In a circle with a radius of 3 ft, an arc is intercepted by a central angle of 2π3 radians. What is the length of the arc?
2π ft
​ 3π ​ ft
​ 6π ​ ft
​ 9π ​ ft

Answer:

Parent function is compressed by a factor of 3/4 and shifted to right by 3 units.

Step-by-step explanation:

We are asked to describe the transformation of function [tex]y=\frac{3}{4x-12}[/tex] as compared to the graph of [tex]y=\frac{1}{x}[/tex].

We can write our transformed function as:

[tex]y=\frac{3}{4(x-3)}[/tex]

[tex]y=\frac{3}{4}*\frac{1}{(x-3)}[/tex]

Now let us compare our transformed function with parent function.

Let us see rules of transformation.  

[tex]f(x-a)\rightarrow\text{Graph shifted to the right by a units}[/tex],

[tex]f(x+a)\rightarrow\text{Graph shifted to the left by a units}[/tex],

Scaling of a function: [tex]a*f(x)[/tex]

If a>1 , so function is stretched vertically.

If 0<a<1 , so function is compressed vertically.

As our parent function is multiplied by a scale factor of 3/4 and 3/4 is less than 1, so our parent function is compressed vertically by a factor of 3/4.

As 3 is being subtracted from x, so our parent function is shifted to right by 3 units or a horizontal shift to right by 3 units.

Therefore, our parent graph is compressed by a factor of 3/4 and shifted to right by 3 units to get our new graph.

Final answer:

The equation y=3/4x-12 represents a linear function with a straight line graph, while y=1/x depicts a rectangular hyperbola with a curved graph showcasing an inverse relationship. These graphs exhibit fundamentally different behaviors, one being constant and the other varying.

Explanation:

The student's question involves comparing the graph of y=3/4x-12 to the graph of y=1/x. The first equation, y=3/4x-12, describes a linear function with a slope of 3/4 and a y-intercept of -12. This means it is a straight line that increases as x increases, crossing the y-axis at (0, -12).

On the other hand, the graph of y=1/x is a rectangular hyperbola. This curve approaches the x-axis and y-axis but never touches them, known as asymptotes. The hyperbola is divided into two parts, one in the first quadrant where both x and y are positive, and another in the third quadrant where both x and y are negative. This graph does not resemble a straight line at any part.

Comparatively, the key difference lies in the nature of their graphs - one is a straight line and the other a curved hyperbola, showcasing completely different behaviors. The linear equation reflects constant change, whereas the hyperbola’s rate of change varies with x, indicating an inverse relationship.

Another way to write the given rule is (x, y) → (x – 3, y + 5).

Given that,

Based on the information, we can conclude that another way to write the given rule is (x, y) → (x – 3, y + 5).

Hence, the other options are incorrect.

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Answer:  The square of the orbital period, T, of a  planet is equal to the cube of the average distance, A, of the planet from the Sun.

Step-by-step explanation:

In mathematics, A number or variable 'a' to its second power is said to be square of 'a' i.e. [tex]\ a^2[/tex] is the square of a.

A number or variable 'b' to its third power is said to be cube of 'b' i.e. [tex]\ b^3[/tex] is the cube of b.

In the given equation: [tex]T^2 = A^3[/tex]

Here,  [tex]T^2 [/tex] is square of T.

And  [tex]A^3[/tex] is cube of A.

Therefore, The square of the orbital period, T, of a  planet is equal to the cube of the average distance, A, of the planet from the Sun.

The Geometric mean of the pair of number 6 and 10 is [tex]7.74[/tex]

The geometric mean of two number m and n is given as,

                       [tex]G.M=\sqrt{m*n}[/tex]

Geometric mean of the pair of number 6 and 10 is,

                    [tex]G.M=\sqrt{6*10} =\sqrt{60} \\\\G.M=7.74[/tex]

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The standard deviation of the customer's book purchases is 0.72.

The standard deviation of a discrete probability distribution can be calculated using the following formula:

σ = √(p[tex](x - \mu)^2[/tex])

where:

σ is the standard deviation

p(x) is the probability of the event x

μ is the mean of the distribution

In this case, the probabilities of the customer purchasing 0, 1, or 2 books are 0.2, 0.3, and 0.5, respectively. So, the mean of the distribution is:

μ = (0 * 0.2) + (1 * 0.3) + (2 * 0.5) = 1.2

The standard deviation is then:

σ = √([tex](0 - 1.2)^2 (0.3)^2[/tex] + [tex](1 - 1.2)^2 (0.3)^2[/tex] + [tex](2 - 1.2)^2 (0.5)^2)[/tex] = 0.72

So, the standard deviation of the customer's book purchases is 0.72.

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To find the number of different routes the delivery driver can take to make the deliveries at 5 out of 6 locations, we can use the concept of permutations.

