use the dual simplex method to find a solution to the linear programming problem formed by adding the constraint 3xi 5x3> 15 to the problem in example 2.

Based on this sample of 23 doors. Yes, there is evidence at the 0.05 level that the doors are too long and need to be trimmed

.Based on the given information, we can conduct a hypothesis test to determine if there is evidence at the 0.05 level that the doors are too long and need to be trimmed.

First, we need to set up our null and alternative hypotheses.

Null hypothesis: The mean height of the doors is equal to or less than 2058.0 millimeters.
Alternative hypothesis: The mean height of the doors is greater than 2058.0 millimeters.

We can use a one-sample t-test to test these hypotheses, since we have a sample mean and standard deviation.

Using a t-test calculator or software, we can find the t-value and the p-value. The t-value is calculated by:

t = (sample mean - null hypothesis mean) / (standard deviation / square root of sample size)

Plugging in the values, we get:

t = (2071.0 - 2058.0) / (29.0 / √23) = 4.06

Using a t-table or software, we can find the p-value associated with this t-value and degrees of freedom (df = n-1 = 22).

At a significance level of 0.05, our critical t-value is 1.717. Since our calculated t-value (4.06) is greater than the critical t-value, we can reject the null hypothesis.

The p-value associated with our t-value is very small, less than 0.0001. This means that there is strong evidence to suggest that the doors are too long and need to be trimmed.

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True. Time series data refers to a collection of observations gathered over time, where the time dimension is a critical component of the data.

Each data point is linked to a specific point in time. In contrast, cross-sectional data is collected at a single point in time and does not have a time dimension. Therefore, the timing of each observation is crucial in time series data but not as important in cross-sectional data.

Time series data is a sequence of data points indexed in time order. It is used to track change over time.

Cross-sectional data is a snapshot of data at a specific point in time. It is used to compare different groups or variables .

Think about how the time at which each observation is made affects the analysis of time series data and cross-sectional data.

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The area of the resulting surface is (2/3)π[tex](e^(3/2) - 2)[/tex], or approximately 58.87 square units.

To find the area of the resulting surface, we need to use the formula for surface area of a solid of revolution.

The formula is:

S = 2π∫[tex]a^b f(x)√(1 + [f'(x)]^2) dx[/tex]

Where a and b are the limits of integration, f(x) is the function being rotated (in this case, y = ex), and f'(x) is the derivative of f(x) with respect to x.

In this case, we have:

f(x) = ex
f'(x) = ex

So, the integral becomes:

S = 2π∫[tex]0^20 ex √(1 + e^2x) dx[/tex]

This integral can be solved using u-substitution, where [tex]u = 1 + e^2x.[/tex]

[tex]du/dx = 2e^2x \\dx = du/(2e^2x)[/tex]

So the integral becomes:

[tex]S = π∫1^e (1/2)(u-1/2) du\\S = π[(2/3)u^(3/2) - (2/3)]_1^e\\S = π[(2/3)e^(3/2) - (2/3) - (2/3)]\\S = (2/3)π(e^(3/2) - 2)\\[/tex]

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Answer:

Step 1: Identify the given solid figure. Step 2: Identify the faces and side lengths of the given solid figure. Step 3: Using the side lengths and shape of the faces, draw each face of the solid figure on a plane and mark the corresponding side length. You will get the net of the solid figure.

Based on the given information, it seems that the number of left-handed desks in each classroom is adequate.

Based on the given information, we can calculate the expected number of left-handed students in a classroom:

Expected number of left-handed students = (total number of students in a classroom) x (proportion of left-handed students in the population)

Expected number of left-handed students = 30 x 0.11

Expected number of left-handed students = 3.3

Since each classroom contains four left-handed desks, it seems that the number of left-handed desks is more than enough to accommodate the expected number of left-handed students. In fact, there may even be some unused left-handed desks in each classroom.

Therefore, based on the given information, it seems that the number of left-handed desks in each classroom is adequate.

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If a two-factor analysis of variance produces a statistically significant interaction, it means that the effect of one independent variable on the dependent variable changes depending on the level of the other independent variable.

