The apothem of a regular polygon is the measure of the line segment from the center of the polygon to the midpoint of one of its sides. A circle is circumscribed about a regular hexagon with an apothem of 4.8 centimeters. a. Find the radius of the circumscribed circle. b. What is the length of a side of the hexagon? c. What is the perimeter of the hexagon?

Answers

Answer 1

The radius of the circumscribed circle is [tex]\( 3.2\sqrt{3} \)[/tex] centimeters.
The length of a side of the hexagon is [tex]\( 3.2\sqrt{3} \)[/tex] centimeters.
The perimeter of the hexagon is [tex]\( 19.2\sqrt{3} \)[/tex] centimeters.

Let's solve each part of the problem step by step:

a. **Find the radius of the circumscribed circle:**

The apothem of a regular polygon and the radius of the circumscribed circle are related by the formula:

[tex]\[ \text{Radius} = \frac{\text{Apothem}}{\cos(180^\circ / \text{number of sides})} \][/tex]

For a regular hexagon (6 sides), the formula becomes:

[tex]\[ \text{Radius} = \frac{4.8}{\cos(180^\circ / 6)} \][/tex]

[tex]\[ \text{Radius} = \frac{4.8}{\cos(30^\circ)} \][/tex]

[tex]\[ \text{Radius} = \frac{4.8}{\frac{\sqrt{3}}{2}} \][/tex]

[tex]\[ \text{Radius} = \frac{4.8 \times 2}{\sqrt{3}} \][/tex]

[tex]\[ \text{Radius} = \frac{9.6}{\sqrt{3}} \][/tex]

[tex]\[ \text{Radius} = \frac{9.6\sqrt{3}}{3} \][/tex]

[tex]\[ \text{Radius} = 3.2\sqrt{3} \][/tex]

b. **What is the length of a side of the hexagon?**

The length of a side of a regular hexagon can be found using the formula:

[tex]\[ \text{Side Length} = 2 \times \text{Apothem} \times \tan(180^\circ / \text{number of sides}) \]\\[/tex]

For a regular hexagon with apothem 4.8 cm:

[tex]\[ \text{Side Length} = 2 \times 4.8 \times \tan(30^\circ) \][/tex]

[tex]\[ \text{Side Length} = 2 \times 4.8 \times \frac{1}{\sqrt{3}} \][/tex]

[tex]\[ \text{Side Length} = \frac{9.6}{\sqrt{3}} \][/tex]

[tex]\[ \text{Side Length} = \frac{9.6\sqrt{3}}{3} \][/tex]

[tex]\[ \text{Side Length} = 3.2\sqrt{3} \][/tex]

c. **What is the perimeter of the hexagon?**

Since a regular hexagon has six equal sides, its perimeter is simply:

[tex]\[ \text{Perimeter} = 6 \times \text{Side Length} \][/tex]

[tex]\[ \text{Perimeter} = 6 \times 3.2\sqrt{3} \][/tex]

[tex]\[ \text{Perimeter} = 19.2\sqrt{3} \][/tex]


Related Questions

In Professor Krugman's economics course, the correlation between the students' total scores prior to the final examination and their final-examination scores is r = 0.5. The pre-exam totals for all students in the course have mean 280 and standard deviation 40. The final-exam scores have mean 75 and standard deviation 8. Professor Krugman has lost Julie's final exam but knows that her total before the exam was 300. He decides to predict her final-exam score from her pre-exam total.

Which of the choices is the least-squares regression line of final-exam scores on pre-exam total scores in this course?

(A) y =-0.1-47x
(B) y=47-0.1x
(C) y=-47 + 0.1x
(D) y= 47 + 0.1x

Answers

Answer:

D

Step-by-step explanation:

The regression equation is

y=a+bx

Now, we have to estimate a and b.

We know that slope can be determine as

b=byx=r*(sy/sx)

=0.5*(8/40)

b=0.1

So, we have the slope of regression equation is 0.1.

Now the intercept "a" can be estimated as

y=a+bx

a=ybar-bxbar

a=75-0.1*280

a=47

So, we have the intercept of regression equation is 47.

y=a+bx

y=47+0.1x

Final answer:

The least-squares regression line of final-exam scores on pre-exam total scores, given the statistics provided, is y = 47 + 0.1x, which corresponds to option (D).

Explanation:

The question involves identifying the least-squares regression line of final-exam scores on pre-exam total scores in an economics course. Given the correlation coefficient (r = 0.5), the mean and standard deviation of both the pre-exam totals and the final-exam scores, we can use the formula for the regression line, which is:

y = b0 + b1*x

where y is the predicted final exam score, x is the pre-exam total score, b0 is the y-intercept, and b1 is the slope of the line. The slope (b1) is calculated by:

b1 = r * (standard deviation of y / standard deviation of x) = 0.5 * (8 / 40) = 0.1

And the y-intercept (b0) is found using the means of x and y:

b0 = mean of y - b1 * mean of x = 75 - 0.1 * 280 = 47

So, the regression line is y = 47 + 0.1x, which corresponds to option (D).

A manufacturer of pharmaceutical products analyzes specimens from each batch of a product to verify the concentration of the active ingredient, required to be .80 grams per liter. Suppose the repeated measurements follow a Normal distribution. Fifteen specimens are taken from a batch and it is found that 0.82 grams per liter and s 0.048 grams per liter. Find a 99% confidence interval for μ, the true average concentration of the active ingredient.

