Assessment Practice
14. Write the unit fraction that represents 1 square.
Then write the fraction that represents the
whole. Select numbers from the box to write
the fractions. Numbers may be used once, more than
once, or not at all. 3.FR.1.1
12 3 4 5 6 7 8
1 square:
Whole:

On solving the provided question, we can say that the equation will be

[tex]N = Fg * cos(15°) = 44.6 N * cos(15°) = 42.8 N[/tex]

A mathematical equation is a formula that links two assertions and signifies equivalence with the equals sign (=). In, a mathematical statement that proves the equality of two mathematical expressions is referred to as an equation. The equal sign, for instance, separates the variables [tex]3x + 5[/tex] and 14 in the equation [tex]3x + 5 = 14.[/tex]There is a mathematical formula that explains the link between the two phrases that appear on either side of a letter. Frequently, the symbol and the single variable are the same. like [tex]2x - 4 = 2[/tex], for example.

Since the weight is not moving, the sum of the three forces acting on it must be zero. This means that the force of gravity acting on the weight is balanced by the normal force and the force due to friction.

The force of gravity on the weight is given by:

[tex]Fg = m * g[/tex]

where m is the mass of the weight (which we can find by dividing the weight in pounds by the acceleration due to gravity, g), and g is the acceleration due to gravity, which is approximately [tex]9.81 m/s^2.[/tex]

[tex]m = 10.0 pounds / (2.205 pounds/kilogram) = 4.54 kg[/tex]

[tex]Fg = 4.54 kg * 9.81 m/s^2 = 44.6 N[/tex]

The normal force N acts perpendicular to the bench, so it can be found using trigonometry. The angle between the bench and the horizontal is 15 degrees, so the angle between the normal force and the horizontal is also 15 degrees.

[tex]N = Fg * cos(15°) = 44.6 N * cos(15°) = 42.8 N[/tex](rounded to one decimal place)

The force due to friction F acts parallel to the bench, in the opposite direction to the component of the weight that is parallel to the bench. This component can be found using trigonometry:

[tex]Fp = Fg * sin(15°) = 44.6 N * sin(15°) = 12.2 N[/tex]

Since the weight is not moving, the force due to friction must be equal and opposite to this component:

[tex]F = -Fp = -12.2 N[/tex] (rounded to one decimal place)

So the magnitude of the normal force is 42.8 N and the magnitude of the force due to friction is 12.2 N.

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Answer:

25

Step-by-step explanation: just like that

Answer: C

asdfasdasdasdfasdf

The appropriate method for the biologist to use for inference to the population is A) A one-sample t-interval for a population mean.

In this situation, the biologist wants to estimate the average weight of a population of dolphins living in a certain region of the ocean. The biologist will collect a random sample of dolphins and use the sample weights to create the estimate.

here, we have,

Therefore, the appropriate method for inference to the population is to use a one-sample t-interval for a population mean.

This method is used when we want to estimate the population mean using a sample mean and the standard deviation of the sample.

The t-interval takes into account the uncertainty of the estimate due to the random sampling process, which makes it an appropriate method for this situation.

Hence, The appropriate method for the biologist to use for inference to the population is A) A one-sample t-interval for a population mean.

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The correct question is:

A marine biologist wants to estimate the average weight of a population of dolphins living in a certain region of the ocean. The biologist will collect a random sample of dolphins and use the sample weights to create the estimate Which of the following is an appropriate method for the biologist to use for inference to the population?

A A one-sample t-interval for a population mean

B A one-sample t-interval for a sample mean

C A one-sample 2-interval for a population proportion

D A matched-pairs t-interval for a mean difference

E A two-sample t-interval for a difference between means

Answer:

he months where Jacy's hometown gets at least 10.5 centimeters of rainfall are July (x = 7) and all the months after July, which are August (x = 8), September (x = 9), October (x = 10), November (x = 11), and December (x = 12).

