The following statistics represent weekly salaries at a construction company Mean Median Mode $535 $615 $505 First quartile $455 Third quartile $735 83rd percentile $875 The most common salary is $505 The salary that half the employees' salaries surpass is $ The percent of employees' salaries that surpassed $735 is %. The percent of employees' salaries that were less than $455 is %. The percent of employees' salaries that surpassed $875 is %. If the company has 100 employees, the total weekly salary of all employees is S$

The values of the function at 9 are 81a + 9 and 81m + c.

We have,

To determine the values of the functions F(x) and g(x) at x = 9, we need to substitute x = 9.

So,

For F(x) = ax² + 9:

F(9) = a(9)² + 9

F(9) = 81a + 9

And,

For g(x) = mx² + c:

g(9) = m(9)² + c

g(9) = 81m + c

Thus,

The values of the function at 9 are 81a + 9 and 81m + c.

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The given equation is a vertical parabola in standard form. To find the focus and directrix, we first need to determine the vertex.

The vertex is (0, -2). The focus is located at a distance of p units vertically above the vertex, where p is the distance from the vertex to the focus. In this case, p = 2. So the focus is at (0, 0). The directrix is located p units vertically below the vertex.

Therefore, the directrix is the horizontal line y = -4. The answer is (b) 2004-06-02-06-00 files/  directrix: 2004-06-02-06-00 files/ .

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The probability that at least one of the ten random numbers drawn from a uniform distribution on [0, 4.62] will exceed 4.62 is approximately 0.450.

In a uniform distribution, the probability of a value falling within a specific range is proportional to the length of that range. Since the range of the uniform distribution is [0, 4.62], the probability of drawing a number less than or equal to 4.62 from this distribution is 1.

Therefore, the probability that at least one number will exceed 4.62 is equal to 1 minus the probability that all ten numbers drawn are less than or equal to 4.62. Since the draws are independent, we can calculate this probability as (1 - 1)^10 = 1^10 = 1.

Rounded to three decimal places, the probability that at least one number will exceed 4.62 is approximately 0.450.

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(2a² + 3a - 5) - (3a² + 3a + 7)

Distributing the negative sign, we have:

2a² + 3a - 5 - 3a² - 3a - 7

Combining like terms, we get:

(2a² - 3a²) + (3a - 3a) + (-5 - 7)

= -a² - 12

The equation (x - 2)² = 13 is not equivalent because it represents a perfect square, not the original quadratic equation.

The equation (x - 2)² = 17 is also not equivalent because the constant term is different from the original equation.

The equation (x - 4)² = 13 is equivalent to the original equation because it represents a perfect square with the correct constant term.

The equation (x - 4)² = 17 is not equivalent because the constant term is different from the original equation.

(2x - 1)(3x + 4)

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Answer:

Exponential Graph - Growth, Decay, Examples | Graphing ...

An exponential graph is a curve that has a horizontal asymptote and it either has an increasing slope or a decreasing slope. i.e., it starts as a horizontal line and then it first increases/decreases slowly, and then the growth/decay becomes rapid.

Step-by-step explanation:

Answer:

12

Step-by-step explanation:

8×9=6x

x=12

That is the answer

To construct a box plot from the given data, which represents the diameters of cans in an assembly line, we need to determine the five-number summary and plot the corresponding box and whisker plot.

The five-number summary consists of the minimum value, the first quartile (Q1), the median (Q2), the third quartile (Q3), and the maximum value.

To construct the box plot, we start by arranging the data in ascending order: 5.1, 5.2, 5.2, 5.2, 5.3, 5.5, 5.5, 5.5, and 5.6. The minimum value is 5.1, and the maximum value is 5.6. The median is the middle value, which in this case is 5.3.

To find the first quartile (Q1) and the third quartile (Q3), we divide the data into two halves. Q1 is the median of the lower half, which consists of 5.1, 5.2, 5.2, and 5.2. Q3 is the median of the upper half, which consists of 5.5, 5.5, 5.5, and 5.6. The box plot will show the minimum value, Q1, Q2 (median), Q3, and the maximum value, giving us a visual representation of the distribution and variability of the data.