Answers

Answer:

[tex]0.82-2.977\frac{0.048}{\sqrt{15}}=0.783[/tex]    

[tex]0.82+2.977\frac{0.048}{\sqrt{15}}=0.857[/tex]    

So on this case the 99% confidence interval would be given by (0.783;0.857)

So then we can conclude that at 99% of confidence the true mean is between (0.783;0.857) and the specification is satisfied since the value of 0.8 is on the confidence interval

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X= 0.82[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

s=0.048 represent the sample standard deviation

n=15 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=15-1=14[/tex]

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that [tex]t_{\alpha/2}=2.977[/tex]

Now we have everything in order to replace into formula (1):

[tex]0.82-2.977\frac{0.048}{\sqrt{15}}=0.783[/tex]    

[tex]0.82+2.977\frac{0.048}{\sqrt{15}}=0.857[/tex]    

So on this case the 99% confidence interval would be given by (0.783;0.857)

So then we can conclude that at 99% of confidence the true mean is between (0.783;0.857) and the specification is satisfied since the value of 0.8 is on the confidence interval

Final answer:

To find a 99% confidence interval for the true average concentration of the active ingredient, we can use the formula: CI = (x - z*(s/sqrt(n)), x + z*(s/sqrt(n))). Plugging in the given values and using the critical value for a 99% confidence level, we get the confidence interval to be (0.799, 0.841) grams per liter.

Explanation:

To find a 99% confidence interval for the true average concentration of the active ingredient, we can use the formula:

CI = (x - z*(s/sqrt(n)), x + z*(s/sqrt(n)))

Where CI is the confidence interval, x is the sample mean, z is the z-score corresponding to the desired confidence level, s is the sample standard deviation, and n is the sample size.

Plugging in the given values: x = 0.82, s = 0.048, and n = 15, and using the critical value for a 99% confidence level (z = 2.576), we get:

CI = (0.82 - 2.576*(0.048/sqrt(15)), 0.82 + 2.576*(0.048/sqrt(15)))

Simplifying the expression, we get the confidence interval to be (0.799, 0.841) grams per liter.

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Convert 9255 cm3 to gallons. Express the volume in gallons to four significant figures.

Answers

Answer:

2.44491234

Step-by-step explanation:

1 gallon = 3785 cm3

Unit conversion is a way of converting some common units into another without changing their real value. The volume of  9255 cm³ in gallons is equal to 2.4451 gallons.

What is Unit conversion?

Unit conversion is a way of converting some common units into another without changing their real value. for, example, 1 centimeter is equal to 10 mm, though the real measurement is still the same the units and numerical values have been changed.

Since 1 gallon is equal to 4546.09 cubic centimeters, therefore, the volume of 9255 cm³ in gallons can be written as,

1 gallons = 3785 cm³

1 cm³ = 1/3785 gallons

The volume of 9255 cm³ in gallons is,

Volume = 9255 × (1/3785 ) gallons

= 2.4451 gallons

Hence, the volume of  9255 cm³ in gallons is equal to 2.4451 gallons.

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how do I solve these three

Answers

Answer:

below

Step-by-step explanation:

2(x+5)=3x+1

2x+10=3x+1

10-1=3x-2x

9=x

x=9

3y-4=6-2y

3y+2y=6+4

5y=10

y=10/5

y=2

3(n+2)= 9(6-n)

3n+6=54-9n

3n+9n=54-6

12n=48

n=48/12

n=4

Answer:

Step-by-step explanation:

1) 2(x + 5) = 3x + 1

Multiplying each term inside the parenthesis by 2, it becomes

2x + 10 = 3x + 1

Subtracting 2x and - 1 from both sides of the equation

2x - 2x + 10 - 1 = 3x - 2x + 1 - 1

3x - 2x = 10 - 1

x = 9

2) 3y - 4 = 6 - 2y

Adding 4 and 2y to the left hand side and the right hand side of the equation, it becomes

3y + 2y - 4 + 4 = 6 + 4 + 2y - 2y

5y = 10

Dividing the both sides of the equation by 5, it becomes

5y/5 = 10/5

y = 10/2 = 5

3) 3(n + 2) = 9(6 - n)

3n + 6 = 54 - 9n

Subtracting 6 from both sides and adding 9n to both sides of the equation, it becomes

3n + 9n + 6 - 6 = 54 - 6 - 9n + 9n

12n = 48

Dividing both sides of the equation by 12, it becomes

12n/12 = 48/12

n = 4

Find the width, in inches, of a 65 mm film. (Enter a number as an integer or decimal.)

Answers

Answer:

The answer to your question is 2.6 in

Step-by-step explanation:

Data

width = ? in

width = 65 mm

Process

1.- To solve this problem, use proportions and cross multiplication

                   1 in ---------------------- 25.4 mm

                   x    ----------------------  65 mm

                   x = (65 x 1) / 25.4

2.- Simplification

                   x = 65 / 25.4

3.- Result

                   x = 2.6 in      

In a class of 125 students, 27 are computer science majors, 51 are mechanical engineering majors, 12 are civil engineers and the rest are general engineering majors. Assume students only have one major.

Suppose six students from the class are chosen at random what is the probability none are mechanical engineering majors?

I tried 74/125 * 73/124 * 72/123 * 71/122 * 70/121 * 69/120 * 68/119 but had incorrect answer

Answers

Answer:

There is a 3.95% probability that none are mechanical engineering majors.

Step-by-step explanation:

[tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula:

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

In this problem, we have that:

There are 125 students.

51 of them are mechanical engineering majors and 125-51 = 74 are not mechanical engineering majors.

Suppose six students from the class are chosen at random what is the probability none are mechanical engineering majors?

The total number of students is 125. So total number of 6 student groups is

[tex]C_{125,6} = \frac{125!}{6!119!}[/tex]

The total number of non mechanical engineering students is 74. So the total number of 6 non mechanical engineering students is.

[tex]C_{74,6} = \frac{74!}{6!68!}[/tex]

The probability is:

[tex]\frac{C_{74,6}}{C_{125,6}} = 74/125 * 73/124 * 72/123 * 71/122 * 70/121 * 69/120 = 0.0395[/tex]

The 68/119 ends up simplified in this exercise, this is your mistake.

There is a 3.95% probability that none are mechanical engineering majors.