Step-by-step explanation:

The given function is:

f(x) = 0.2x^2 - 1.8x + 6.5

where x represents the number of months passed and f(x) represents the average monthly rainfall in centimeters.

To find the months where the average rainfall is at least 10.5 centimeters, we need to set the function f(x) greater than or equal to 10.5 and solve for x:

0.2x^2 - 1.8x + 6.5 ≥ 10.5

0.2x^2 - 1.8x - 4 ≥ 0

Multiplying both sides by 5, we get:

x^2 - 9x - 20 ≥ 0

We can factor the left-hand side of the inequality as:

(x - 5)(x - 4) ≥ 0

The solution to this inequality is the set of values of x that make the inequality true. This includes the intervals where:

(x - 5) ≥ 0 and (x - 4) ≥ 0, which gives x ≥ 5

OR

(x - 5) ≤ 0 and (x - 4) ≤ 0, which gives x ≤ 4

Thus, the months where Jacy's hometown gets at least 10.5 centimeters of rainfall are July (x = 7) and all the months after July, which are August (x = 8), September (x = 9), October (x = 10), November (x = 11), and December (x = 12).

Jacy's hometown gets at least 10.5 centimeters of rainfall in the months of September, September of the following year, September of the year after that, and so on.

An inequality is a relation between two numbers or expressions that are not equal.

It can show which of them is greater or smaller by using symbols like < or >. It can also be a statement of fact about the order relationship of quantities.

According to the question given,

We need to solve the inequality:

A(t) ≥ 10.5

Substituting the given function, we get:

2.3sin(π/6)t ≥ 10.5

sin(π/6)t ≥ 4.57.

Since the sine function has a maximum value of 1, the inequality is only satisfied when:

sin(π/6)t = 1

Solving for t, we get:

π/6t = π/2 + 2πk or π/6t = 3π/2 + 2πk

where k is any integer.

Simplifying each equation, we get:

t = 3 + 12k or t = 9 + 12k, where k is any integer.

Therefore, the only solution is given by:

t = 9 + 12k

where k is any integer.

Hence, Jacy's hometown gets at least 10.5 centimeters of rainfall in the months of September, September of the following year, September of the year after that, and so on.

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Answer:

-2

Step-by-step explanation:

Standard equation of a proportional relationship:

y = kx

We use the given point to find k.

4 = k × (-16)

k = 4/(-16)

k = -1/4

The equation of this proportional relationship is

y = -1/4 x

Now we use x = 8 in the equation just above to find its corresponding y value.

x = 8

y = -1/4 × 8

y = -2

Answer: -2

16%  percentage of the students have grade point averages that are greater than 2.11

Using the empirical rule, we know that for a bell-shaped distribution, approximately:

68% of the data falls within one standard deviation of the mean

95% of the data falls within two standard deviations of the mean

99.7% of the data falls within three standard deviations of the mean

To find the percentage of students with a grade point average greater than 2.11

We first need to calculate how many standard deviations away from the mean 2.11 is:

z = (2.11 - 2.56) / 0.45

= -1

This tells us that 2.11 is 1 standard deviation below the mean.

Since the distribution is symmetric.

The percentage of students with a GPA greater than 2.11 is the same as the percentage of students with a GPA less than 2.56 + 1*0.45, which is:

68% + 95% = 163%

So, approximately

100% - 163%

= 16% of the students have a GPA greater than 2.11.

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The area of the quadrilateral is 140 m^2.

For closed figure made by 4 line segments joined end to end in series is called a quadrilateral.

A parallelogram in which adjacent sides are perpendicular to each other is called a rectangle.

A rectangle is always a parallelogram and a quadrilateral but reverse statement could be not be true.

