Question 2
Given the following information, find the probability that a randomly selected student will be short or very short. Number of students who are very short: 45, short: 60, tall: 82, very tall: 21
2DB. Probabilities from a given distribution of frequencies (Links to an external site.) (DOCX)
Group of answer choices
50.5%
21.0%
49.5%
28.8%

The relative frequency of yellow is 25%, which means that yellow was the outcome of 25 out of every 100 selections.

What is the relative frequency?

Relative frequency is the ratio of the number of times an event occurs to the total number of data points.

The relative frequency of landing on yellow is 100/400 = 0.25 or 25%.

This is found by dividing the number of times yellow is landed on (100) by the total number of trials (400).

This means that out of the 400 times that Amelia selected a counter, 25% of those times the counter was yellow.

It is a measure of how often a particular outcome occurs in comparison to the total number of trials or selections.

Hence, the relative frequency of yellow is 25%, which means that yellow was the outcome of 25 out of every 100 selections.

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Answer:

A) 540

B) x = 115/3

Step-by-step explanation:

A) to find the sum of the interior angles of any gon, use this formula: 180(n-2); n = the number of sides of the gon you are given (ex. octagon will be n=8). Therefore, a pentagon will be: 180(5-2) = 540

B) To find x, we can add all the angles and make them equal to 540.

(3x+50) + (2x+60) + (3x-30) + 2x + 2x = 540

12x = 460

x = 115/3

C) Plug-in x for every angle

To draw a box plot for the quiz scores of Professor A's class, you would start by drawing a number line along the horizontal axis and marking the minimum, Q1, median, Q3, and maximum scores.

Then, you would draw a box from Q1 to Q3 to represent the middle 50% of the data in professor A's class. The median score would be marked within the box, and a line would be drawn from either end of the box to the minimum and maximum scores to represent the lower and upper extremes of the data.

The highest quiz score is 20, and the lowest quiz score is 3.

The score that separates the bottom 75% from the top 25% is Q3, which is 16.

By definition, half of the class should have scored more than the median, so 20 students scored more than 14 points.

By definition, the median is 14, so half of the class should have scored more than 14 points, so 20 students scored more than 14 points.

Using the min value and Q1 value, we know that 25% of the students scored less than 12 points.

To determine if 4 is an outlier, we can use the IQR (interquartile range) method, which is calculated by subtracting Q1 from Q3. In this case, the IQR is 4 (16 minus 12). We can then use the IQR to find the lower and upper bounds of the data, which are determined by subtracting 1.5 IQR from Q1 and adding 1.5 IQR to Q3. In this case, the lower bound is 8 (12 minus 1.54) and the upper bound is 20 (16 plus 1.54). 4 is lower than the lower bound; therefore, it is considered an outlier.

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The required sample size for the confidence interval is given as follows:

n = 267.

A confidence interval of proportions has the bounds given by the rule presented as follows:

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which the variables used to calculated these bounds are listed as follows:

The margin of error is defined as follows:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

The confidence level is of 95%, hence the critical value z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so the critical value is z = 1.96.

The parameters for this problem are given as follows:

[tex]M = 0.06, \pi = 0.5[/tex]

The estimate is of 0.5 as it is when the highest sample size is needed, and the sample size is obtained as follows:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.06 = 1.96\sqrt{\frac{0.5(0.5)}{n}}[/tex]

[tex]0.06\sqrt{n} = 1.96 \times 0.5[/tex]

[tex]\sqrt{n} = \frac{1.96 \times 0.5}{0.06}[/tex]

[tex](\sqrt{n})^2 = \left(\frac{1.96 \times 0.5}{0.06}\right)[/tex]

n = 267.

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The areas of the shaded region of the given polygons is as follows

1. Area = 18.84cm²

2. Area = 67cm²

3. Area = 602.88cm²

4. Area = 1.14cm²

5. Area = 182.72cm²

6. Area = 33 cm²

The area of the shaded area is the difference between the total area of the polygon and the area of the portion of the polygon that is not shaded. In polygons, the area of the shaded component might appear in two different ways. A polygon's sides or its center are both potential locations for the shaded area.

Here, we have

1. r= 6 cm, θ = 60°

Area of shaded region = πr² × θ/360

Area = 3.14 × 6× 6 × 60/360

Area = 18.84cm²

2. r= 8 cm

θ = 360-240 = 120

Area = 3.14 × 8× 8 × 120/360

Area = 67cm²

3. r = 16 cm

θ = 360 - 90 = 270

Area = 3.14 × 16× 16 × 270/360

Area = 602.88cm²

4. r= 2 cm

θ =  90

Area = 3.14 × 2× 2 × 90/360

Area = 1.14cm²

5. r= 8 cm

θ =  270

Area  = πr² × θ/360 + 1/2(bh)

Area = 3.14 × 8× 8 × 270/360 + 1/2(8×8)