As part of the National Health and Nutrition Examination Survey, subjects were asked how long they slept the preceding night, and the following times (hours) were reported: 8, 7, 5, 7, 4, 7, 6, 7, 8, 8, 8, 6
The standard deviation is 1.3 hours.
What is the variance?

Answers

Final answer:

The variance for the given data is 1.69 hours squared.

Explanation:

To find the variance, we need to square the standard deviation. The standard deviation is given as 1.3 hours, so when we square it, we get 1.69 hours squared. Therefore, the variance for the given data is 1.69 hours squared.

What is the correct negation of "There is someone in Spain who speaks at least 10 languages"?1) "Everyone in Spain speaks at least 10 languages."2) "Everyone outside of Spain speaks at least 10 languages."3) "There is someone in Spain who speaks at most 9 languages."4) "Everyone in Spain speaks at most 9 languages."5) "Everyone outside of Spain speaks at most 9 languages."6) "There is someone outside of Spain who speaks at most 9 languages."7) "There is someone outside of Spain who speaks at least 10 languages."

Answers

Answer:

4) "Everyone in Spain speaks at most 9 languages."

Step-by-step explanation:

A negation is a statement that contradicts the original statement. Since the original statement regards people in Spain, options 2, 5, 6 and 7 can be ruled out as they refer to speakers outside of Spain.

Evaluating the remaining statements.

1) "Everyone in Spain speaks at least 10 languages."

If that is true, then someone in Spain speaks at least 10 languages, it doesn't negate the original statement.

3) "There is someone in Spain who speaks at most 9 languages."

It only refers to one specific individual, it doesn't negate the original statement.

4) "Everyone in Spain speaks at most 9 languages."

This means that nobody in Spain speaks more than 9 languages, contradicting the original statement.

The answer is statement 4.

Listed below are the annual tuition amounts of the 10 most expensive colleges in a country for a recent year.
What does this "Top 10" list tell about the population of all that country's college tuitions?
Find the mean, midrange, median, and mode of the data set. Round to two decimal places as needed.

$52,384 $54,079 $52,762 $54,364 $52,584

$53,987 $53,108 $51,588 $50,554 $53,987

Answers

Answer:

[tex]\bar X = \frac{52384 +54079 +52762 +54364 +52584+53987 +53108 +51588 +50554 +53987}{10}=52939.7[/tex]

[tex] Mode= 53987[/tex]

[tex] Median = \frac{52762+53108}{2}=52935[/tex]

[tex] Midrange=\frac{50554+54364}{2}=52459[/tex]

Step-by-step explanation:

We can calculate the mean with the following formula:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

And if we replace the values given we got:

[tex]\bar X = \frac{52384 +54079 +52762 +54364 +52584+53987 +53108 +51588 +50554 +53987}{10}=52939.7[/tex]

The mode is the most repeated value and for this case with a frequency of 2 the mode is:

[tex] Mode= 53987[/tex]

In order to find the median we need to order the dataset on increasing way like this:

50554, 51588, 52384  ,52584, 52762

53108, 53987 , 53987 , 54079, 54364

Since we have 10 values an even number the median is calculated from the average between positions 5 and 6 on the data ordered, and we got:

[tex] Median = \frac{52762+53108}{2}=52935[/tex]

The mid range is defined like this:

[tex] Midrange = \frac{Max +Min}{2}[/tex]

And if we replace we got:

[tex] Midrange=\frac{50554+54364}{2}=52459[/tex]

The 'Top 10' list gives insights into the population of college tuitions in a country. The mean, midrange, median, and mode of the data set are $53,138.50, $52,459, $52,584., and  $53,987 respectively.

The "Top 10" list of annual tuition amounts for the most expensive colleges in a country provides insights into the range and distribution of college tuitions within the country's higher education system.

Mean: The mean (average) tuition is calculated by summing up all the tuition amounts and dividing by the total number of colleges (in this case, 10).

Mean = ($52,384 + $54,079 + $52,762 + $54,364 + $52,584 + $53,987 + $53,108 + $51,588 + $50,554 + $53,987) / 10 = $53,138.50 (rounded to two decimal places).

Midrange: The midrange is the average of the minimum and maximum values in the dataset.

Midrange = ($50,554 + $54,364) / 2 = $52,459 (rounded to two decimal places).

Median: The median is the middle value when the data is arranged in ascending or descending order. In this case, the data is already provided in ascending order, so the median is the fifth value.

Median = $52,584.

Mode: The mode is the value that appears most frequently in the dataset.

Mode = $53,987.

In summary, the "Top 10" list of college tuitions in the country indicates that there is a range of tuition amounts, with the mean being around $53,138.50. The midrange falls around $52,459, the median is $52,584, and the mode is $53,987.

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A grocery store manager must decide how to best present a limited supply of milk and cookies to its customers. Milk can be sold by itself for a profit of $1.50 per gallon. Cookies can likewise be sold at a profit of $2.50 per dozen. To increase appeal to customers, one gallon of milk and a dozen cookies can be packaged together and are then sold for a profit of $3.00 per bundle. The manager has 100 gallons of milk and 150 dozen cookies available each day. The manager has decided to stock at least 75 gallons of milk per day and demand for cookies is always 140 dozen per day. To maximize profits, how much of each product should the manager stock.Which of the following is the objective function for the grocer's problem?A) Max P = 1.5M + 2.5C + 3BB) Min P = 1.5M + 2.5C + 3BC) Max P = 2.5M + 1.5C + 3BD) Max P = 2.5M + 3C + 1.5BE) Min P = 1.5M + 1.5C + 3B

Answers

Answer:

Option A.

Step-by-step explanation:

Let M be the milk per gallon.

C be the cookies per dozen.

B be the bundle (one gallon of milk and a dozen cookies ).

Milk can be sold by itself for a profit of $1.50 per gallon. Cookies can likewise be sold at a profit of $2.50 per dozen and bundle is sold for a profit of $3.00 per bundle.