Area of the quadrilateral = 2 x area of the traingle

Thus, Area of the triangle = 1/2 x base  x height

Area of 1 triangle = 1/2 x 10 x 16

Area of 1 triangle = 80 m^2

Area of the quadrilateral = 2 x 80

Area of the quadrilateral = 140 m^2

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The complete question is;

What is the area of the quadrilateral if the dimensions of the quadrilateral  are 10cm, 16cm and 22cm

(Part 1)  The expression [tex]18\sqrt{8}[/tex] can be written in the form [tex]a\sqrt{b}[/tex] as [tex]36\sqrt{2}[/tex], where a and b are integers and b is as small as possible.

(Part 2)  The expression [tex]8\sqrt{18}[/tex] can be written in the form [tex]a\sqrt{b}[/tex] as [tex]16\sqrt{3}[/tex].

(Part 3)  The expression [tex]18\sqrt{8}[/tex] - [tex]8\sqrt{18}[/tex] + [tex]\sqrt{n}[/tex] is equal to ([tex]36\sqrt{2}[/tex] - [tex]16\sqrt{3}[/tex]) + [tex]\sqrt{n}[/tex].

(Part 1) Expressing [tex]18\sqrt{8}[/tex] in the form [tex]a\sqrt{b}[/tex], where a and b are integers and b is as small as possible:

We can factor [tex]\sqrt{8}[/tex] as [tex]\sqrt{2}[/tex]* [tex]\sqrt{4}[/tex]

So, [tex]18\sqrt{8}[/tex]= 18 * [tex]\sqrt{2}[/tex] *[tex]\sqrt{4}[/tex] = 18 *[tex]\sqrt{2}[/tex] * 2 = [tex]36\sqrt{2}[/tex]

So the expression [tex]18\sqrt{8}[/tex] can be written in the form [tex]a\sqrt{b}[/tex] as [tex]36\sqrt{2}[/tex], where a = 36 and b = 2.

(Part 2) Expressing [tex]8\sqrt{18}[/tex] in the form [tex]a\sqrt{b}[/tex], where a and b are integers and b is as small as possible:

We can factor [tex]\sqrt{18}[/tex] as [tex]\sqrt{3}[/tex] * [tex]\sqrt{6}[/tex]

So, [tex]8\sqrt{18}[/tex] = 8 * [tex]\sqrt{3}[/tex] * [tex]\sqrt{6}[/tex]

[tex]8\sqrt{18}[/tex]= 8 * [tex]\sqrt{3}[/tex] * [tex]\sqrt{2}[/tex] * [tex]\sqrt{3}[/tex]

[tex]8\sqrt{18}[/tex]= 8 *[tex]\sqrt{6}[/tex]* [tex]\sqrt{2}[/tex]

[tex]8\sqrt{18}[/tex]= 8 * [tex]\sqrt{2}[/tex] * [tex]\sqrt{6}[/tex]

[tex]8\sqrt{18}[/tex]= 8 * 2 * [tex]\sqrt{3}[/tex]

[tex]8\sqrt{18}[/tex]= [tex]16\sqrt{3}[/tex]

So the expression [tex]8\sqrt{18}[/tex] can be written in the form [tex]a\sqrt{b}[/tex] as [tex]16\sqrt{3}[/tex], where a = 16 and b = 3.

(Part 3) [tex]18\sqrt{8}[/tex] - [tex]8\sqrt{18}[/tex] + [tex]\sqrt{n}[/tex]:

We can substitute the simplified forms of [tex]18\sqrt{8}[/tex] and [tex]8\sqrt{18}[/tex] from parts 1 and 2 into this expression:

[tex]18\sqrt{8}[/tex] - [tex]8\sqrt{18}[/tex] + [tex]\sqrt{n}[/tex] = [tex]36\sqrt{2}[/tex] - [tex]16\sqrt{3}[/tex] + [tex]\sqrt{n}[/tex] = ([tex]36\sqrt{2}[/tex] - [tex]16\sqrt{3}[/tex]) + [tex]\sqrt{n}[/tex].

So, the expression [tex]18\sqrt{8}[/tex] - [tex]8\sqrt{18}[/tex] + [tex]\sqrt{n}[/tex] is equal to ([tex]36\sqrt{2}[/tex] - [tex]16\sqrt{3}[/tex]) + [tex]\sqrt{n}[/tex] where n is an unknown constant.