In other words, the effect of one independent variable on the dependent variable is not the same across all levels of the other independent variable. Therefore, the interaction term is necessary to properly model the relationship between the independent variables and the dependent variable. It is important to interpret the interaction carefully to fully understand the relationship between the variables being studied.

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The length of the hypotenuse is 4.

Let the three sides of the right triangle be a, b, and c, where a is the shortest side, b is the middle side, and c is the hypotenuse. Since the sides form a geometric progression, we know that:

b/a = c/b

Multiplying both sides by b, we get:

[tex]b^2/a = c[/tex]

Since a = 2, we can substitute this into the above equation to get:

[tex]b^2/2 = c[/tex]

We also know that the Pythagorean theorem applies, so we have:

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

Substituting a = 2 and [tex]c = b^2/2[/tex] , we get:

[tex]2^2 + b^2 = (b^2/2)^2[/tex]

Simplifying this equation, we get:

[tex]4 + b^2 = b^4/4[/tex]

Multiplying both sides by 4, we get:

[tex]16 + 4b^2 = b^4[/tex]

Rearranging and factoring, we get:

[tex](b^2 - 4)(b^2 - 12) = 0[/tex]

Since b is a positive length, we can discard the solution[tex]b^2 = 12.[/tex] Therefore, we have:

[tex]b^2 = 4[/tex]

Taking the square root of both sides, we get:

[tex]b = 2\sqrt{2 }[/tex]

Finally, we can use the equation [tex]c = b^2/2[/tex] to find the length of the hypotenuse:

[tex]c = (2\sqrt{2} )^2/2 = 4.[/tex]

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2.23% is the probability of a city with a population of 200,000, and there will be exactly 2 accidental drownings per year.

The Poisson distribution, is useful for modeling the number of events (in this case, accidental drownings) in a fixed interval (here, a year). Given the average rate of 3.0 drownings per 100,000 people per year, we first need to find the expected number of drownings in a city with a population of 200,000.

Expected drownings per year = (3.0 drownings / 100,000 people) * 200,000 people = 6 drownings

Now, we can use the Poisson probability formula to calculate the probability of exactly 2 accidental drownings per year in this city:

P(X = k) = (λ^k * e^(-λ)) / k!

Here, X represents the number of drownings, k is the desired number of drownings (2 in this case), λ (lambda) is the expected number of drownings per year (6), e is the base of the natural logarithm (approximately 2.718), and k! is the factorial of k.

P(X = 2) = (6^2 * e^(-6)) / 2! = (36 * e^(-6)) / 2 = (36 * 0.002478) / 2 = 0.04461 / 2 = 0.022305

Therefore, the probability that in a city with a population of 200,000, there will be exactly 2 accidental drownings per year is approximately 0.0223 or 2.23%.

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The value of {x} is 8 and the measure of angle A is 30°.

Refer to the image attached. We can see that ∠A and ∠B form a pair of co - interior angles. This means that we can write the relation as -

∠A + ∠B = 180°

So, we can write that -

6x - 18 + 14x + 38 = 180°

20x + 20 = 180°

20x = 160°

{x} = 8

∠A = 6x - 18

∠A = 6 x 8 - 18

∠A = 30°

So, the value of {x} is 8 and the measure of angle A is 30°.

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To design a good test for this hypothesis, we need to first understand the hypothesis itself. Let's say that our hypothesis is that a certain medication reduces the symptoms of a particular disease. To test this hypothesis, we would need to design a study that meets the conditions specified in the question.

First, we need to ensure that our study is designed in a way that allows us to test the hypothesis in a controlled environment. This means that we would need to recruit participants who have the disease and randomly assign them to either the medication group or a placebo group. This will help us to control for potential confounding variables such as age, gender, and disease severity.

Next, we would need to measure the symptoms of the disease in both groups at the beginning of the study and at regular intervals throughout the study. This will allow us to compare the symptom scores between the two groups and determine whether the medication is having an effect.

To meet the conditions specified in the question, we would need to ensure that the evidence we collect is far more likely given that the hypothesis is true than it would be if the hypothesis were false. This means that we would need to calculate the probability of the evidence (i.e. symptom scores) given that the hypothesis is true (Pr(E|H)) and compare it to the probability of the evidence given that the hypothesis is false (Pr(E|~H)).