[tex]Profit= 1.5M + 2.5C + 3B[/tex]

We need to maximize the profits. So, our objective function is

Therefore, the correct option is A.

Determine if the statement is true or false:

1. If two matrices are equivalent, then one can be transformed into the other with a sequence of elementary row operations.
2.Different sequences of row operations can lead to different echelon forms for the same matrix.
3.Different sequences of row operations can lead to different reduced echelon forms for the same matrix.
4.If a linear system has four equations and seven variables, then it must have infinitely many solutions.

Answers

Answer:

Step-by-step explanation:

given are four statements and we have to find whether true or false.

.1 If two matrices are equivalent, then one can be transformed into the other with a sequence of elementary row operations.

True  

2.Different sequences of row operations can lead to different echelon forms for the same matrix.

True in whatever way we do the reduced form would be equivalent matrices

3.Different sequences of row operations can lead to different reduced echelon forms for the same matrix.

False the resulting matrices would be equivalent.

4.If a linear system has four equations and seven variables, then it must have infinitely many solutions.

True, because variables are more than equations.  So parametric solutions infinite only is possible

Final answer:

Statements 1 and 2 are true, and statements 3 and 4 are false. While equivalent matrices can be transformed into each other and different row operations can yield different echelon forms, the reduced echelon form is unique, and a linear system with more variables than equations does not necessarily have infinitely many solutions.

Explanation:

1. The statement is true. If two matrices are equivalent, one can indeed be transformed into the other through a sequence of elementary row operations. These operations include swapping two rows, multiplying a row by a non-zero scalar, or adding a multiple of one row to another row.

2. The statement is true. Different sequences of row operations can yield different echelon forms of the same matrix as the operations can redistribute information about the system of equations in different ways.

3. The statement is false. Regardless of the sequence of row operations performed, the reduced echelon form of a matrix is unique. This is because the reduced echelon form is a canonical form, meaning there's only one possible reduced echelon form for a given matrix.

4. The statement is false. Even though a linear system has more variables than equations, this does not guarantee that it will have infinitely many solutions. It could have no solutions or, under certain conditions, even a unique solution. Further analysis is required in these cases.

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Math problem is on the attached picture. Thanks!

Answers

Answer:

the consecutive integars is 11,12,13,14,15

Step-by-step explanation:

let the first integar  be described as y since they are consecutive ( following each other)the second will y + 1the third is y + 2the fourth is y + 3the fifth y + 4

then it says the sum of the first and 4 times the third = 60 less than 3 times the sum of the second, fourth and fifth

60 less than 3 times the sum of the second, fourth and fifth means that 3 times the sum of the second, fourth and fifth minus 60

y+ 4(y+2) = 3{ (y +1) +(y+3) + (y+4)} -60

open the bracket

y + 4y + 8 =3[ y +1 + y + 3 +y+4} - 60

5y +8=3(y+y+y+1+3+4)-60

5y+8=3(3y+8)-60

simplify and open the brackets

5y+8=9y+24-60

5y+8=9y-36

subtract 8 from both sides

5y+8-8=9y-36-8

5y=9y-44

subtact 9y from both sides

5y-9y=9y-9y-44

-4y=-44

divide -4 from both sides

-4y/-4=-44/-4 ( minus divided by minus is plus

y = 11

the first integar is y = 11the second will y + 1 = 11+1 = 12the third is y + 2 = 11 +2 = 13the fourth is y + 3 = 11 + 3 = 14the fifth y + 4 = 11 + 4 = 15the consecutive integars is 11,12,13,14,15

Two cards are drawn one at a time from a full deck of cards. a) What’s the probability they are an ace or a heart? b) What’s the probability they are neither ace nor a heart?

Answers

Final answer:

The probability of drawing an ace or a heart from a full deck of cards is approximately 5.88%. The probability of drawing a card that is neither an ace nor a heart is approximately 45.89%.

Explanation:

This question is about probability in Mathematics. A full deck has 52 cards, 4 of which are aces, and 13 of which are hearts. The total number of favorable outcomes are the scenarios where we draw an ace or a heart.

For part a) There are 16 favorable outcomes (4 aces + 12 remaining hearts). For 2 draws, that's 16/52 * 15/51 = 0.0588 or 5.88% chance of drawing an ace or a heart.

For part b) There are 36 non-favorable outcomes (52 total - 16 favorable). The chance of drawing a card that is neither ace nor heart is 36/52 * 35/51 = 0.4589 or 45.89%.

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Your city is represented in a coordinate plane where each unit represents 1 kilometer. The library is at (−2, −2), the post office is at (2, 2), and your house is at (−3, 2). You ride your bike from your house to the library, then the post office, and then back home. What is the minimum distance that you can ride your bike? Round your answer to the nearest tenth.

Answers

Final answer:

By using the distance formula, we calculate the distances between the house, library, and post office, finding the total minimum distance to be approximately 14.8 kilometers.

Explanation:

This question requires the use of the distance formula in mathematics, which is derived from the Pythagorean theorem. The formula is √[(x₂ - x₁)² + (y₂ - y₁)²]. To find the total minimum distance you ride your bike, you calculate the distance from your house to the library, then the library to the post office, and finally, the post office back to your house.

First, calculate the distance from your house at (-3, 2) to the library at (-2, -2): √[(-2 - (-3))² + (-2 - 2)²] = √[1² + -4²] = √[1 + 16] = √17 ≈ 4.1 kilometers.

Second, calculate the distance from the library at (-2, -2) to the post office at (2, 2): √[(2 - (-2))² + (2 - (-2))²] = √[4² + 4²] = √[16 + 16] = √32 ≈ 5.7 kilometers.

Finally, calculate the distance from the post office at (2, 2) back to your house at (-3, 2): √[(-3 - 2)² + (2 - 2)²] = √[-5² + 0²] = √[25 + 0] = √25 = 5 kilometers.