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(a) Average for a student with 99 attendance is 81.1% and (b) The average for a student with 21 attendance is 78.8%.

Professor Cramer determines a final grade based on attendance, two papers, three major tests, and a final exam, the activities carry different weights.

Attendance is worth 3%,

each paper is worth 5%,

each test is worth 13%,

and the final is worth 48%.

(a) The average for a student with 99 on attendance, 75 on the first paper, 61 on the second paper, 94 on test 1, 86 on test 2, 77 on test 3, and 79 on the final exam

= [(99×3%) + (75×5%) + (61×5%) + (94×13%) + (86×13%) + (77×13%) + (79×48%)] ÷ [0.03 + 2(0.05) + 3(0.13) + 0.48]

= [2.97 + 3.75 + 3.05 + 12.22 + 11.18 + 10.01 + 37.92] ÷ 1

= 81.1 %

(b) The average for a student with the above scores on the papers, tests, and final exam, but with a score of only 21 on attendance

= [(21×3%) + (75×5%) + (61×5%) + (94×13%) + (86×13%) + (77×13%) + (79×48%)] ÷ [0.03 + 2(0.05) + 3(0.13) + 0.48]

= [0.63 + 3.75 + 3.05 + 12.22 + 11.18 + 10.01 + 37.92] ÷ 1

= 78.8 %

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Answer:

  $6400.42

Step-by-step explanation:

You want the value after 10 years of $5000.00 invested at 2.5%, compounded annually.

The formula for an amount earning compound interest is ...

  A = P(1 +r/n)^(nt)

where P is the amount invested at annual rate r compounded n times per year for t years.

Here, we have P=$5000, r=0.025, n=1, t=10, and the amount is ...

  A = $5000(1 +0.025)^(1·10) ≈ $6400.42

The final value after 10 years is $6400.42.

Mean value for the given data is 8.129, median = 7.7, Standard deviation is 1.699, Probability for MPa value greater than 10 is 0.148 and Coefficient of variation = 20.9

a) To estimate the mean value, we have to calculate all the observations and divide it by number of observations.

  x = ∑ xi / n = 219.5/27 = 8.129

So here the estimator used is x

b) To estimate the strength value or to separate weakest 50% from strongest 50%, we have to calculate the median. For that we have to arrange in ascending order and find the middle most term.

Since  n is odd, middle term is (n+1)/2 th term = (27+1)/2 = 28/2 = 14th term. Ascending order is given as image. the 14th term is 7.7.

c) Next we have to calculate standard deviation

 σ =  √(∑xi² - (∑x)²/n)/(n-1) = [1859.53 - (219.5²/27)] /(27-1)

                                          = (1859.53- 1784.454) / 26

                                         = 1.699

d) We have to calculate the probability flexural strength greater than 10. Here there are 4 values above 10.

So, p = 4/27 = 0.148

e) Coefficient of variation = (σ/x) ×100

                                           = [tex]\frac{1.699}{8.129} * 100[/tex]

                                            = 20.9

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Answer: a

Step-by-step explanation:

The set B' ∩ D' represents the event "the numbers do not add to 6." It contains 20 elements.

The complement of event B is B' = {2, 3, 4, 5, 7, 8, 9, 10, 11, 12}. The complement of event D is D' = {2, 3, 4, 5, 6, 7, 8, 10, 11, 12}. Taking the intersection of their complements, we get B' ∩ D' = {2, 3, 4, 5, 7, 8, 10, 11}. This set represents the event "the numbers do not add to 6." It contains 8 elements out of the 36 possible outcomes of rolling two dice, so the probability of this event is 8/36 or 2/9.

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It is determined that she has a sebaceous cyst. This encounter should be reported with code(s)  L72.9. So option B is correct.