If the medication is truly effective, we would expect to see a significant difference between the symptom scores of the medication group and the placebo group. In this case, the evidence we collect would be far more likely given that the hypothesis is true than it would be if the hypothesis were false, meeting the first condition specified in the question.

To avoid potential confounders, we would need to ensure that our study is designed in a way that controls for other factors that could affect symptom scores. For example, we could control for diet and exercise habits, as well as other medications that the participants may be taking.

In conclusion, to design a good test for this hypothesis, we would need to design a controlled study that measures symptom scores in both the medication group and the placebo group, and calculate the probability of the evidence given that the hypothesis is true and false. We would also need to control for potential confounders to ensure that our results are valid.

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If the sample selection was perfect, the difference between the sample and population averages would be zero.

We have,

The sample would be representative of the population, and the statistics calculated from the sample would accurately reflect the characteristics of the population.

However, in practice, it is often difficult to achieve a perfectly representative sample.

Sampling bias, which occurs when some members of the population are more likely to be selected than others, can lead to differences between the sample and population averages.

Therefore, it is important to use appropriate sampling techniques and methods to minimize sampling bias and ensure that the sample is as representative of the population as possible.

Thus,

If the sample selection was perfect, the difference between the sample and population averages would be zero.

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The statement that best describes the result of the data that Maya collected, would be D. Small - size drinks cost more than $ 1. 00 at restaurants in Maya's city.

We see that in Maya's city, there was no entry in the $ 0.91 to $ 1.00 category. This shows that there are no restaurants (according to the sample) that sell small - sized drinks for less than $ 1.00

The other statements are false because the largest restaurants fall in the $ 1. 51 to $ 1. 60 category and Maya conducted the sampling at 50 restaurants because there were 10 restaurants per sample.

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The true population mean of monthly incomes for workers in the factory falls between $1806 and $2263 with 98% of confidence intervals.

To calculate the 98% confidence interval for the mean of the population, we need to use the formula:

CI = sample mean ± Zα/2 * (σ/√n)

Where the sample mean, Zα/2 is the critical value for the given level of confidence (98% in this case), σ is the population standard deviation, and n is the sample size.

First, we need to find the sample mean from the given data:

sample mean = (2120 + 1980 + 2300 + ... + 1940) / 20 = 2034.5

Next, we need to find the critical value (Zα/2) using a standard normal distribution table or calculator. For a 98% confidence interval, the critical value is approximately 2.33.

Now, we can plug in the values and calculate the confidence interval:

CI = 2034.5 ± 2.33 * (518/√20) = (1806, 2263)

Therefore, we can say with 98% confidence that the true population mean of monthly incomes for workers in the factory falls between $1806 and $2263. This means that if we were to take many random samples of the same size from the population and calculate their confidence intervals, approximately 98% of these intervals would contain the true population mean.

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It takes approximately 52.7 hours for the activity of a sample of gallium-67 to fall to 6.73 percent of its original value.

The decay of radioactive substances is governed by the following equation:

[tex]N(t) = N_{0} e^{(-\lambda t)[/tex]

where:

N(t) is the amount of the substance at time t

N₀ is the initial amount of the substance

λ is the decay constant

t is time

The half-life of gallium-67 is 78.2 hours, which means that:

λ = ln(2)/t₁/₂ = ln(2)/78.2 = 0.00887 h⁻¹

Let N be the amount of gallium-67 remaining after time t, and N₀ be the initial amount. Then, we can use the above equation to find the time t required for the activity of the sample to fall to 6.73 percent of its original value:

N/N₀ = 0.0673

[tex]0.0673 = e^{(-\lambda t)}[/tex]

Taking the natural logarithm of both sides:

ln(0.0673) = -λt

t = ln(1/0.0673)/λ

t = ln(14.84)/0.00887

t ≈ 52.7 hours

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Dakota rode his bike for 60 km. He understood that he could have been there 16 hours earlier if he had traveled 12 mph quicker. Dakota's actual speed during the bike ride was approximately 41.8 mph.

Let's assume that Dakota's actual speed during the bike ride was x miles per hour.