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VE[X] is -10001999900.

By comparing the values, we can see that E[VF] = E[X] ≥ 1, and E[VE[X]] = -10001999900.

To prove that for a concave function g, we have 9(E[X]) > E[g(X)], we can use Jensen's inequality. Jensen's inequality states that for a concave function g and a random variable X, we have:

g(E[X]) ≥ E[g(X)]

Let's start the proof:

Since g is a concave function, we have g''(x) < 0 for all x.

By Jensen's inequality, we have g(E[X]) ≥ E[g(X)].

Now, let's compare E[X] and E[g(X)]:

E[X] = ∑[x] x * P(X = x) (where ∑[x] denotes the sum over all possible values of X)

E[g(X)] = ∑[x] g(x) * P(X = x)

Since g''(x) < 0 for all x, g(x) is a concave function. By applying Jensen's inequality to g(x), we have:

g(E[X]) ≥ E[g(X)]

Now, we can multiply both sides of the above inequality by 9 (a positive constant):

9 * g(E[X]) ≥ 9 * E[g(X)]

Since g(E[X]) ≥ E[g(X)], we can replace g(E[X]) on the left-hand side:

9 * g(E[X]) ≥ E[g(X)]

Therefore, we have 9(E[X]) > E[g(X)].

This proves that for a concave function g, we always have 9(E[X]) > E[g(X)].

Moving on to the second part of the question:

a. To determine the distribution of Y = X, we can simply use the given probability mass function of X.

P(Y = y) = P(X = y) (since Y = X)

Therefore, the distribution of Y is the same as the distribution of X.

b. We need to compare E[VF] and E[VE[X]]. Using the given function g(x) = -1, we can see that it is a convex function.

By Jensen's inequality for convex functions, we have:

g(E[X]) ≤ E[g(X)]

Substituting g(x) = -1, we have:

-1 * E[X] ≤ E[-1]

-E[X] ≤ -1

E[X] ≥ 1

This implies that E[VF] = E[X] ≥ 1.

To compare E[VF] and E[VE[X]], we need to compute E[VE[X]]. Using Exercise 8.11 (which is not provided in the question), or by directly calculating, we find:

E[VE[X]] = E[X * X] = ∑[x] (x * x) * P(X = x)

c. To compute VE[X], we need to find the variance of X. Using the formula for variance, we have:

VE[X] = E[X^2] - (E[X])^2

Substituting the given probability mass function of X, we can calculate:

E[X^2] = ∑[x] (x^2) * P(X = x)

E[X^2] = (0^2 * 2) + (1^2 * 100) + (10^2 * 10000)

= 0 + 100 + 1000000

= 1000100

E[X] = ∑[x] x * P(X = x)

E[X] = (0 * 2) + (1 * 100) + (10 * 10000)

= 100010

VE[X] = E[X^2] - (E[X])^2

= 1000100 - (100010)^2

= 1000100 - 10002000100

= -10001999900

Therefore, VE[X] is -10001999900.

By comparing the values, we can see that E[VF] = E[X] ≥ 1, and E[VE[X]] = -10001999900.

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The answer to the question is 56. The sequence appears to alternate between adding and multiplying by a certain equation number.

From 20 to 5, we multiplied by 0.25 (or divided by 4). From 5 to 30, we added 25. From 30 to 6, we multiplied by 0.2 (or divided by 5). From 6 to 42, we added 36.  Therefore, to continue the pattern, we need to multiply 7 by a certain number and then add another number. It turns out that if we multiply 7 by 8, we get 56. Adding 49 to 56 gives us the next number in the sequence: 105.

So, the long answer is that the number which best completes the sequence is 56, and the next number in the sequence after that would be 105. The sequence can be split into two sub-sequences: 20, 30, 42 and 5, 6, 7. The first sub-sequence represents an increasing series of even numbers (20, 30, 42) with a difference of 10, 12, and so on. The second sub-sequence represents a consecutive series of odd numbers (5, 6, 7). Combining these sub-sequences forms the original sequence.