The ICD-10-CM code for sebaceous cysts is unspecified. The other options are not specific to a sebaceous cyst or are not relevant to the given scenario.

L72.1 is for the epidermal cyst, L30.9 is for unspecified dermatitis, and L72.3 is for the inflamed sebaceous cyst.

ICD-10-CM codes are used for medical diagnosis coding and are important for medical record-keeping, insurance reimbursement, and data analysis.

Each code corresponds to a specific medical condition, and it is essential to choose the correct code that accurately represents the patient's diagnosis.

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Answer:

Step-by-step explanation:

Therefore, 4 people have been to Europe but not Asia.

A Venn diagram is a visual representation of sets and their relationships. It is used to show the similarities and differences between different sets of data. A Venn diagram consists of overlapping circles or other shapes, each representing a set. The size of each circle is proportional to the number of elements in the set it represents. The overlapping parts of the circles represent the elements that are shared by two or more sets. The non-overlapping parts represent the elements that are unique to each set. By analyzing the intersections and differences between sets in a Venn diagram, we can gain insights into the relationships between them. Venn diagrams are often used in mathematics, logic, statistics, and other fields to represent complex relationships between sets of data. They are also used in education to teach critical thinking and problem-solving skills.

Here,

Let E be the set of people who have been to Europe, A be the set of people who have been to Asia, and N be the set of people who have been to neither.

From the problem, we know that:

|E| = 5 (5 people have been to Europe)

|A| = 3 (3 people have been to Asia)

|E ∩ A| = 1 (1 person has been to both Europe and Asia)

To find the number of people who have been to Europe but not Asia, we need to find |E \ A|. We can use the formula:

|E \ A| = |E| - |E ∩ A|

Substituting the values we know:

|E \ A| = 5 - 1 = 4

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1.20 If the system has three unknowns and R has three nonzero rows, then the system has at least one solution.

This statement is true.

1.21 It is only when the rank of the coefficient matrix is less than the number of unknowns that the system can have infinitely many solutions

1.22 The system below has an infinite number of solutions:

2x + 3y + 5z + 6 - 7 - 8v = 0

3x - 4y + 7z + + 8 + 5y = 0

-7x + 9y - 2z -- 4w - 5u + 2y = 0

--5x - 5y +92 +3w + 2u + 7y = 0

-9x + 3y - 9z+5w - 3u - 4y = 0

This statement is true.

When we perform row reduction on a system of linear equations, the resulting reduced row echelon form (R) will have the same number of nonzero rows as the rank of the coefficient matrix.

In other words, if R has three nonzero rows, then the rank of the coefficient matrix is also 3.

If the rank of the coefficient matrix is equal to the number of unknowns, then the system has a unique solution.

However, if the rank of the coefficient matrix is less than the number of unknowns, then the system has either no solution or infinitely many solutions.

But in this case, since the rank is equal to the number of unknowns, the system must have at least one solution.

1.21 If the system has three unknowns and R has three nonzero rows, then the system can have an infinite number of solutions.

This statement is false. If R has three nonzero rows, then the rank of the coefficient matrix is also 3.

If the rank of the coefficient matrix is equal to the number of unknowns, then the system has a unique solution.

It is only when the rank of the coefficient matrix is less than the number of unknowns that the system can have infinitely many solutions.

1.22 The system below has an infinite number of solutions:

2x + 3y + 5z + 6 - 7 - 8v = 0

3x - 4y + 7z + 8 + 5y = 0

-7x + 9y - 2z - 4w - 5u + 2y = 0

-5x - 5y + 92 + 3w + 2u + 7y = 0

-9x + 3y - 9z + 5w - 3u - 4y = 0

This statement is true.

To check if the system has infinitely many solutions, we need to check the rank of the coefficient matrix and the rank of the augmented matrix. In this case, the rank of the coefficient matrix is 3, which is less than the number of unknowns (5).