Using the distance formula:

distance = speed x time

We can calculate the time it took Dakota to complete the ride at his actual speed as:

60 = x * t

where t is the time in hours.

Now, according to the problem, if he had gone 12 mph faster, he would have arrived 16 hours sooner. This means that he would have covered the same distance in less time.

So we can set up another equation:

60 = (x + 12) * (t - 16)

Simplifying this equation:

60 = xt - 16x + 12t - 192

76 = xt - 16x + 12t

Substituting the value of t from the first equation, we get:

76 = 60 - 16x + 12(60/x)

Simplifying this equation:

[tex]16x^2 - 720x + 7200 = 0[/tex]

Dividing both sides by 16:

[tex]x^2 - 45x + 450 = 0[/tex]

Solving this quadratic equation using the quadratic formula:

[tex]$x = \frac{45 \pm \sqrt{45^2 - 41450}}{2}$[/tex]

[tex]$x = 22.5 \pm 7.5\sqrt{5}$[/tex]

We can reject the negative root because it doesn't make sense in the context of the problem.

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The dimensions of the cylindrical mailing tube of greatest volume that may be mailed using the US Postal Service are approximately:

Diameter: 3.45 inches

Length: 18.7 inches
Volume: 123.8 cubic inches.

To find the dimensions of the cylindrical mailing tube of greatest volume that may be mailed using the US Postal Service, we need to consider the dimensional restrictions set by the Postal Service. According to their regulations, the length and girth (circumference plus the diameter) of a package must not exceed 108 inches combined.

Let's assume that the tube has a diameter of "d" and a length of "L". The girth of the package is the sum of the circumference of the circular ends and the diameter of the tube, so we have:

Girth = 2π(d/2) + d = πd + d

The length and girth must satisfy the inequality:

L + Girth ≤ 108

Substituting the expression for Girth, we get:

L + πd + d ≤ 108

Solving for L in terms of d, we get:

L ≤ 108 - (π + 1)d

Now, let's consider the volume of the cylindrical tube, which is given by:

V = πr²L = (πd²/4) L

Substituting the expression for L, we get:

V = (πd²/4) (108 - (π + 1)d)

To find the maximum volume, we can take the derivative of V with respect to d, set it equal to zero, and solve for d:

dV/dd = 0

d[πd²/4 (108 - (π + 1)d)]/dd = 0

πd/2 - 3π²d/4 - π/4 + 27/4 = 0

Solving for d, we get:

d = 27/(4π - 6π²)

Substituting this value of d back into the expression for L and the equation for the maximum volume, we get:

L = 108 - (π + 1)d ≈ 18.7 inches

V = (πd²/4) (108 - (π + 1)d) ≈ 123.8 cubic inches

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The amount of money in the account after 15 years is approximately $596.69.

Compounding refers to the process of earning interest not only on the original principal amount but also on the interest that accumulates over time.

Using the formula V = Pert for continuous compounding, where V is the value of the account in t years, P is the principal initially invested, e is the base of a natural logarithm, and r is the rate of interest, we have:

[tex]V = Pe^{(rt)[/tex]

Plugging in the given values, we get:

[tex]V = 352e^{(0.035\times 15)[/tex]

Simplifying and evaluating, we get:

V ≈ $596.69

Therefore, the amount of money in the account after 15 years is approximately $596.69.

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The expected score for the optimal strategy is 4.25.

To determine the optimal strategy, we need to calculate the expected value of each option and choose the one with the higher expected value. Let's consider the two options:

Option 1: Keep the first roll

The expected score of this option is simply the average of all the possible outcomes of rolling the die once, which is (1+2+3+4+5+6)/6 = 3.5.

Option 2: Roll again

If we decide to roll again, we need to consider two possibilities:

We roll a value less than or equal to the first roll: In this case, we keep the first roll, and our score is the value of the first roll.

We roll a value greater than the first roll: In this case, we keep the second roll, and our score is the value of the second roll.