Adding these distances together, 4.1 km + 5.7 km + 5 km = 14.8 kilometers.

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A supermarket has determined that daily demand for milk containers has an approximate bell shaped distribution, with a mean of 55 containers and a standard deviation of six containers. How often can we expect between 49 and 61 containers to be sold in a day?

Answers

Answer:

[tex] \\ P(49<x<61) = 0.8413 - 0.1587 = 0.6826 [/tex] or 68.26%.

Step-by-step explanation:

The daily demand for milk containers has a Normal (or Gaussian) distribution, and we can use values from the cumulative distribution function and z-scores to solve the question.

We know from the question that the mean of the distribution is:

[tex] \\ \mu = 55 [/tex]

And a standard deviation of:

[tex] \\ \sigma = 6 [/tex]

The z-scores permit calculates the probabilities for any case whose values have a Normal o Gaussian distribution. Then, for this, we need to calculate the z-scores for 49 containers and 61 containers to establish the corresponding probabilities, as well as the differences between these two values to determine the probability between them.

These z-scores are given by:

[tex] \\ z = \frac{x-\mu}{\sigma} [/tex]

Thus,

The z-scores for 49 and 61 containers are:

[tex] \\ z_{49} = \frac{49 - 55}{6} = \frac{-6}{6} = -1 [/tex] [1]

[tex] \\ z_{61} = \frac{61 - 55}{6} = \frac{6}{6} = 1 [/tex] [2]

Well, this is a special case when in both cases the values are one standard deviation from the mean, but in one case ([tex] \\ z_{49} = -1 [/tex]) the values are smaller than the mean and in the other case ([tex] \\ z_{61} = 1 [/tex]) the values are greater than the mean.

In other words, the cumulative probability for ([tex] \\ z_{61} = 1 [/tex]), obtained from any Table of the Normal Distribution available on the Web, is: 0.8413 (or 84.13%) and the cumulative probability for ([tex] \\ z_{49} = -1 [/tex]) is: 1 - 0.8413 = 0.1587 (or 15.87%), because of the symmetry of the Normal Distribution.

Then, the probability of expecting to sell between 49 and 61 containers in a day is the difference of both obtained probabilities:

[tex] \\ P(49<x<61) = 0.8413 - 0.1587 = 0.6826 [/tex] or 68.26%.

See the graph below.

Final answer:

Approximately 68% of the time, the number of milk containers sold in a day at the supermarket is expected to be between 49 and 61 containers, following the Empirical Rule for normal distribution.

Explanation:

The question asks us about the probability of daily sales being between certain values when they follow a bell-shaped distribution, specifically between 49 and 61 containers. Since we know the distribution is approximately normal with a mean of 55 containers and a standard deviation of six containers, we can use the Empirical Rule or the Standard Normal Distribution to find the probability.

In the context of a normal distribution, we know that:

About 68% of the data falls within one standard deviation of the mean. (Mean ± 1SD)

About 95% falls within two standard deviations. (Mean ± 2SD)

For this supermarket, one standard deviation from the mean (55 ± 6) gives us a range from 49 to 61 containers. Therefore, approximately 68% of the time, we can expect the number of milk containers sold to be between 49 and 61 containers in a day, based on the Empirical Rule.

Which of the following points represents -2 1/4?

Answers

Point E represents -2 1/4

Drug x, a drug that claims to treat male pattern scalp hair loss, was administered for 12 months to over 1800 men aged 18 to 41 with mild to moderate amounts of ongoing hair loss. Whether they were receiving drug x or a placebo ( a pill containing no medication), all men were given a medicated shampoo. In general, men who took drug x maintained or increased the number of visible scalp hairs; while scalp hairs counts in men who took the placebo continued to decrease. This concluded that drug x is effective in maintaining or increasing the amount of scalp hair in men.a. Which statement in this example can be referred to as descriptive statistics?b. Which statement in this example can be referred to as inferential statistics?

Answers

The following statement

'In general, men who took drug x maintained or increased the number of visible scalp hairs; while scalp hairs counts in men who took the placebo continued to decrease'

can be referred as descriptive statistics, since we are saying that the msot of the men who took the drug gain scalp hairs while the contrary happened from those men who took placebo.

On the other hand, the statement

'drug x is effective in maintaining or increasing the amount of scalp hair in men'

Is a deduction you make, so it can be referred as inferential statistics

Membership in Mensa requires an IQ score above 131.5. Nine candidates take an IQ test, and they have a mean IQ score of 133. IQ scores are normally distributed with a mean of 100 and a standard deviation of 15 (this should be used as the baseline for the questions below.) (a) If one person is randomly selected from the general population, what is the probability of getting someone with an IQ of at least 133?

Answers

Answer:

a) [tex]P(X\geq 133)=P(\frac{X-\mu}{\sigma}\geq \frac{133-\mu}{\sigma})=P(Z\geq \frac{133-100}{15})=P(Z\geq 2.2)[/tex]

And we can find this probability using the complement rule:

[tex]P(Z\geq 2.2)=1- P(z<2.2)=1-0.986=0.0139 [/tex]

b) [tex]P(\bar X\geq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\geq \frac{133-\mu}{\sigma_{\bar x}})=P(Z\geq \frac{133-100}{5})=P(Z\geq 6.6)[/tex]

And we can find this probability using the complement rule:

[tex]P(Z\geq 6.6)=1- P(z<6.6)=1-0.999999=2.06x10^{-11} [/tex]

c) No. The mean can be lower than 131.5 if we find the probability:

[tex]P(\bar X\leq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\leq \frac{131.5-\mu}{\sigma_{\bar x}})=P(Z\leq \frac{131.5-100}{5})=P(Z\leq 6.3)[/tex]

[tex]P(Z\leq 6.3) \approx 1 [/tex]

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

A.    If 1 person is randomly selected from thegeneral population, find the probability of getting someone with anIQ score of at least 133.