Also, when we perform row reduction on the augmented matrix, we get the following reduced row echelon form:

1 0 -1 0 1 0

0 1 2 0 -1 0

0 0 0 1 2 0

0 0 0 0 0 1

0 0 0 0 0 0

Since the rank of the augmented matrix is less than the number of unknowns, the system has infinitely many solutions.

The variables with free parameters are z, u, and y, which can take any value.

The other variables can be expressed in terms of these free parameters.

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14 boxes of erasers should Mr. Franklin need to buy.

A division is a process of splitting a specific amount into equal parts.

Given that Franklin wants to buy an eraser for every fourth-grade student

There are 12 erasers in each box.

There are 7 fourth-grade classes with 24 students in each class

The total number of students=7×24

=168

We need to find the number of boxes are required to buy an eraser for each student.

Since there are 12 erasers in each box, we can divide the total number of students  by 12 to find the number of boxes Mr. Franklin needs to buy:

number of boxes = total students / erasers per box

number of boxes = 168 / 12

number of boxes = 14

Hence, 14 boxes of erasers should Mr. Franklin need to buy.

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In a Quincy College class consisting of 15 students 12 or more have blue eyes.

If half the students at Quincy College have blue eyes, then the most probable event is that a randomly selected student from the college will have blue eyes.

This is because, if half the students have blue eyes, then there is a greater chance that a randomly selected student will have blue eyes than any other eye color.

For example, if we were to consider the event of a randomly selected student having green eyes, this would be less probable because fewer than half the students are likely to have green eyes.

Therefore, the most probable event is that a randomly selected student from Quincy College will have blue eyes.

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The completed table is,

Input    (x)  -1       0    1   2

Output (y)    -6   -2   2    6

Function rule is the rule of writing the relationship between the two variables, one is dependent and another is independent.

We are given that, the output is 4 less than the input.

The table given in the problem is;

Input    (x)  -1   0  1  2

Output (y)

Thus we need to write such a function, which gives the value of (y) is 4 less than the value of (x), when we put this into the function.

y = 4x - 2

Complete the table using the above function rule;

At (x) equal to -1,

y = 4(-1)-2

y = -6

At (x) equal to 0,

y = 4(0)-2

y = -2

At (x) equal to 1,

y = 4(1)-2

y = 2

At (x) equal to 2,

y = 4(2)-2

y = 6

Hence, the function rule for the statement, "the output is 2 less than 4 times x the input is y = 4x - 2 and the completed table is,

Input    (x)  -1       0    1   2

Output (y)    -6   -2   2    6

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Answer: The measure of angle J is 40 degrees, 25 minutes, and 16 seconds, which can be expressed in decimal form as 40 + 25/60 + 16/3600 = 40.42111... degrees.

Since ∠J and ∠K are complementary, their measures add up to 90 degrees.

So, the measure of ∠K can be found by subtracting the measure of ∠J from 90 degrees:

∠K = 90 - 40.42111... = 49.57888... degrees

So, the measure of ∠K is approximately 49 degrees, 34 minutes, and 44 seconds.

Step-by-step explanation:

The required time hen will the rock be 9 meters from ground level is 6.40 seconds.

We can use the kinematic equations of motion to solve this problem. The equation we need is:

h = vi(t) + (1/2)at²

where h is the height of the rock above the ground at time t, vi is the initial velocity (positive when upward), a is the acceleration due to gravity (negative), and t is the time elapsed.

At the top of the cliff, the initial height h₀ = 45 meters and the initial velocity vi = 22 meters per second. When the rock is 9 meters above ground level, its height h = 9 meters. We want to find the time t at which this occurs.