The probability of rolling a value less than or equal to the first roll is 1/2, and the probability of rolling a value greater than the first roll is also 1/2. Therefore, the expected score of this option is:

(1/2)(3.5) + (1/2)(1+2+3+4+5+6)/6 = (7/2 + 3.5)/2 = 5.25/2 = 2.625

Comparing the expected values of the two options, we can see that the expected value of rolling again is higher than the expected value of keeping the first roll. Therefore, the optimal strategy is to roll again if the first roll is less than or equal to 2.625, and keep the first roll otherwise.

The expected score for this optimal strategy can be calculated as follows:

If we roll a 1 or 2 on the first roll, we will roll again, and the expected score is (1+2+3+4+5+6)/6 = 3.5.

If we roll a 3, 4, 5, or 6 on the first roll, we will keep the first roll, and the expected score is the value of the first roll.

Therefore, the expected score for the optimal strategy is:

(1/6)(3.5) + (1/6)(3) + (1/6)(4) + (1/6)(5) + (1/6)(6) = 4.25

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The number of divisors counted twice is equal to the number of positive integers that have the form [tex]$2^a\cdot 3^b\cdot 5^c$[/tex] for some nonnegative integers a, b, and c such that and [tex]0 \le c \le 5[/tex]. There are [tex]11\cdot 11\cdot 6 = 726[/tex] such divisors.

We will first find the prime factorization of each of the given numbers: [tex]\begin{align*}12^{12} &= (2^2\cdot 3)^{12} = 2^{24} \cdot 3^{12}, \10^{10} &= 2^{10} \cdot 5^{10}, \15^{15} &= (3\cdot 5)^{15} = 3^{15}\cdot 5^{15}.\end{align*}[/tex]

For a positive integer n to be a divisor of at least one of these numbers, it must contain some combination of the prime factors 2, 3, and 5.

Any divisor of [tex]12^{12}[/tex] must have the form [tex]2^a\cdot 3^b[/tex]for some nonnegative integers a and b such that [tex]0 \le a \le 24[/tex] and [tex]0 \le b \le 12.[/tex] Therefore, there are [tex]25\cdot 13 = 325[/tex]possible divisors of [tex]12^{12}.[/tex]

Similarly, any divisor of [tex]$10^{10}$[/tex]must have the form[tex]$2^a\cdot 5^b$[/tex]for some nonnegative integers a and b such that [tex]$0 \le a \le 10$[/tex] and [tex]$0 \le b \le 10$[/tex]. Therefore, there are [tex]$11\cdot 11 = 121$[/tex]possible divisors of[tex]$10^{10}$.[/tex]

Finally, any divisor of [tex]$15^{15}$[/tex]must have the form for some nonnegative integers a and b such that [tex]$0 \le a \le 15$[/tex] and [tex]$0 \le b \le 15$[/tex]. Therefore, there are [tex]$16\cdot 16 = 256$[/tex] possible divisors of [tex]$15^{15}$[/tex].

We want to count the total number of positive integers that are divisors of at least one of these numbers. Notice that we have counted some divisors twice (for example, [tex]$2^1 \cdot 3^1$[/tex]is a divisor of both [tex]$12^{12}$[/tex] and[tex]$10^{10}$)[/tex], so we need to subtract the number of divisors that we have counted twice. Similarly, we have counted some divisors three times, so we need to add the number of divisors that we have counted three times. We have not counted any divisor four or more times, so we do not need to consider those cases.

The number of divisors counted twice is equal to the number of positive integers that have the form [tex]$2^a\cdot 3^b\cdot 5^c$[/tex] for some nonnegative integers a, b, and c such that [tex]$0 \le a \le 10$[/tex], [tex]$0 \le b \le 5$[/tex], and [tex]$0 \le c \le 10$[/tex]. There are [tex]$11\cdot 6\cdot 11 = 726$[/tex] such divisors (we chose the exponents for 2, 3, and 5 separately).

Finally, the number of divisors counted three times is equal to the number of positive integers that

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The area of the original trapezoid is B. half the area of the rectangle in step 4.

Given a trapezoid which has the lengths of the parallel bases as b₁ and b₂.

We know that,

Area of a trapezoid = (b₁ + b₂) h / 2, where h is the height between the two bases.

Two trapezoids are joined to get a parallelogram and then rearrange it to form a rectangle.