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The interval(s) of θ where tan(θ) > 0 are 0 < θ < π/2 and π < θ < 3π/2.

To determine the interval(s) of θ where tan(θ) > 0, we need to consider the sign of the tangent function in different quadrants of the unit circle.

Recall that the tangent function is positive in the first and third quadrants of the unit circle.

In the first quadrant (0 < θ < π/2), tan(θ) > 0.

In the third quadrant (π < θ < 3π/2), tan(θ) > 0.

Therefore, the correct answer is:

a. 0 < θ < π/2

c. π < θ < 3π/2

So, the interval(s) of θ where tan(θ) > 0 are 0 < θ < π/2 and π < θ < 3π/2.

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a) Confidence Interval = (X₁ - X₂) ± t x √[(s1² / n1) + (s2² / n2)]

b) The sample sizes are large enough (typically considered to be at least 30) or the populations are normally distributed.

c) C.I. = -11854.4100434 and -44145.5899566

a) The formula that should be used to compute the interval for the difference between the average salaries of male and female neurologists is:

Confidence Interval = (X₁ - X₂) ± t x √[(s1² / n1) + (s2² / n2)]

where:

X₁ and X₂ are the sample means of the salaries for female and male neurologists, respectively.

s1 and s2 are the sample standard deviations of the salaries for female and male neurologists, respectively.

n1 and n2 are the sample sizes for female and male neurologists, respectively.

t is the critical value from the t-distribution based on the desired confidence level and the degrees of freedom.

b) The assumptions that need to be met in order to use the above formula are:

The samples are simple random samples from their respective populations.

The populations from which the samples are drawn are approximately normally distributed.

The standard deviations of the populations are unknown.

The sample sizes are large enough (typically considered to be at least 30) or the populations are normally distributed.

c) To compute the interval, we need to calculate the critical value (t) based on the desired confidence level and the degrees of freedom, which is the sum of the sample sizes minus 2 (n1 + n2 - 2).

Given that we want a 99% confidence interval, the corresponding significance level (α) is 0.01. Degrees of freedom = n1 + n2 - 2 = 18 + 21 - 2 = 37.

Using a t-table or a statistical software, the critical value for a 99% confidence level with 37 degrees of freedom is approximately 2.708.

Plugging in the values into the formula:

Confidence Interval = ($175,000 - $203,000) ± 2.708 x √[($15,000² / 18) + ($22,000² / 21)]

= -28000 ± 16145.5899566

= -28000 + 16145.5899566 and -28000 - 16145.5899566

= -11854.4100434 and -44145.5899566

d) Assuming that both populations are normally distributed and that the two population variances are unequal, the formula used to compute the interval is the one described in part (a):

Confidence Interval = (X₁ - X₂) ± t x √[(s1² / n1) + (s2² / n2)]

This formula takes into account the sample means, sample standard deviations, and sample sizes for both groups.

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False, if an overall F-test in an ANOVA model is significant, it is important to conduct follow-up comparisons to test for differences between pairs of means.

When the overall F-test in an ANOVA model is found to be significant, it indicates that there is evidence of at least one significant difference among the group means. However, it does not provide specific information about which particular group means are different from each other. Therefore, follow-up comparisons, such as post hoc tests or pairwise comparisons, are necessary to determine the specific pairs of means that are significantly different.

These follow-up comparisons allow for a more detailed understanding of the group differences and help identify which specific groups are driving the significant overall F-test result. By conducting these additional tests, researchers can gain insights into the specific pairwise differences and make more accurate and informed interpretations of their data.

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we have 90% confidence interval for μ so the answer will be: B) μ=40±1.638(4)

To calculate the confidence interval, we use the formula:

Confidence Interval = sample mean ± (critical value) × (standard error)

The critical value is determined based on the desired confidence level and the sample size. In this case, with a 90% confidence level and a sample size of 4, the critical value is 1.638.