First, we can use the equation of motion to find the time it takes for the rock to reach its maximum height. At the highest point, the velocity of the rock is zero, so vi = 22 m/s, a = -9.8 m/s^2, and h = h₀ + 0. We can solve for the time t1:

[tex]h=v_{i}t_{1}+\frac{1}{2}at_{1} ^{2}[/tex]

[tex]0=22t_{1}-4.9t_{1}^{2}[/tex]

[tex]t_{1} = 4.49 seconds[/tex]

Next, we can use the same equation of motion to find the time it takes for the rock to reach a height of 9 meters on the way back down. This time, the initial height is h0 = 45 - (maximum height) = 45 - 22.45 = 22.55 meters, and the initial velocity is -vi = -22 m/s. We can solve for the time t2:

[tex]h=v_{i}t_{2}+\frac{1}{2}at_{2} ^{2}[/tex]

[tex]9 = 22t_{2} + 4.9*t_{2}^{2}[/tex]

[tex]t_{2} = 1.91 seconds[/tex]

The total time for the rock to reach a height of 9 meters on the way back down is t = [tex]t_{1}+t_{2}[/tex] = 6.40 seconds (rounded to two decimal places).

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6 cos(π/2 t) is the best model for the position of the weight.

The motion of the weight on the spring can be modeled by a sine or cosine function because it oscillates back and forth around its equilibrium position.

We know that, the weight is initially pulled down 6 inches from its equilibrium position, so the function should have an amplitude of 6.

The time it takes for the spring to complete one oscillation is 4 seconds, so the period of the function is 4 seconds.

The general form of a sine or cosine function with amplitude A and period T is:

f(t) = A sin(2πt/T) or f(t)

= A cos(2πt/T)

Substituting the given values,

we get:

f(t) = 6 sin(2πt/4) or f(t)

= 6 cos(2πt/4)

Simplifying, we get:

f(t) = 6 sin(π/2 t) or f(t)

= 6 cos(π/2 t)

Therefore,

the function that best models the position of the weight is

s(t) = 6 cos(π/2 t).

So, the answer is option (D).

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To calculate the number of containers that Heavey Compressors should be using, we need to determine the number of parts that need to be produced per day and divide it by the number of parts that can fit in each container.

100 parts per 8-hour day / 7 parts per container

= 14.2857 containers (round up to 15)

=15 containers

Therefore,  Heavey Compressors be using 15 containers.

Maximum inventory levels = reorder point + reorder quantity – [minimum consumption × minimum lead time].

= 100+15-[12.5x8]

= 115-100

= 15

Therefore, the maximum system inventory for this part is 15.

The maximum stock position is the largest number of goods a company can store to give its guests with service at the smallest possible cost. It's vital to keep force control in line with demand.

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Complete question:

Heavey Compressors uses a lean production assembly line to make its compressors. In one assembly area, the demand is 100 parts per eight-hour day. It uses a container that holds seven parts. It typically takes about five hours to round-trip a container from one work center to the next and back again. Heavey also desires to hold 20 percent safety stock of this part in the system. a. How many containers should Heavey Compressors be using? Do not round intermediate calculations. Round your answer up to the nearest whole number. containers b. Calculate the maximum system inventory for this part. Use the rounded value of the number of containers from part answer to the nearest whole number. parts c. If the safety stock percentage is reduced to zero, how would this impact the number of containers, all else being equal? calculations. Round your answer up to the nearest whole number. The number of containers will to containers.

[tex]{\Large \begin{array}{llll} y=ab^x \end{array}} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} x=2\\ y=360 \end{cases}\implies 360=ab^2\implies 360=abb\implies \cfrac{360}{b}=ab \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} x=3\\ y=216 \end{cases}\implies 216=ab^3\implies 216=abb^2\implies \stackrel{\textit{substituting from above}}{216=\left( \cfrac{360}{b} \right)b^2} \\\\\\ 216=360b\implies \cfrac{216}{360}=b\implies \boxed{\cfrac{3}{5}=b} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\stackrel{\textit{since we know that}}{360=ab^2}\implies 360=a\left( \cfrac{3}{5} \right)^2\implies 360=\cfrac{9a}{25} \\\\\\ \cfrac{25}{9}\cdot 360=a\implies \boxed{1000=a}~\hfill {\Large \begin{array}{llll} y=1000\left( \frac{3}{5} \right)^x \end{array}}[/tex]

The probabilities are given as follows:

It is a distribution with two bounds, given by a and b, in which each possible outcome is equally as likely.

