A^(-1))^T * A.

Now, multiply both sides by A^(-1) on the left: (A^(-1))^T * A^T * A^(-1) = (A^(-1))^T * A * A^(-1).

By the properties of matrix transpose, (AB)^T = B^T * A^T, we can rewrite the equation as (A^(-1) * A)^T * A^(-1) = A^(-1)^T * A * A^(-1).

Since A^(-1) * A is the identity matrix I, we have I^T * A^(-1) = A^(-1)^T * A * A^(-1).

Since I is symmetric (the identity matrix is always symmetric), we can simplify the equation to A^(-1) = A^(-1)^T * A * A^(-1).

Now, we have shown that A^(-1) = A^(-1)^T * A * A^(-1), which implies (A^(-1))^T = A^(-1).

Therefore, we have proved that the inverse of an invertible symmetric matrix A, denoted as A^(-1), is also a symmetric matrix.

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To prove that the statement using mathematical induction we will verify the base case and then show that if the statement holds for k, it also holds for k + 1 in the inductive step. This establishes that the statement is true for all positive integer values greater than 10.

To prove that for any positive integer number n > 10, (n⁴ + 3n - 8) is even using induction, we need to follow the steps of mathematical induction:

Step 1: Base Case

We start by checking the base case, which is n = 11, the smallest value greater than 10.

For n = 11:

(n⁴ + 3n - 8) = (11⁴ + 3(11) - 8) = (14641 + 33 - 8) = 14666

The result is indeed an even number since it is divisible by 2. Hence, the base case holds.

Step 2: Inductive Hypothesis

Assume that for some positive integer k > 10, (k⁴ + 3k - 8) is even. This is our inductive hypothesis.

Step 3: Inductive Step

We need to prove that if the hypothesis holds for k, it also holds for k + 1.

For k + 1:

((k + 1)⁴ + 3(k + 1) - 8) = (k⁴ + 4k³ + 6k² + 4k + 1 + 3k + 3 - 8)

                          = (k⁴ + 4k³ + 6k² + 7k - 4)

Now, let's consider the difference between the two expressions:

[(k⁴ + 3k - 8) + 4k³ + 6k² + 7k - 4]

From the inductive hypothesis, we know that (k⁴ + 3k - 8) is even.

Moreover, the expression (4k³ + 6k² + 7k - 4) can be rewritten as 2(2k³ + 3k² + 3.5k - 2), which is also even since it is divisible by 2.

Adding an even number to another even number always results in an even number.

Hence, the sum [(k⁴ + 3k - 8) + 4k³ + 6k² + 7k - 4] is even.

Therefore, by mathematical induction, we can conclude that for any positive integer number n > 10, (n⁴ + 3n - 8) is even.

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a) There are 4368 strings of hexadecimal digits consisting of one through three digits.

b) There are 17909080 strings of hexadecimal digits consisting of two through six digits.

(a) To determine the number of strings of hexadecimal digits consisting of one through three digits, we can calculate the total number of possibilities for each case and then sum them up.

For one-digit strings, there are 16 options (0 through F).

For two-digit strings, each digit can be one of the 16 options independently. So, there are 16 options for the first digit and 16 options for the second digit, resulting in a total of 16 * 16 = 256 possibilities.

For three-digit strings, we apply the same logic as for two-digit strings. Each digit can be one of the 16 options independently, so there are 16 * 16 * 16 = 4096 possibilities.

By summing up the possibilities for each case, we have 16 + 256 + 4096 = 4368 strings of hexadecimal digits consisting of one through three digits.

(b) To calculate the number of strings of hexadecimal digits consisting of two through six digits, we need to consider the possibilities for each case.

For two-digit strings, we already determined that there are 256 possibilities.

For three-digit strings, we have 4096 possibilities.

For four-digit strings, the logic is the same as for two-digit strings, so there are 16 * 16 * 16 * 16 = 65536 possibilities.

For five-digit strings, we have 16 * 16 * 16 * 16 * 16 = 1048576 possibilities.