The standard error is calculated as the square root of the sample variance divided by the square root of the sample size. Since the sample variance is given as 16 and the sample size is 4, the standard error is 2.

Plugging in the values, we get:

Confidence Interval = 40 ± 1.638 × 2

Simplifying, we have:

Confidence Interval = 40 ± 3.276

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From the cosine function Transformation, the amplitude and period range for cosine function y= -4 cos 8x, are -4 and [0,2π] respectively.

The changes to the amplitude, period, and midline are called transformations of the basic sine and cosine function form. The standard forms for the cosine function is y = a cos(x − h) + k , where a is the amplitude,

The period of a periodic function is called a interval of x-values on which the cycle of the graph which repeated in both directions lies.

The amplitude of the graph of y= acos(bx) is the amount by which it varies above and below the x -axis. We have a cosine function, y = -4 cos 8x --(1). We have to determine the period range and amplitude of this function. Let's see the graph of function y, present in attached figure. From all discussion, the amplitude of function y is - 4. The periodic range for cosine function is 0 to 2π. Hence, required interval value is [0, 2π].

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Answer:

period [tex]\frac{1}{4}[/tex]π, amplitude 4

Step-by-step explanation:

Answer:

Step-by-step explanation:

(-5,8)

(2,1)

(0,1)

(-3,-20)

2. Evaluate the following expression: -1(5,-1) + 2(1,1)

(5,-2)

(-3,3)

(-5,2)

(7,1)

3. Evaluate the following expression (1,4) -5(1,1)

4. Let v=(8,-4) and w=(-4,2). Which of the following is true?

V * V =40

The x-component of V is 4

v= -2w

The y-component of w is 2

5. Which of the following vectors are orthogonal to (-1,3)? Check all that apply.

(1,3)

(-2,-3)

(3,1)

(-6,-2)

The correct statement is option D: "The p-value is less than or equal to 0.05."

The significance level, also known as the alpha level, is the threshold used to determine whether the results of a statistical test are statistically significant. In this case, the test resulted in the rejection of the null hypothesis at a significance level of 0.05.

The p-value is a measure of the strength of evidence against the null hypothesis. It represents the probability of observing a test statistic as extreme as, or more extreme than, the one obtained if the null hypothesis is true. If the p-value is less than or equal to the chosen significance level (0.05 in this case), it indicates that the evidence is statistically significant and supports the rejection of the null hypothesis.

Therefore, the correct statement is that the p-value is less than or equal to 0.05. Option A is not necessarily true because the results may not be statistically significant at the 10% level. Option B is also not necessarily true because the results may not be statistically significant at the 1% level. Option C is incorrect as it contradicts the fact that the null hypothesis was rejected at the 0.05 significance level.

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the diagonal length is approximately 0.0686 units.

What are Polynomials?

Polynomials are mathematical expressions consisting of variables, coefficients, and exponents. They are widely used in various fields of mathematics, science, engineering and even computer science.

To determine the dimension of the subspace H, we need to find the number of linearly independent vectors that span the subspace. The dimension of a subspace is equal to the number of vectors in any basis for that subspace.

First, let's check if the vectors in H are linearly independent by setting up a system of equations:

a(10x^2 - 12x - 13) + b(13x - 4x^2 + 9) + c(5x^2 - 7x - 7) = 0

Expanding and collecting like terms:

(5c - 4b + 10a)x^2 + (-7c + 13b - 12a)x + (-7c + 9b - 13a) = 0

For this equation to hold true for all values of x, the coefficients of each power of x must be zero. We can set up a system of equations:

5c - 4b + 10a = 0 (1)

-7c + 13b - 12a = 0 (2)

-7c + 9b - 13a = 0 (3)

We can solve this system of equations to determine if there are any non-trivial solutions. However, we can also observe that the determinant of the coefficient matrix is non-zero:

| 10 -4 5 |

| -12 13 -7 | = 76

| -13 9 -13 |

Since the determinant is non-zero, the system of equations has a unique solution, which means the vectors in H are linearly independent.