To find the number of different routes the delivery driver can take, we can use the concept of permutations. Since the delivery driver has to make deliveries at 5 out of the remaining 6 locations, we need to calculate the number of ways to arrange these locations.

We can use the formula for permutations to find the number of different routes:

nPr = n! / (n - r)!

Using this formula, we can calculate the number of different routes as:

6P5 = 6! / (6 - 5)! = 6! / 1! = 6 × 5 × 4 × 3 × 2 × 1 / 1 = 720

Therefore, there are 720 different routes the delivery driver can take to make the deliveries.

The inequality -4(x+3)<-2-2x simplifies to x > -5. Therefore, the number line representing this solution should shade the values greater than -5, thus, highlighting that x is any value greater than -5.

Let's solve the inequality -4(x+3)<-2-2x step by step:

As a result, the number line that represents this solution is one where the values greater than -5 are shaded. This means the value of x is anything greater than -5. Any number line visualization should start at -5 and include everything to the right of that point.

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The original loan amount for which the final amount payable is $130.50 over nine years at an interest rate of 10% was approximately $55.39.

To find out the original amount of the loan for which the final amount payable is $130.50 over nine years at an interest rate of 10%, we will use the formula for the present value of a loan.

The formula given is P = A / (1 + r/n)^nt, where P is the principal (initial loan amount), A is the final amount, r is the annual interest rate (expressed as a decimal), n is the number of times that interest is compounded per year, and t is the time the money is borrowed for in years.

As the question doesn't specify how many times the interest is compounded per year, we will assume it is compounded once a year (n=1).

We have:

Plugging these values into the formula:

P = $130.50 / (1 + 0.10/1)^(1*9)

P = $130.50 / (1 + 0.10)^9

P = $130.50 / (1.10)^9

P = $130.50 / 2.35794769

P = $55.39 approximately

Therefore, the original loan amount was approximately $55.39.

a. Probability defective hard drive from F2 ≈ 0.3226.

b. Probability defective hard drive from F4 ≈ 0.7742.

c. Probability non-defective hard drive ≈ 0.9785.

To solve this problem, we can use Bayes' theorem, which states:

[tex]\[ P(A|B) = \frac{P(B|A) \times P(A)}{P(B)} \][/tex]

Where:

- [tex]\( P(A|B) \)[/tex] is the probability of event A happening given that event B has occurred.

- [tex]\( P(B|A) \)[/tex] is the probability of event B happening given that event A has occurred.

- [tex]\( P(A) \)[/tex] and [tex]\( P(B) \)[/tex] are the probabilities of events A and B, respectively.

Let's solve each part of the problem:

a. If a defective HDC hard drive is picked at random, what is the probability that it was produced at F2?

Let:

- A be the event that the hard drive is defective.

- B be the event that the hard drive was produced at F2.

We need to find [tex]\( P(B|A) \)[/tex], the probability that the hard drive was produced at F2 given that it is defective.

[tex]$\begin{aligned} & P(B \mid A)=\frac{P(A \cap B)}{P(A)} \\ & P(A \cap B)=P(A \mid B) \times P(B)=0.02 \times 0.25=0.005 \\ & P(A)=P(A \cap F 1)+P(A \cap F 2)+P(A \cap F 3)+P(A \cap F 4) \\ & P(A)=0.015 \times 0.20+0.02 \times 0.25+0.01 \times 0.15+0.03 \times 0.40=0.0155 \\ & P(B \mid A)=\frac{0.005}{0.0155} \approx 0.3226\end{aligned}$[/tex]

b. If a defective HDC hard drive is picked at random, what is the probability that it was produced at F4?

We need to find [tex]\( P(F4|A) \)[/tex], the probability that the hard drive was produced at F4 given that it is defective.

[tex]\[ P(F4|A) = \frac{P(A \cap F4)}{P(A)} \][/tex]

[tex]\[ P(A \cap F4) = P(A|F4) \times P(F4) = 0.03 \times 0.40 = 0.012 \][/tex]

[tex]\[ P(F4|A) = \frac{0.012}{0.0155} \approx 0.7742 \][/tex]

c. If an HDC hard drive is picked at random, what is the probability that it is non-defective?

Let P(non-defective) = P(non-defective at F1) + P(non-defective at F2) + P(non-defective at F3) + P(non-defective at F4)

P(non-defective) = [tex](1 - 0.015) \times 0.20 + (1 - 0.02) \times 0.25 + (1 - 0.01) \times 0.15 + (1 - 0.03) \times 0.40[/tex]

P(non-defective) = [tex]0.985 \times 0.20 + 0.98 \times 0.25 + 0.99 \times 0.15 + 0.97 \times 0.40[/tex]

P(non-defective) = 0.197 + 0.245 + 0.1485 + 0.388 = 0.9785

So, the probability that an HDC hard drive picked at random is non-defective is approximately [tex]\( 0.9785 \)[/tex].

There are 86 lilies he did plant in 12th row of the garden.