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(100,15)[/tex]  

Where [tex]\mu=100[/tex] and [tex]\sigma=15[/tex]

We are interested on this probability

[tex]P(X\geq 133)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this:

[tex]P(X\geq 133)=P(\frac{X-\mu}{\sigma}\geq \frac{133-\mu}{\sigma})=P(Z\geq \frac{133-100}{15})=P(Z\geq 2.2)[/tex]

And we can find this probability using the complement rule:

[tex]P(Z\geq 2.2)=1- P(z<2.2)=1-0.986=0.0139 [/tex]

And the probability is calculated from the normal standard table or with excel.

B.     If 9 people are randomly selected,find the probability that their mean IQ score is at least 133.

For this case since the distribution for the random variable X is normal then the distribution for the sample mean is also normal and given by:

[tex] \bar X = \sim N(\mu= 100 ,\sigma_{\bar x}= \frac{15}{\sqrt{9}}=5)[/tex]

The new z score is defined as

[tex]z=\frac{x-\mu}{\sigma_{\bar x}}[/tex]

If we apply this formula to our probability we got this:

[tex]P(\bar X\geq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\geq \frac{133-\mu}{\sigma_{\bar x}})=P(Z\geq \frac{133-100}{5})=P(Z\geq 6.6)[/tex]

And we can find this probability using the complement rule:

[tex]P(Z\geq 6.6)=1- P(z<6.6)=1-0.999999=2.06x10^{-11} [/tex]

And the probability is calculated from the normal standard table or with excel.

C.     Although the results are available,the individual IQ test scores have been lost. Can it be concluded that all 9 candidates have IQ scores above 131.5 so that they allare eligible for Mensa membership?

No. The mean can be lower than 131.5 if we find the probability:

[tex]P(\bar X\leq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\leq \frac{131.5-\mu}{\sigma_{\bar x}})=P(Z\leq \frac{131.5-100}{5})=P(Z\leq 6.3)[/tex]

[tex]P(Z\leq 6.3) \approx 1 [/tex]

Answer:

0.0139

Step-by-step explanation:

We have to find P(X≥133)

IQ scores normally distributed with mean=100 and standard deviation=sd=15

[tex]P(X\geq 133)=P(\frac{x-mean}{sd} \geq \frac{133-100}{15}[/tex]

[tex]P(Z\geq 2.2)=P(0<Z<infinity)-P(0<z<2.2)[/tex]

[tex]P(Z\geq 2.2)=0.5-0.4861[/tex]

[tex]P(Z\geq 2.2)=0.0139[/tex]

P(X≥133)=0.0139.

So, if one person is selected at random the probability of getting someone with at least 133 IQ score is 1.39%

What is the additive inverse of the complex number -8 + 3i?
-8-3i
-8+3i
8-3i
8+3i

Answers

Answer:

cc

Step-by-step explanation:

The additive inverse of the complex number -8 + 3i is 8 - 3i.

What are complex Numbers?

The complex number is basically the combination of a real number and an imaginary number.

The additive inverse of a complex number is the number that, when added to the original number, results in zero.

Or it is the negative of the original number.

To find the additive inverse of the complex number -8 + 3i

we need to negate both the real and imaginary parts. So:

Additive inverse = -( -8 + 3i)

= 8 - 3i

Therefore, the additive inverse of the complex number -8 + 3i is 8 - 3i.

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Suppose Q is the midpoint of line segment PR, PQ = x + 10, and QR = 4x - 2

What is the value of PR?
A. 4
B. 8
C. 14
D. 28

Answers

answer and step by step in the attachment

The value of PR is 28.

PQ = x + 10

QR = 4x - 2

To solve the question, we'll equate both equations which will be:

PQ = QR.

x + 10 = 4x - 2

Collect like terms

4x - x = 10 + 2.

3x = 12

x = 12/3

x = 4

Therefore,

PQ = x + 10 = 4 + 10 = 14

QR = 4x - 2 = 4(4) - 2 = 16 - 2 = 14

Therefore, PR = 14 + 14 = 28

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Find the third Taylor polynomial P3(x) for the function f(x) = (x-1)ln(x) about x0 = 1.

(a) Use P3(0.5) to approximate f(0.5). Find an upper bound for error |f(0.5) - P3(0.5)| using the error formula, and compare it to the actual error.
(b) Find a bound for the error |f(x) - P3(x)| in using P3(x) to approximate f(x) on the interval [0.5,1.5].
(c) Approximate \int_{0.5}^{1.5}f(x)dx using \int_{0.5}^{1.5}P_{3}(x)dx .
(d) Find an upper bound for the error in (c) using \int_{0.5}^{1.5}|R_{3}(x)dx| , and compare the bound to the actual error.

Answers

Answer:

A) The actual error |f(0.5) - P3(0.5)| is 0.0340735. The error using Taylor's error formula R₃(0.5) is 0.2917.

B) Te error in the interval [0.5,1.5] is 0.0340735.

C) The approximate integral using P₃(x) is 1/12

D) The real error between the two integrals is 4.687·10^(-3) while the error using \int_{0.5}^{1.5}|R_{3}(x)dx| is 0.018595

Step-by-step explanation:

A) To determine the error we first have to write the third grade Taylor polynomial P3(x):

[tex]P_n(x)=\displaystyle\sum_{k=0}^N \frac{f^{k}(a)}{k!}(x-a)[/tex] with N=3

[tex]P_3(x)=\displaystyle\sum_{k=0}^3 \frac{f^{k}(1)}{k!}(x-1)=(x-1)^2-\frac{1}{2} (x-1)^3[/tex]

The error for the third grade Taylor polynomial R₃(x) is represented as the next term non-written in the polynomial expression near to the point x:

[tex]R_3(x)=\displaystyle \frac{f^{4}(x)}{4!}(x-a)=\frac{1}{24} (\frac{2}{x^3}+\frac{6}{x^4})(x-1)^4[/tex]

Therefore the real error is:

[tex]Err(0.5)=|f(0.5)-P_3(0.5)|=0.0340735[/tex]

The error of the Taylor polynomial R₃(0.5) is:

[tex]R_3(0.5)=\displaystyle \frac{1}{24} (\frac{2}{0.5^3}+\frac{6}{0.5^4})(0.5-1)^4=0.2917[/tex]

B) The error in the interval [0.5,1.5] is the maximum error in that interval.