Area of a rectangle = Length × width

Here, length = b₁ + b₂

Width = height of the trapezoid = h

Area of rectangle in step 4 = (b₁ + b₂) h

Area of trapezoid is half of this.

Hence the correct option is B.

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It is nonsensical to talk about the direction of specific differences in ANOVA because it is an omnibus test.

An omnibus test is a statistical test that evaluates whether there is a difference between groups or conditions, without specifying which groups or conditions differ from each other. In the case of ANOVA (Analysis of Variance), it tests whether the means of two or more groups are equal or not. However, it does not tell us which specific groups are different from each other.

Instead, to identify the specific group differences, post-hoc tests such as Tukey's test, Bonferroni's test, or Scheffe's test can be used. These tests compare the means of each group and identify which groups differ significantly from each other.

Therefore, talking about the direction of specific differences in ANOVA is nonsensical because ANOVA does not provide information about which groups differ from each other. It only tells us whether there is a significant difference between groups or not.

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Answer:

Step-by-step explanation:

Assuming that the giant sugar cube is a perfect cube, we can calculate its volume by using the formula:

Volume = edge length³

To find the edge length, we need to first calculate the amount of wood used per unit of volume, which is given by:

wood per unit volume = 213.5 m³ / 1 unit of volume

We don't know the unit of volume of the sugar cube, but we can use any consistent unit, such as meters, centimeters, or millimeters. Let's use meters for consistency with the given material amount.

Now, we can find the edge length by solving the following equation for x:

wood per unit volume = 213.5 m³ / x³

x³ = 213.5 m³ / wood per unit volume

x³ = 213.5 m³ / (213.5 m³ / 1 unit of volume)

x³ = 1 unit of volume

Therefore, the edge length of the giant sugar cube is 1 meter.

Finally, we can calculate the volume inside the giant sugar cube by using the formula:

Volume = edge length³

Volume = 1³ = 1 cubic meter

Therefore, the volume inside the giant sugar cube is 1 cubic meter.

Step-by-step explanation:

C  rate =   1 chapter / 9 hr   = 1/ 9    chap/hr

P rate = 1/6

together  :

1 chapter / ( c rate + p rate)  =   1 /( 1/9 + 1/6) =  1/ ( 2/18 + 3/18) = 1/ (5/18) =

    18/5 hr   = 3  3/5 hr

It would take Cynthia and Phillip 3.6 hours working together to proof a chapter of Hawkes Learning Systems' Introductory Algebra book.

To find out how long it would take Cynthia and Phillip working together to proof a chapter of Hawkes Learning Systems' Introductory Algebra book, we can use the work formula:

Work = Rate × Time

First, we'll find the individual rates for Cynthia and Phillip:
- [tex]Cynthia's rate:\frac{1 chapter}{9 hours}[/tex]
- [tex]Phillip's rate:\frac{1 chapter}{6 hours}[/tex]
Now, we'll add their rates together to find their combined rate:
[tex]Combined rate = \frac{1}{9}  + \frac{1}{6}[/tex]

To add these fractions, we need a common denominator, which is 18:
[tex]Combined rate = \frac{2}{18}  + \frac{3}{18}=\frac{5}{18}[/tex]

Now, we'll use the work formula to find the time it would take for them to complete the proofreading together. Since they're working on 1 chapter, we can set Work equal to 1:
[tex]1 = \frac{5}{18} (time)[/tex]

Next, we'll solve for Time:
[tex]Time = 1 (\frac{5}{18})[/tex]
[tex]Time=1 (\frac{18}{5})[/tex]
[tex]Time = \frac{18}{5} = 3.6 hours[/tex]

So, it would take Cynthia and Phillip 3.6 hours working together to proof a chapter of Hawkes Learning Systems' Introductory Algebra book.

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According to the flood recurrence interval equation, a flood that occurs in 2021, after 149 years of record keeping, and is the third largest ever recorded has a recurrence interval of approximately 50 years.

The flood recurrence interval equation is a statistical method used to estimate the likelihood of a flood of a certain magnitude occurring in any given year. It is based on historical records of floods and takes into account the size and frequency of floods that have occurred in the past.