Therefore, a basis for the subspace H is {10x^2 - 12x - 13, 13x - 4x^2 + 9, 5x^2 - 7x - 7}.

the diagonal length is approximately 0.0686 units.

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The Cartesian product of two C¹ surfaces, denoted as M x M1 by M2, where M1 and M2 are surfaces of dimensions m1 and m2 in [tex]R^{n_{1} }[/tex]and [tex]R^{n_{2} }[/tex]respectively, is a C¹ surface of dimensions m1 * m2 in [tex]R^{n_{1} +n_{2} }[/tex]. The tangent space of M x M1 by M2 at a point can be expressed in terms of the tangent space.

Consider two C¹ surfaces, M1 in [tex]R^{n_{1} }[/tex] and M2 in [tex]R^{n_{2} }[/tex], with dimensions m1 and m2 respectively. The Cartesian product of these surfaces, denoted as M x M1 by M2, is obtained by taking every point (p, q) where p belongs to M1 and q belongs to M2. This results in a new surface of dimensions m1 * m2.

To understand the tangent space of M x M1 by M2 at a specific point, we need to consider the tangent spaces of M1 and M2 at their respective points. Let's denote the tangent space of M1 at a point p as Tp(M1), and the tangent space of M2 at a point q as Tq(M2).

The tangent space of M x M1 by M2 at a point (p, q) can be expressed as the Cartesian product of Tp(M1) and Tq(M2). In other words, it can be written as Tp(M1) x Tq(M2). This means that the tangent space of the Cartesian product surface is obtained by taking every combination of tangent vectors from Tp(M1) and Tq(M2).

Overall, the Cartesian product of two C¹ surfaces, M x M1 by M2, is a C¹ surface of dimensions m1 * m2 in [tex]R^{n_{1} +n_{2} }[/tex] . The tangent space of M x M1 by M2 at a point (p, q) is expressed as the Cartesian product of the tangent spaces of M1 and M2 at points p and q, respectively.

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If the number of units expected to be sold is uniformly distributed between 78 and 120, we can use the formula for generating a random number within a given range to express the sales.

Let's denote the random number between 0 and 1 as r. We can calculate the sales using the following expression:

Sales = (120 - 78) * r + 78

In this expression, (120 - 78) represents the range of the uniform distribution (42), and we multiply it by the random number r. Then we add the lower bound of the distribution (78) to obtain the sales value.

By substituting different values of r between 0 and 1, we can generate a random sales value within the range of 78 to 120, following a uniform distribution.

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A Poisson random variable represents the number of occurrences of an event in a fixed interval of time or space. It is a discrete random variable, which means that it can only take on integer values, starting from zero. Therefore, the smallest numerical value that a Poisson random variable can be is zero.

This means that there is a possibility that the event will not occur at all during the given interval. For example, if we are counting the number of customers who visit a store in an hour, it is possible that no customers show up during that hour, resulting in a Poisson random variable of zero.

However, the probability of this occurring depends on the average rate of the event occurring, which is denoted by the parameter λ in the Poisson distribution. The larger the value of λ, the smaller the probability of a Poisson random variable being zero.

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Therefore, according to the squeeze theorem, the limit of x^2 cos(1/x^2) as x approaches 0 is also 0: lim(x→0) x^2 cos(1/x^2) = 0.

To prove that lim(x→0) x^2 cos(1/x^2) = 0, we can use the squeeze theorem.

First, we establish the following inequalities:

-1 ≤ cos(1/x^2) ≤ 1

Since -1 ≤ cos(1/x^2) ≤ 1 for all values of x, we can multiply each side of the inequality by x^2 to obtain:

-x^2 ≤ x^2 cos(1/x^2) ≤ x^2

Now, we need to evaluate the limits of the lower and upper bounds as x approaches 0:

lim(x→0) -x^2 = 0

lim(x→0) x^2 = 0

Since both lower and upper bounds approach 0 as x approaches 0, we can conclude that the function x^2 cos(1/x^2) is "squeezed" between these two functions.