For six-digit strings, we have 16 * 16 * 16 * 16 * 16 * 16 = 16777216 possibilities.

By summing up the possibilities for each case, we have 256 + 4096 + 65536 + 1048576 + 16777216 = 17909080 strings of hexadecimal digits consisting of two through six digits.

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The probability that the random variable in a uniform distribution is equal to 7.91, given that the distribution is defined over the interval from 6 to 10, is zero.

In a uniform distribution, the probability is evenly spread across the entire interval. The probability of any specific value within the interval is determined by the width of the interval. In this case, the interval is from 6 to 10.
Since the random variable is continuous and the probability is spread evenly, the probability of any specific value within the interval is infinitesimally small. Therefore, the probability of the random variable being equal to 7.91, which falls within the interval from 6 to 10, is zero.
In conclusion, in a uniform distribution defined over the interval from 6 to 10, the probability of the random variable being equal to any specific value, including 7.91, is zero.

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The zero polynomial is a polynomial of degree at most 3 with integer coefficients, and it belongs to the given set.

Thus, the given set is a subspace of P for n = 3.

The set P of all polynomials of degree at most 3 with integer coefficients.

The given set is a subspace of P for an appropriate value of n.

It can be justified by the following explanation:

A subspace is a subset of the vector space such that it has three properties, that are:

It is closed under addition, It is closed under scalar multiplication, and It contains the zero vector.

A polynomial is an expression consisting of variables and coefficients which involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.

The given set is a subspace of P with n = 3 because it satisfies all the three properties of a subspace.

i) The sum of two polynomials is a polynomial of degree at most 3 with integer coefficients.

ii) Multiplication of a polynomial by a scalar is a polynomial of degree at most 3 with integer coefficients.

iii) The zero polynomial is a polynomial of degree at most 3 with integer coefficients, and it belongs to the given set.

Thus, the given set is a subspace of P for n = 3.

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The dimensions of the rectangle with maximum area are 58 for the length and 29 for the width, resulting in a maximum area of 1,682 square units.

Let's consider the construction of the rectangle within the semicircle. The diameter of the semicircle is twice the radius, which is 58. Thus, the length of the rectangle should be equal to this diameter. To find the width of the rectangle, we need to analyze the relationship between the rectangle and the semicircle. The two other vertices of the rectangle lie on the semicircle. As a rectangle has opposite sides equal in length, the width of the rectangle will be equal to the radius of the semicircle, which is 29.

Therefore, the dimensions of the rectangle with maximum area are 58 for the length and 29 for the width. To maximize the area of the rectangle, we use the formula for the area of a rectangle, which is given by length multiplied by width. Substituting the dimensions we found, the maximum area of the rectangle is 58 * 29 = 1,682 square units.

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The EOQ (Economic Order Quantity) is approximately 3953 disks.

To calculate the EOQ (Economic Order Quantity), we can use the formula EOQ = sqrt((2 * D * S) / H), where D represents the annual demand, S represents the setup or ordering cost, and H represents the holding or carrying cost per unit.

Given the following information:

Annual demand (D) = 35200 disks

Setup cost (S) = $0.87 per disk

Discount price = $5.08

Quantity needed to qualify for the discount = 5900 disks

First, we need to calculate the holding cost per unit (H) by subtracting the discount price from the regular price: H = $5.08 - $0.87 = $4.21

Plugging these values into the EOQ formula, we get EOQ = sqrt((2 * 35200 * $0.87) / $4.21). After calculating this expression, and rounding the result to the nearest whole number, we find that the EOQ is approximately 3953.

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The probability of getting 2 consecutive heads when flipping a coin 4 times is 3/16, or 0.1875

To determine the probability of getting 2 consecutive heads when flipping a coin 4 times, we need to consider the possible outcomes that satisfy this condition.

When flipping a coin, there are 2 possible outcomes for each flip: heads (H) or tails (T). Since we are interested in getting 2 consecutive heads, we need to identify the sequences that meet this criterion.