Arithmetic Progression (AP) is a sequence of numbers in order, in which the difference between any two consecutive numbers is a constant value. It is also called Arithmetic Sequence.

Here, The number of lilies plants in first, second third,..., and last row are respectively.

20,26,32,........

the number of rows of lilies plants is 12.

The sequence 20,26,32,........ is an A.P. with first term a =20, common difference d = 6 and n =12

formula for nth term.

∴aₙ = a+(n−1)d

aₙ=20+ (12-1).6

aₙ = 20 + 11 X 6

aₙ = 86

Thus, there are 86 lilies he did plant in 12th row of the garden.

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Answer: We want to evaluate g(x) in x = n -5, where g(x) = (x^2-6)/2x

then, doing it step by step: [tex]g( n- 5)= \frac{(n-5)^{2} - 6 }{2*(n-5)}= \frac{(n^{2} - 2*5*n + 25) - 6 }{2n - 10} = \frac{n^{2} - 10n + 19 }{2n - 10}[/tex]

where i used that (a - b)^2 = a^2 -2ab + b^2

So the correct option is C.

Answer:

  x = 5 or x = -1

Step-by-step explanation:

x^2 – 4x = 5

Solve it by completing the square method

In completing the square method, we take the coefficient of x that is -4, divide it by 2 and then square it

-4/2 = -2  

square it (-2)^2 = 4

Now add 4 on both sides

x^2 – 4x +4 = 5+4

x^2 – 4x +4 = 9

Now factor the left hand side

(x-2)(x-2)=9

(x-2)^2 = 9

Take square root on both sides

x-2 = +-3

x-2 = 3 , so x= 5

x-2 = -3, so x= -1

The Mean of the combined class is 70.19.

Mean is the average of a set of two or more numbers. The arithmetic and geometric mean are the types of mean that can be calculated.

[tex]\rm Mean = \dfrac{sum\ of\ the\ term}{total\ number\ of\ term}[/tex]

Given

The test scores of a class of 34 students have a mean of 73.1 and

The test scores of another class of 26 students have a mean of 66.4.

To find

The mean of the combined group.

The sum of the total of the test score = Total number of students x Mean

Class 1 sum will be

The sum of the total test score = 34 x 73.1

The sum of the total test score = 2485.4

Class 2 sum will be

The sum of the total test score = 26 x 66.4

The sum of the total test score = 1726.4

Than Mean for the combine will be

[tex]\rm Mean = \dfrac{sum\ of\ the\ term}{total\ number\ of\ term}\\\\Mean = \dfrac{2485.4+1726.4}{34+26} \\\\Mean = \dfrac{4211.8}{60} \\\\Mean = 70.19[/tex]

Thus the Mean of the combined class is 70.19.

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Final answer:

To represent the decimal numbers from 0 to 999 in binary code, we require 10 bits. This is determined by finding the binary equivalent of the largest number, 999, which necessitates at least 10 powers of 2 (from [tex]2^0 \ to \ 2^9[/tex]).

Explanation:

To determine how many bits are required to represent the decimal numbers in the range from 0 to 999 in straight binary code, we need to find the binary equivalent of the largest number in that range, which is 999. In binary, this would require the highest power of 2 that is less than or equal to 999. The largest power of 2 less than 999 is 512 (29), and we continue to add powers of 2 to represent the number:

Adding these up, we see that we need at least 10 bits to represent the number 999 in binary, because adding powers of 2 from 20 to 29 will give us the required range. Therefore, we need 10 bits to represent any number from 0 to 999 in binary.

The mean of the values in the stem-and-leaf plot is 42.

To find the mean of the values in the stem-and-leaf plot, we first need to interpret the plot correctly. Here's the stem-and-leaf plot description:

Stem | Leaves

 1  | 5, 8

 2  |

 3  |

 4  | 6

 5  | 0, 0, 0, 0, 7

 6  |

 7  |

 8  |

 9  |

Let's list the individual values from the plot:

- From stem 1: Leaves are 5 and 8

- From stem 4: Leaf is 6

- From stem 5: Leaves are 0, 0, 0, 0, 7

Now, we calculate the mean (average) of these values.

Step-by-step calculation:

1. List of values:

  - Values from stem 1: 15, 18

  - Values from stem 4: 46

  - Values from stem 5: 50, 50, 50, 57

2. Count the number of values:

  - There are 2 values from stem 1, 1 value from stem 4, and 5 values from stem 5.

  - Total count = 2 + 1 + 5 = 8

3. Calculate the sum of all values:

  - Sum = 15 + 18 + 46 + 50 + 50 + 50 + 50 + 57

  - Sum = 336

4. Calculate the mean:

  - Mean = Sum / Count

  - Mean = 336 / 8

  - Mean = 42

Therefore, the mean of the values in the stem-and-leaf plot is [tex]{42} \).[/tex]