Thus, the statement is proven.

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Therefore, the equation x^3 - 15x + c = 0 has at most one root in the interval (-2, 2).

To show that the equation x^3 - 15x + c = 0 has at most one root in the interval (-2, 2), we can use the concept of the Intermediate Value Theorem and Rolle's Theorem.

Let's assume that the equation has two distinct roots, denoted as a and b, in the interval (-2, 2). Without loss of generality, we assume a < b.

Since the function is continuous on the closed interval [-2, 2] and differentiable on the open interval (-2, 2), we can apply Rolle's Theorem. According to Rolle's Theorem, there exists a point c in the open interval (a, b) such that the derivative of the function at c is zero.

Consider the derivative of the function f(x) = x^3 - 15x + c:

f'(x) = 3x^2 - 15

Setting f'(c) = 0, we have:

3c^2 - 15 = 0

c^2 - 5 = 0

c^2 = 5

Taking the square root of both sides, we get:

c = ±√5

Now, let's consider the function values at the endpoints of the interval (-2, 2):

f(-2) = (-2)^3 - 15(-2) + c = -8 + 30 + c = 22 + c

f(2) = (2)^3 - 15(2) + c = 8 - 30 + c = -22 + c

If c = √5, then f(-2) = 22 + √5 and f(2) = -22 + √5.

If c = -√5, then f(-2) = 22 - √5 and f(2) = -22 - √5.

In either case, the function values at the endpoints have different signs. This implies that there exists at least one value, say k, in the interval (-2, 2) such that f(k) = 0, according to the Intermediate Value Theorem.

However, we assumed at the beginning that there are two distinct roots in the interval (-2, 2), denoted as a and b. This contradicts our finding that there is at most one root in the interval. Hence, our assumption of having two distinct roots is false.

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According to the Question the area of one of the penny's faces increases by 0.135% when its temperature is increased by 100°C.

When the temperature of a copper penny is increased by 100°C, its diameter increases by 0.17%. However, to determine the change in the area of one of its faces, we need to use the formula for the area of a circle, which is πr². Since the radius of the penny changes with the increase in temperature, we can use the formula for the change in area of a circle, which is 2πrΔr. Using the percentage change in diameter (0.17%), we can find the corresponding percentage change in radius (which is half the diameter) by dividing 0.17 by 2, which gives us 0.085%. We can then use this percentage to calculate the change in the area of one of the penny's faces as follows:

Change in area = 2πrΔr = 2π(0.5r)(0.085% of 0.5r)

= 0.00135πr²

Therefore, the area of one of the penny's faces increases by 0.135% when its temperature is increased by 100°C.

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The order that the friends are walking in from the first person on the trail is; Jamie, Chris, Tony, Garbanzo, Mr. Peculis

The list of the friends in order from oldest to youngest is; Jamie, Tony, Mr. Peculis, Chris, Garbanzo

An order is an arrangement or disposition of items in relation to each other.

The data in the table indicates, that the word problem can be analyzed as follows;

The number of friends Mr. Peculis has = Four friends, therefore, there are five people on the hiking trip

The number of older people in front of Garbanzo  = 3

The number of older people behind Garbanzo = 1

Therefore, Garbanzo is the youngest of the five people walking in the second to the last position.

The number of older people in front of Jamie = 0

The number of older people in behind Jamie = 0

Therefore, Jamie is the oldest person, of the five friends, and is the only person that has no one walking in front of him, therefore, Jamie is in the first position

The number of older people in front of Chris = 1

The number of older people in behind Chris = 2

Therefore;

Chris is the second youngest person, and Chris is in front of Garbanzo, such that Chris is in the second position, on the hiking trail

The number of people older than Tony, indicates that Tony is the second oldest person

The 2 number of people older than Mr. Peculis, indicates that he is the third oldest person, and the other two older people are in front of him along with the two younger people, such that Mr. Peculis is walking at the back of the line, which indicates that Tony is in the third position

The order is therefore; Jamie, Chris, Tony, Garbanzo, and Mr. Peculis

The order from oldest to youngest is; Jamie, Tony, Mr. Peculis, Chris, Garbanzo

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The total surface area of the given triangular prism is 204 cm².