Out of the total number of possible outcomes when flipping a coin 4 times (2⁴ = 16), there are 3 sequences that have 2 consecutive heads: HHTT, THHT, and TTHH. These sequences have consecutive heads occurring in the first two flips, second and third flips, and third and fourth flips, respectively.

Therefore, the probability of getting 2 consecutive heads when flipping a coin 4 times is 3/16, or 0.1875, which can be expressed as a decimal or a fraction.

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The standard normal probability table can only be used to compute probabilities for normal random variables with parameters μ = 0 and σ = 1. The correct statement among the given options is e.

a. The statement in option a is incorrect. While the standard normal distribution is commonly used as a model in various statistical analyses and is often used as an approximation for naturally arising populations, it does not always perfectly represent the characteristics of all naturally occurring populations.

b. The statement in option b is incorrect as not all given statements are correct.

c. The statement in option c is incorrect. If a random variable X is normally distributed with parameters μ and σ, then the mean of X is indeed μ, but the variance of X is σ², not "o" as stated in the option.

d. The statement in option d is incorrect. The cumulative distribution function (CDF) of a standard normal random variable Z is denoted as P(Z ≤ z), not P(Z = z). The CDF provides the probability that Z takes on a value less than or equal to a given value z.

Therefore, the correct statement is e, which states that the standard normal probability table can only be used to compute probabilities for normal random variables with parameters μ = 0 and σ = 1.

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The inspector can use this information to evaluate whether the rail company has breached the terms of its license by comparing the actual number of canceled trains to the maximum number of canceled trains allowed under the terms of its franchise license.

The government has required the rail company to make sure that no more than 5% of all train journeys are canceled. The inspector can use this information to evaluate whether the rail company has breached the terms of its license in the following ways:

Since there are 12500 train journeys scheduled to run, the 5% threshold for canceled trains would be:

5% of 12500 = (5/100) x 12500 = 625

For the rail company to adhere to the terms of its license, no more than 625 trains should be canceled. 680 trains were canceled, according to the inspector's findings. The rail company has, therefore, breached the terms of its license by having a higher number of canceled trains.

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They have failed to meet the government's requirement of ensuring that no more than 5% of all train journeys are cancelled and are at risk of losing their franchise license. The inspector can report this finding to the government, which can then take appropriate action.

The inspector should use this information to assess whether the rail company is in breach of the terms of their license by comparing the percentage of train journeys cancelled to the government's requirement of no more than 5%.Here's how the inspector can calculate the percentage of train journeys cancelled:

Percentage of train journeys cancelled = (Number of train journeys cancelled / Total number of train journeys scheduled) x 100%Substituting the values given in the question,

Percentage of train journeys cancelled =

(680 / 12500) x 100%≈ 5.44%

Since the percentage of train journeys cancelled is more than 5%, the rail company is in breach of the terms of their license.

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The probability that the mean weight of the sample babies would be less than 3174 grams is 0.1237 (rounded to four decimal places).

Given that the mean weight of babies born in a large hospital is 3215 grams and the variance is 84681. A sample of 67 babies is chosen at random from the hospital. We need to find the probability that the mean weight of the sample babies is less than 3174 grams.

To solve this, we can use the central limit theorem, which states that the sample means of a large sample (n > 30) taken from a population with a mean μ and a standard deviation σ will be approximately normally distributed with a mean μ and a standard deviation σ / √n.

Here,

n = 67,

μ = 3215 and

σ² = 84681.

σ = √σ² = √84681 = 290.8191

σ / √n = 290.8191 / √67 = 35.4465

To find the probability that the sample mean weight of the babies is less than 3174 grams, we need to find the z-score.

z = (x - μ) / (σ / √n) = (3174 - 3215) / 35.4465 = -1.1572

From the standard normal distribution table, we find that the probability of z being less than -1.1572 is 0.1237.

Therefore, the probability that the mean weight of the sample babies would be less than 3174 grams is 0.1237 (rounded to four decimal places).