The probability of finding a value between c and d is:

[tex]P(c \leq X \leq d) = \frac{d - c}{b - a}[/tex]

Uniformly distributed between 11 minutes and 14 minutes, hence the bounds are given as follows:

a = 11, b = 14.

As the uniform distribution is a continuous distribution, the probability of an exact value, such as X = 11, is of zero.

The probability of a value between 11 and 13 is given as follows:

p = (13 - 11)/(14 - 11) = 2/3.

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Answer:

hope this helps

The Corollary to the polygon is explained below and the  formula to find the measure of each interior angle is  " (n - 2)×180°/n "   .

The Corollary to the polygon Angle Sum Theorem states that : the sum of the interior angles of a regular n gon is written as :

that means , ⇒ Sum of interior angles = (n - 2) × 180° ...equation(1)

In a regular "n-gon" , all the interior angles are said to be congruent.

Let "x" be  measure of each interior angle of a regular n-gon.

So , we can write ;

⇒ Sum of interior angles = (n)×(x)  ;

Equating the above expressions with equation(1),

⇒ (n)×(x) = (n - 2) × 180°  ;

On Solving for x,

⇒ x = (n - 2)×180°/n  ;

Therefore, the measure of each interior angle of a regular "n-gon" is x = (n - 2)×180°/n  .

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(a) Expression for  P(3x−4<5) is ∫[from -∞ to 3] fx(x) dx

(b) Expression for P(X>Y)(c) P(X+Y=5) is ∬[over (x,y) satisfying x > y] fx(x) fY(y) dxdy

(c) Expression for E(cov(XY)) is ∬xyfx(x) fY(y) dxdy - E(X)E(Y)

(d) Expression for Mx+y(t) (the mgf of X +Y) is Mx(t) My(t)

(a) P(3x-4<5) can be written as:

P(3x < 9)

P(x < 3)

The expression involving fx(x) for this probability would be:

∫[from -∞ to 3] fx(x) dx

(b) P(X>Y) can be written as:

P(X-Y > 0)

The expression involving fx(x) and fY(y) for this probability would be:

∬[over (x,y) satisfying x > y] fx(x) fY(y) dxdy

(c) P(X+Y=5) can be written as:

P(Y = 5 - X)

The expression involving fx(x) and fY(y) for this probability would be:

∬[over (x,y) satisfying x+y=5] fx(x) fY(y) dxdy

(d) The covariance of X and Y is defined as:

cov(X,Y) = E(XY) - E(X)E(Y)

So, E(cov(X,Y)) can be written as:

∬xyfx(x) fY(y) dxdy - E(X)E(Y)

(e) The MGF of X+Y can be written as:

Mx+y(t) = E(e^(t(X+Y)))

Since X and Y are independent, we can write this as:

Mx+y(t) = E(e^(tX) e^(tY))

Using the fact that X and Y have their own MGFs,

We can write this as:

Mx+y(t) = Mx(t) My(t)

Where Mx(t) and My(t) are the MGFs of X and Y, respectively.

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The slope is 3 which represents the cost per mile and the y-intercept is 5 which represents the fixed cost of the taxi.

A connection between a number of variables results in a linear model when a graph is displayed. The variable will have a degree of one.

The linear equation is given as,

y = mx + c

Where m is the slope of the line and c is the y-intercept of the line.

The equation is given below.

C = 3m + 5

Where 'C' represents the cost and 'm' represents the number of miles.

The slope is 3 which represents the cost per mile and the y-intercept is 5 which represents the fixed cost of the taxi.

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The complete question is given below.

The equation c=3m+5 represents a line of the best fit for a scarer plot where c represents the total cost of a taxi in dollars and m represents the number of the trip. What do the numbers 5 and 3 represent?