This is found in the extremes of the intervals. We analyze what happens in X=1.5:

[tex]Err(1.5)=|f(1.5)-P_3(1.5)|=0.01523[/tex]

[tex]R_3(1.5)=\displaystyle \frac{1}{24} (\frac{2}{1.5^3}+\frac{6}{1.5^4})(1.5-1)^4=\frac{1}{216}=4.63\cdot 10^{-3}[/tex]

Both of these errors are smaller than Err(0.5) and R₃(0.5). Therefore the error in this interval is Err[0.5,1.5] is Err(0.5).

C) The approximation of the integral and the real integral is:

[tex]I_f=\displaystyle\int_{0.5}^{1.5} f(x)\, dx=\int_{0.5}^{1.5} (x-1)ln(x)\, dx=0.08802039[/tex]

[tex]I_{p3}=\displaystyle\int_{0.5}^{1.5} P_3(x)\, dx=\int_{0.5}^{1.5} (x-1)^2-0.5(x-1)^3\, dx=1/12=0.0833333[/tex]

D) The error in the integrals is:

[tex]Errint=|I_f-I_{p3}|=4.687\cdot10^{-3}[/tex]

[tex]Errint(R_3)=\displaystyle\int_{0.5}^{1.5} R_3(x)\, dx=\int_{0.5}^{1.5} \frac{1}{24} (\frac{2}{x^3}+\frac{6}{x^4})(x-1)^4\, dx=0.018595[/tex]

With its radar, an aircraft spots another aircraft 8000 feet away at a 12 degree angle of depression Determine the vertical distance and horizontal distance between the two aircraft.

Answers

Answer:the vertical distance is 1663.2 feet.

The horizontal distance is 7824.8 feet.

Step-by-step explanation:.

The vertical distance between the two aircrafts is represented by x.

To determine x, we would apply trigonometric ratio

Sin θ = opposite side/hypotenuse

Sine 12 = x/8000

x = 8000Sin12 = 8000 × 0.2079

x = 1663.2 feet

The horizontal distance between the two aircrafts is represented by y.

To determine y, we would apply trigonometric ratio

Cos θ = opposite side/hypotenuse

Cos 12 = y/8000

x = 8000Cos12 = 8000 × 0.9781

x = 7824.8 feet

Suppose that the average and standard deviation of the fine for speeding on a particular highway are 111.12 and 13.04, respectively. Calculate an interval that is symmetric around the mean such that it contains approximately 68% of fines. Assume that the fine amount has a normal distribution. 1) (124.16, 98.08) 2) (111.12, 13.04) 3) (98.08, 124.16) 4) (85.04, 137.2) 5) (137.2, 85.04)

Answers

Answer:

3) (98.08, 124.16)

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 111.12

Standard deviation = 13.04

Calculate an interval that is symmetric around the mean such that it contains approximately 68% of fines.

68% of the fines are within 1 standard deviation of the mean speed. So

From 111.12 - 13.04 = 98.08 to 111.12 + 13.04 = 124.16

The interval notation in the smallest value before the highest value.

So the correct answer is:

3) (98.08, 124.16)

3. when the amount paid for land is $50,000 and the amount paid for expenses is $18,000, the balance in total assets after transaction (b) is $

Answers

Answer:

68,000

Step-by-step explanation:

$50,000 + $18,000 = $68,000

in the function fx=a(x+2)(x-3)^b, a and b are bothj integer constants and b is positive, find the value of x for f(x)=0

Answers

Answer:

f(x) = 0 , when x=(-2) and x=3

Step-by-step explanation:

for a≠0 and b positive then for the function

f(x) = a*(x+2)*(x-3)ᵇ

f(x) = 0 when

1) (x+2)=0 → x=(-2)

2) (x-3)=0 → x=-3

then for x=3 and x=(-2) , f(x)=0

Construct the cumulative frequency distribution for the given data. Daily Low ​(degrees°​F) Frequency ​35-39 11 ​ 40-44 22 ​ 45-49 44 ​ 50-54 13 55-59 66 ​ 60-64 88 ​ 65-69 11 Construct the cumulative frequency distribution.

Answers

Final answer:

The cumulative frequency distribution of a data set is found by adding the frequency of each category to the sum of the frequencies of all previous categories. Using this method, the cumulative frequency distribution for the given data set is as follows: 35-39: 11, 40-44: 33, 45-49: 77, 50-54: 90, 55-59: 156, 60-64: 244, 65-69: 255.

Explanation:

To construct the cumulative frequency distribution for the given data, we need to take into account the sum of all frequencies to the current one in addition to its own frequency. Let's look at how this works using the provided data:

35-39: 11 (there are no previous categories, so the cumulative frequency is 11)40-44: 22 + 11 = 3345-49: 44 + 33 = 7750-54: 13 + 77 = 9055-59: 66 + 90 = 15660-64: 88 + 156 = 24465-69: 11 + 244 = 255

Therefore, the cumulative frequency distribution for the given data set is:

35-39: 1140-44: 3345-49: 7750-54: 9055-59: 15660-64: 24465-69: 255

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A cumulative frequency distribution adds each frequency to the cumulative total of the previous ones. The sequence of cumulative frequencies for the given temperature data is 11, 33, 77, 90, 156, 244, and 255 respectively.