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The numbers in this problem are classified as follows:

A. Rational.

B. Irrational.

C. Rational.

D. Rational.

E. Rational.

F. Irrational.

G. Rational.

H. Not real.

The square root of negative numbers are the numbers that are classified as not real.

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Answer:

w = 2, w = -4

Step-by-step explanation:

5w2 + 10w -40 = 0

5w2 + 20w - 10w - 40 = 0

5w(w + 4) - 10(w + 4) = 0

(5w - 10)(w + 4)=0

w= 2 , w = -4

As per the given information and the mean, the possible set of scores for the five students could be: 1, 3, 4, 5, 7

Lowest score = 1

Highest score = 7

Average = Mean = 4

When all the numbers in a data collection are added up, the average, or mean, is obtained by dividing the total by the total number of data points. The sequence of the supplied students indicating the scores attained from lowest to highest is 1, 3, 4, 5, 7, under the condition that the average (mean) is 4, after carefully analysing the provided data and executing a series of calculations.

The explanation for the series of action is that there is one possible set of scores for the five students that satisfy the given conditions (lowest score of 1, highest score of 7, and an average of 4) is 1, 3, 4, 5, 7

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All real numbers except x = -4, x = 2, and x = -3.  there are no removable discontinuities in this function. Since the degrees are equal, there are no horizontal asymptotes.

(a) The domain of the given rational function is all real numbers except the values that would make the denominator zero. In this case, the denominator is (x + 4)(x - 2)(x + 3). So, the domain of the function is all real numbers except x = -4, x = 2, and x = -3.

(b) To find the removable discontinuities, we need to determine if there are any common factors between the numerator and denominator that can be canceled out. In this case, there are no common factors between (x + 4)(x - 5)(x + 1) and (x + 4)(x - 2)(x + 3). Therefore, there are no removable discontinuities in this function.

(c) To find the equation(s) of horizontal asymptotes, we need to compare the degrees of the numerator and denominator. In this case, both the numerator and denominator are of degree 3. Since the degrees are equal, there are no horizontal asymptotes.

(d) To find the equation(s) of vertical asymptotes, we need to determine the values of x that make the denominator zero. In this case, the vertical asymptotes occur at x = -4, x = 2, and x = -3, as these are the values that would make the denominator (x + 4)(x - 2)(x + 3) equal to zero.

Solving the equation algebraically: √3 - 6x - 4 = x

To solve the equation, we can isolate the square root term and the x term on opposite sides: √3 - 4 = x + 6x

Simplifying: √3 - 4 = 7x

Now, we can isolate x by dividing both sides by 7: x = (√3 - 4) / 7

The solution to the equation is x = (√3 - 4) / 7.

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When k = 1/6, the solution to the equation log9(x) - log9(x + 8) = log9(k) is x = 8/5.

Let's start by simplifying the equation log9(x) - log9(x + 8) = log9(k). Applying the logarithmic property of subtraction, we can rewrite it as a single logarithm:

log9(x/(x + 8)) = log9(k).

Now, to solve for x, we can equate the expressions inside the logarithm:

x/(x + 8) = k.

Next, we substitute k = 1/6 into the equation:

x/(x + 8) = 1/6.

To solve this equation for x, we can cross-multiply:

6x = x + 8.

Simplifying further:

6x - x = 8,

5x = 8,

x = 8/5.

Therefore, when k = 1/6, the corresponding value of x is x = 8/5.

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After considering the given data we conclude that the coefficient of variation is 64.28% which is option C , and the interquartile range is 10 which is option B.