In this case, the flood that occurred in 2021 is the third largest ever recorded. Based on the historical records of floods over the past 149 years, this flood has a recurrence interval of approximately 50 years. This means that there is a 2% chance of a flood of this magnitude occurring in any given year.

It is important to note that the flood recurrence interval equation is not a perfect predictor of future floods. It is only an estimation based on historical records. Other factors such as climate change and changes in land use can also impact the likelihood and severity of floods.

In conclusion, the recurrence interval for a flood that occurred in 2021, after 149 years of record keeping and is the third largest ever recorded is approximately 50 years. However, it is important to remember that the accuracy of this estimation may vary based on several factors.

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8.) The surface area of the given shape would be =532ft²

9.) The surface area of the given circle = 24.62cm²

For question 8.)

To calculate the surface area of the square based pyramid the formula given below is used;

S.A = b² + 2bs

where;

b = 14ft

s = 12 ft

S.A = 14² + 2(14×12)

= 196+ 2(168)

= 196+336

= 532ft²

For question 9:

To calculate the area of the circle, the formula that should be used is given as follows:

S.A = πr²

where:

r = 2.8cm

π = 3.14

S.A = 3.14×2.8×2.8

= 24.62cm²

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Answer:

0.2515384615

Step-by-step explanation:

she makes 5/13

5/13*0.654

Thus, to calculate the probability of Derek selecting two striped socks at random, we need to know the number of striped socks and the total number of socks in his drawer. The formula to find the probability is (x/y) * ((x-1)/(y-1)).

To answer this question, we must first know the total number of socks in Derek's drawer and the number of striped socks among them.

The probability of selecting a striped sock can be calculated by dividing the number of striped socks by the total number of socks. Let's assume there are "x" striped socks and "y" total socks in the drawer.

For the first selection, the probability of choosing a striped sock is P1 = x/y.

After selecting one striped sock, there will be (x-1) striped socks and (y-1) total socks remaining in the drawer.
For the second selection, the probability of choosing another striped sock is P2 = (x-1)/(y-1).

To find the probability of selecting two striped socks in a row, we multiply the probabilities of each event occurring: P(both striped socks) = P1 * P2 = (x/y) * ((x-1)/(y-1)).

In summary, to calculate the probability of Derek selecting two striped socks at random, we need to know the number of striped socks and the total number of socks in his drawer. The formula to find the probability is (x/y) * ((x-1)/(y-1)).

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The value of an 8 in the tens place as a fraction of an 8 in the hundreds place is 1 / 10.

In the number system we use, each digit's value is determined by its position and is ten times the value of the digit to its right. Therefore, the value of an 8 in the tens place is 10 times the value of an 8 in the ones place.

Likewise, the value of an 8 in the hundreds place is 10 times the value of an 8 in the tens place, which in turn is 10 times the value of an 8 in the ones place. Therefore, the value of an 8 in the hundreds place is 10 x 10 times the value of an 8 in the ones place.

So in equation form, we have:

= 8 / 80

= 1 / 10

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To create a 90% confidence interval with a margin of error of 4%, the researcher should survey at least 424 students.

To calculate the sample size needed for a 90% confidence interval with a margin of error of 4% (0.04) when estimating the true proportion of students who have a job in addition to attending school, without a preliminary estimate, follow these steps:

1. Identify the desired confidence level (90%) and find the corresponding z-score (z-value). For a 90% confidence level, the z-score is 1.645.

2. Determine the margin of error, which is given as 4% or 0.04.

3. Since no preliminary estimate is available, use 0.5 (50%) as the proportion. This is the most conservative estimate and will result in the largest required sample size.

4. Use the formula for calculating the sample size (n) for proportions:

  n = (Z^2 * P * (1-P)) / E^2

  Where:
  - n is the required sample size
  - Z is the z-score (1.645 for a 90% confidence level)
  - P is the estimated proportion (0.5)
  - E is the margin of error (0.04)

5. Plug the values into the formula:

  n = (1.645^2 * 0.5 * (1-0.5)) / 0.04^2

6. Calculate the result:

  n ≈ 423.06

Since you cannot have a fraction of a student, round up the result to the nearest whole number:

  n = 424

Therefore, to create a 90% confidence interval with a margin of error of 4%, the researcher should survey at least 424 students.

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