To construct a cumulative frequency distribution, we add up the frequencies as we go down the list of daily low temperatures. We will begin with the first frequency and add each subsequent frequency to the cumulative total of the previous frequencies. Here is how it would look for the data provided:

35-39: 11 (11)

40-44: 22 (11 + 22 = 33)

45-49: 44 (33 + 44 = 77)

50-54: 13 (77 + 13 = 90)

55-59: 66 (90 + 66 = 156)

60-64: 88 (156 + 88 = 244)

65-69: 11 (244 + 11 = 255)

The numbers in brackets represent the cumulative frequencies. Remember, to calculate Heating Degree Days (HDD) and Cooling Degree Days (CDD), you sum the number of days when the average temperature is below or above 65°F, respectively, multiplied by the difference from 65°F.

Determine whether the statement is true or false. You toss a coin and roll a die. The event​ "tossing tails and rolling a 4 or 6​" is a simple event.

Answers

Answer: False.

Step-by-step explanation:

The probability for any event = [tex]\dfrac{\text{favorable outcomes}}{\text{Total outcomes}}[/tex]

Definition : A simple event is an event that has only one possible favorable outcomes.

If we toss a coin and roll a die .

The possible outcomes for event​ "tossing tails and rolling a 4 or 6​" = {(T, 4) , (T,6)}

Since this event has 2 possible favorable  outcomes , Therefore it is not an simple event.

Hence, the given statement is "false".

Final answer:

The statement 'tossing tails and rolling a 4 or 6' is not a simple event. A simple event only consists of a single outcome, whereas this case includes multiple outcomes: getting tails in the coin toss and either a 4 or 6 in the die roll. As such, the statement is false.

Explanation:

The statement 'tossing tails and rolling a 4 or 6' is false. This is because it is not a simple event. In probability theory, a simple event is an event that consists of a single outcome. However, in this case, there are multiple outcomes - getting a tails on the coin toss, and getting either a 4 or 6 on the die roll.

For example, the possible sets are: {Tails, 4} and {Tails, 6}. This indicates that there are multiple outcomes within this event, thus it cannot be considered a simple event.

Hence, the given statement is false because a single outcome in this scenario could only be {Tails, 4} or {Tails, 6}, not both of them combined.

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The volume of a cube is 7.14 cubic meters. What is the cross-sectional area that is parallel to one of its faces?
A = ________

Answers

Answer:

A = 3.706 square meters

Step-by-step explanation:

the length of a cube has equal side.

therefore, the volume of a cube is given by S³

V = S³ = 7.14

where S is the length of a side

the surface area of a cube is = 6S²

where the area a aside is calculated, then it is multiply by 6.

if S³ = 7.14

S = ∛(7.14) = 1.925 m

What is the cross-sectional area that is parallel to one of its faces?

this is saying we should calculate for the cross-sectional area of one face. All faces in a cube is equal and parallel to each other.

the crossectional  area of one side of a cube is = S² = 1.925² = 3.706 square meters

A = 3.706 square meters

If cos theta = -2/3 which of the following are possible?

A. sin theta = - (sq5)/3 and tan theta = (sq5)/2

B. csc theta = 3/(sq5) and tan theta = -(sq5)/2

C. csc theta = -3/2 and tan theta = (sq5)/2

D. sin theta = (sq5)/3 and tan theta = (sq5)/2

Answers

Answer:

sin theta = opp/hyp = √5/3 = (sq5)/3

csc theta = 1 / sin theta = 1 / (√5/3) =  1 x 3/√5 = 3/√5 = 3/(sq5)

csc theta = 3/(sq5) and tan theta = -(sq5)/2 = B

Step-by-step explanation:

if cos theta = -2/3

the -2 is adjacent

3 is the hypothenus

using pythagoras theorem

(hypothenus)² = (opposite)² + (adjacent)²

3² = opp²+ (-2)² = opp²+ (4)

9-4 = opp² = 5

opp = √5

sin theta = opp/hyp = √5/3 = (sq5)/3

tan theta = opp/adj = √5/-2  = (sq5)/2

csc theta = 1 / sin theta = 1 / (√5/3) =  1 x 3/√5 = 3/√5 = 3/(sq5)

csc theta = 3/(sq5) and tan theta = -(sq5)/2 = b

Answer:

A and B are correct!

Step-by-step explanation:

Opposite side= - or +(sq5)

Adjacent side= -2

Hypotenuse= 3

Billy is walking from the front door of his house to his bus stop, which is 960 feet away from his front door. As Billy walks out his front door, he walks in a straight path toward his bus stop at a constant rate of 7.5 feet per second.

(A) Illustrate the situation with a diagram and define variables to represent the values of the relevant varying quantitities. (Label the variables on your picture.)
(B) Define a function f to determine Billy's distance from his bus stop in terms of the number of seconds he has been walking.
(C) What is the independent quantity and what is the domain of f (the values the independent quantity can take on)?
(D) What is the dependent quantity and what is the range of f (the values the dependent quantity can take on)?
(E) What do each of the following represent: f(0) and f(60.25)?

Answers

Answer:

b) 690 - 7.5*t

c) 0 < t < 92s time (t) is independent quantity

d) 0 < s < 690ft distance from bus stop (s) is dependent quantity

e) f(0) = 690 ft away from bus stop , f(60.25) = 238.125 ft away from bus stop

Step-by-step explanation:

Part a - see diagram

part b

initial distance from bus stop s0 = 690 ft

distance covered = 7.5*t

s = s0 - distance covered

s = 690 - 7.5*t = f(t)

part c

s = 0 or s = 690

0 = 690 -7.5*t

t = 92 s

Hence domain : 0 < t < 92s time (t) is independent quantity

part d

s = 0 or s = 690

Hence range : 0 < s < 690ft distance from bus stop (s) is dependent quantity because it depends on time (t)

part e

f(0) is s @t = 0

f(0) = 690 ft away from bus stop

f(60.25) is s @t = 60.25

f(60.25) = 690 - 7.5*60.25 = 238.125 ft away from bus stop.

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