The mean of a sample is 5 and the standard deviation of the sample is 4.062. The sample data is: 5, 3, 2, 3, 12, 13. To evaluate the coefficient of variation, we can apply the formula:
coefficient of variation = [tex](standard deviation / mean) * 100%[/tex]
First, we need to evaluate the mean of the sample:
mean = (5 + 3 + 2 + 3 + 12 + 13) / 6 = 6.33
Next, we can evaluate the standard deviation of the sample:
standard deviation = [tex]\sqrt(((5-6.33)^2 + (3-6.33)^2 + (2-6.33)^2 + (3-6.33)^2 + (12-6.33)^2 + (13-6.33)^2) / 5) = 4.062[/tex]
Now, we can evaluate the coefficient of variation:
coefficient of variation = (4.062 / 6.33) × 100% = 64.28%

Then, the coefficient of variation is 64.28%.
To evaluate the interquartile range, we need to first find the first and third quartiles.
First, we need to order the sample data:
2, 3, 3, 5, 12, 13
The median of the sample is (3 + 5) / 2 = 4.
The first quartile [tex](Q_1)[/tex] is the median of the lower half of the sample data:
2, 3, 3
[tex]Q_1 = (2 + 3) / 2 = 2.5[/tex]
The third quartile [tex](Q_3)[/tex] is the median of the upper half of the sample data:
5, 12, 13
[tex]Q_3 = (12 + 13) / 2 = 12.5[/tex]
Now, we can evaluate the interquartile range:
interquartile range = [tex]Q_3 - Q_1 = 12.5 - 2.5 = 10[/tex]
Therefore, the interquartile range is 10
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The complete question is
A researcher has collected the following sample data. The mean of the sample is 5 and the standard deviation of sample is 4.062. 5 3 2 3 12 13 The coefficient of variation is a. 80.00% I b. 125.00% c. 64.28% d. 33.0% 14 The interquartile range is a. 1 b. 10 C. 2 d. 12

The transformation undergone by the triangle is: a horizontal translation by 4 units to the right

There are different types of transformations of shapes in geometry such as:

Translation

Reflection

Rotation

Dilation

Now, we are told that the triangle was moved by 4 units to the left.

We know that translation in transformation simply means moving an object from one point to another without any of the dimensions being affected and as such, we can easily say that:

The transformation undergone by the triangle is a horizontal translation by 4 units to the right

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The normal vector to the plane given by the equation 5(x z) = 6(x + y) is (-6, -6, 5).

To find a normal vector to the given plane equation, let's first rewrite the equation in a simplified form. The equation 5(x z) = 6(x + y) can be expanded to 5xz = 6x + 6y. Rearranging the terms, we have 5xz - 6x - 6y = 0.

Now, we can identify the coefficients of x, y, and z in the equation. The coefficient of x is 5z - 6, the coefficient of y is -6, and the coefficient of z is 5x. These coefficients form the components of the normal vector to the plane.

To find the normal vector, we can write it as a vector with the components (A, B, C). From the equation, we have A = 5z - 6, B = -6, and C = 5x.

However, since there is no specific value given for x or z, we can express the normal vector in terms of x and z. Therefore, the normal vector to the plane is (5z - 6, -6, 5x).

It's important to note that the normal vector represents a direction perpendicular to the plane. Any scalar multiple of the normal vector would also be a valid normal vector to the plane. Therefore, we could multiply the components of the normal vector by a constant if desired.

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The average rate of Superman can be calculated by dividing the change in distance by the change in time.

Average rate = (final distance - initial distance) / (final time - initial time)

Average rate = (1164 - 1372) / (18 - 14)

Average rate = -208 / 4

Average rate = -52 meters per second

To find how far Superman flies every 15 seconds, we can use the concept of proportionality. Since we know the rate at which Superman is flying, we can set up a proportion to find the distance.

Rate = Distance / Time

-52 meters per second = Distance / 4 seconds

Distance = -52 * 15

Distance = -780 meters (Note: Distance cannot be negative, so we consider the magnitude)

C. To determine how close Superman is to Lois after 33 seconds, we can use the average rate to calculate the distance traveled.

Distance = Average rate * Time

Distance = -52 * 33

Distance = -1716 meters (Note: Distance cannot be negative, so we consider the magnitude)

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The expression at the limits of integration [tex]∫[C2] (x + 5y) dx + x^2 dy = [5(1) - (1/3)(1)^3[/tex]

To evaluate the line integral ∫(x + 5y) dx + x^2 dy along the curve C, which consists of line segments from (0, 0) to (5, 1) and from (5, 1) to (6, 0), we will calculate the integral along each segment separately and then sum the results.