To find the total surface area of a triangular prism, we need to calculate the areas of each individual face and then sum them up.

Given that the dimensions are not to scale, we'll consider the following measurements:

Base of the triangular face: 10 cm

Height of the triangular face: 6 cm

Length of the prism: 8 cm

First, let's find the area of the triangular faces:

Area of one triangular face = (1/2) × base × height

= (1/2) × 10 cm × 6 cm

= 30 cm²

Since there are two triangular faces, the total area of the triangular faces is 2 × 30 cm² = 60 cm².

Next, let's find the area of the rectangular faces:

Area of one rectangular face = length * height

= 8 cm × 6 cm

= 48 cm²

Since there are three rectangular faces, the total area of the rectangular faces is 3 × 48 cm² = 144 cm².

Finally, to find the total surface area of the prism, we add the areas of the triangular and rectangular faces:

Total surface area = Area of triangular faces + Area of rectangular faces

= 60 cm² + 144 cm²

= 204 cm²

Therefore, the total surface area of the given triangular prism is 204 cm².

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Question

The diagram shows the sketch of a net of a triangular prism . 10 cm not to scale  6 cm 8 cm 15 work out the total surface area of the prism. X10-Dhx L xb2%x.

Answer:

(2,3)

Step-by-step explanation:

y = x/2 + 2 = 0.5x + 2.

also y = x +1.

so 0.5x + 2 = x + 1.

2 -1 = x - 0.5x

1 = 0.5x

x = 2.

y?

y = x + 1 = 2 + 1 = 3.

so (2, 3) is the coordinate solution

The SI for travel to St. Kitts using the ratio to centered moving average method are: Sep 0.12,Oct 0.10,Nov 0.08,Dec 0.06.

The ratio to centered moving average method is a simple moving average method that uses a centered moving average. The centered moving average is calculated by taking the average of the current value and the two values before and after it. The SI is then calculated by dividing the current value by the centered moving average. In the Excel file, the data for travel to St. Kitts is in the range A2:B13. The centered moving average is calculated in the range C2:C13. The SI is calculated in the range D2:D13. The following steps were used to calculate the SI for travel to St. Kitts using the ratio to centered moving average method: The centered moving average was calculated for each month. The SI was calculated for each month by dividing the current value by the centered moving average. The following are the results of the calculation:  Sep 0.12,Oct 0.10,Nov 0.08,Dec 0.06.

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Step-by-step explanation:

- 1/2 x + 6 > - 12        add 12 to both sides of the equation

 - 1/2x + 18 > 0          add 1/2 x to both sides

        18 > 1/2 x          multiply both sides by two

         36 > x        or     x < 36          Done.

The correct answer is B) 11.30 ounces. To find the mean setting for the soft drink dispenser so that a 12-ounce cup will overflow less than 1% of the time, we need to determine the z-score corresponding to a cumulative probability of 0.99.

Since we assume a normal distribution, we can use the z-score formula:

z = (x - μ) / σ

where:

z is the z-score

x is the value we want to find the z-score for (in this case, 12 ounces)

μ is the mean setting of the dispenser

σ is the standard deviation of the dispenser (0.3 ounce)

We want to find the z-score that corresponds to a cumulative probability of 0.99, which is 1% of the time.

Using a standard normal distribution table or calculator, we can find that the z-score corresponding to a cumulative probability of 0.99 is approximately 2.33.

Now, let's plug in the values into the z-score formula and solve for μ:

2.33 = (12 - μ) / 0.3

Rearranging the formula:

12 - μ = 2.33 * 0.3

12 - μ = 0.699

μ = 12 - 0.699

μ ≈ 11.301

Rounding to two decimal places, the mean setting of the dispenser should be approximately 11.30 ounces.

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Answer:

That's 43 cents ( D ).