Let's start by evaluating the line integral along the first segment of the curve, which goes from (0, 0) to (5, 1). We can parametrize this segment as:

x(t) = 5t, where t varies from 0 to 1,

y(t) = t, where t varies from 0 to 1.

Using the parametric equations, we can express dx and dy in terms of dt:

dx = 5dt,

dy = dt.

Substituting these expressions into the line integral, we have:

∫[C1] (x + 5y) dx + x^2 dy = ∫[0 to 1] [(5t + 5t) * 5dt + (5t)^2 * dt].

Simplifying the integral, we get:

[tex]∫[C1] (x + 5y) dx + x^2 dy = ∫[0 to 1] (10t + 25t^2) dt[/tex].

Integrating each term separately, we obtain:

[tex]∫[C1] (x + 5y) dx + x^2 dy = [5t^2 + (25/3)t^3][/tex] evaluated from 0 to 1.

Evaluating the expression at the limits of integration, we get:

[tex]∫[C1] (x + 5y) dx + x^2 dy = [5(1)^2 + (25/3)(1)^3] - [5(0)^2 + (25/3)(0)^3][/tex].

Simplifying further, we find:

[tex]∫[C1] (x + 5y) dx + x^2 dy = 5 + 25/3[/tex].

Now, let's evaluate the line integral along the second segment of the curve, which goes from (5, 1) to (6, 0). We can parametrize this segment as:

x(t) = 5 + t, where t varies from 0 to 1,

y(t) = 1 - t, where t varies from 0 to 1.

Using the parametric equations, we can express dx and dy in terms of dt:

dx = dt,

dy = -dt.

Substituting these expressions into the line integral, we have:

[tex]∫[C2] (x + 5y) dx + x^2 dy = ∫[0 to 1] [(5 + t) * dt + (5 + t)^2 * (-dt)][/tex].

Simplifying the integral, we get:

[tex]∫[C2] (x + 5y) dx + x^2 dy = ∫[0 to 1] (5dt - t^2 + 10t + 25) dt[/tex].

Integrating each term separately, we obtain:

[tex]∫[C2] (x + 5y) dx + x^2 dy = [5t - (1/3)t^3 + 5t^2 + 25t][/tex] evaluated from 0 to 1.

Evaluating the expression at the limits of integration, we get:

[tex]∫[C2] (x + 5y) dx + x^2 dy = [5(1) - (1/3)(1)^3[/tex]

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To draw a less than type ogive for the given weight data and determine the median weight, we can plot the cumulative frequency against the upper class boundaries. Here's a step-by-step approach:

Create a table with two columns: "Weight (in kg)" and "Cumulative Frequency."

Weight (in kg) Cumulative Frequency

Less than 38 0

Less than 40 3

Less than 42 5

Less than 44 9

Less than 46 14

Less than 48 28

Less than 50 32

Less than 52 35

Plot the cumulative frequency against the upper class boundaries on a graph.

The upper class boundaries are: 38, 40, 42, 44, 46, 48, 50, 52.

The corresponding cumulative frequencies are: 0, 3, 5, 9, 14, 28, 32, 35.

Connect the plotted points to form a less than type ogive.

To find the median weight from the graph, draw a horizontal line at the cumulative frequency value of N/2, where N is the total number of students (35 in this case).

The median weight can be determined by the intersection of this horizontal line with the less than type ogive.

To verify the result using the formula, we can use the cumulative frequency distribution.

Median weight = L + ((N/2 - CF) * w) / f

Where:

L = lower class boundary of the median class

N = total number of students

CF = cumulative frequency of the class before the median class

w = width of the median class

f = frequency of the median class

By following these steps and using the graph and formula, you can determine the median weight from the given data and verify the result.

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Answer: 1,215

Step-by-step explanation:

5 times 3 = 15 1st term

15 times 3 = 45 2nd term

45 times 3 = 135 3rd term

135 times 3 = 405 4th term

405 times 3 = 1,215 5th term

Hope this helps