Area = 182.72cm²

6. R = 7 cm

r = 4 cm

Area = πR² - πr²

= 22/7 × 7² - 22/7 × 4²

= 49 - 16

= 33 cm²

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Full question:

Answer:  [tex]y=\frac{1}{5} x+6[/tex]

Step-by-step explanation:

(-10,4) **substitute these points into the equation below**

y = [tex]\frac{1}{5} x+c[/tex]

4   =  [tex]\frac{1}{5}[/tex][tex](-10)[/tex][tex]+c[/tex] (make c the subject)

c  = 4 + 2

 c = 6 (this is the y-intercept that is needed when creating the equation of the line)

therefore: [tex]y=\frac{1}{5} x+6[/tex]

[tex]\sf y =\dfrac{1}{5} x+6.[/tex]

We're given the slope and an ordered pair the function passes through.

[tex]\sf Slope (m)=\dfrac{1}{5}[/tex]

[tex]\sf Point: (-10,4)\\ \\Therefore:\\x_{1}=-10\\y_{1} =4[/tex]

Here's that formula: [tex]\sf y-y_{1} =m(x-x_{1} )\\ \\[/tex]

[tex]\sf y-(4) =(\dfrac{1}{5} )(x-(-10) )\\ \\[/tex]

[tex]\sf y-(4) =(\dfrac{1}{5} )(x+10 )\\ \\\\\sf y-(4) =(\dfrac{1}{5} )(x)+(\dfrac{1}{5} )(10)\\ \\ \\\sf y-(4) =\dfrac{1}{5} x+(\dfrac{10}{5} )\\ \\ \\\sf y-4 =\dfrac{1}{5} x+2\\ \\ \\\sf y-4+4 =\dfrac{1}{5} x+2+4\\ \\ \\\sf y =\dfrac{1}{5} x+6[/tex]

From plain sight we can tell that the slope of this equation is indeed 1/5, because the value that multiplies "x" is 1/5 when the equation is solved for "y". Now, to make sure it passed through the given point, substitute "x" by "-10" and make sure it returns a value of "4" for "y".

[tex]\sf y =\dfrac{1}{5} (-10)+6\\ \\ \\\sf y =\dfrac{-10}{5}+6\\ \\ \\\sf y =-2+6\\ \\ \\y=4[/tex]

The answer is correct!

Therefore, the equation of the line that has a slope or 1/5 and passes through (-10,4) is: [tex]\sf y =\dfrac{1}{5} x+6[/tex].

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This Polynomial expression x² + 4 could be added to the expression 2x² - x to result in a sum that contains only a constant term 4.

What is the polynomial expression?

Any expression which consists of variables, constants, and exponents, and is combined using mathematical operators like addition, subtraction, multiplication, and division is a polynomial expression. Polynomial expressions can be classified as monomials, binomials, and trinomials according to the number of terms present in the expression.

A polynomial expression that could be added to 2x² - x to result in a sum that contains only a constant term 4 is x² + 4.

When adding x² + 4 to 2x² - x, the x² terms will cancel out and the constant term 4 will be left.

(x² + 4) + (2x² - x) = (x² + 2x²) + (4 - x) = 3x² -x + 4

The resultant polynomial expression is 3x² -x + 4 in which the only constant term is 4.

Hence, This Polynomial expression x² + 4 could be added to the expression 2x² - x to result in a sum that contains only a constant term 4.

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Answer:

$20.44

Step-by-step explanation:

The average of a set of numbers is defined as:

[tex]\dfrac{\textrm{sum of values}}{\textrm{number of values}}[/tex]

In this problem, we are shown a table where each row has a coupon value in the left column and the number of that coupon value in the right column (e.g., if we look at the top row, we can see there are 70 coupons each valued at $10).

So, the sum of values in this problem (i.e., the total number of dollars given out by the store in the form of coupons) is defined as the sum of the product of each row.

[tex]\textrm{sum of values} = (\$10 \times 70) + (\$20 \times 40) + (\$40 \times 20) + (\$60 \times 4) + (\$120 \times 2)[/tex]

[tex]\textrm{sum of values} = \$700 + \$800 + \$800 + \$240 + \$240[/tex]

[tex]\textrm{sum of values} = \$2780[/tex]

The number of values in this problem is just the sum of the numbers in the right column (i.e., the number of coupons given out).

[tex]\textrm{number of values} = 70 + 40 + 20 + 4 + 2[/tex]

[tex]\textrm{number of values} = 136[/tex]

Finally, to answer the problem, we can plug the two numbers that we just solved for into the formula for the average of a set.

[tex]\textrm{average savings} = \dfrac{\textrm{value of all tickets}}{\textrm{number of tickets}}[/tex]

[tex]\textrm{average savings} = \dfrac{\$2780}{136}[/tex]

[tex]\textrm{average savings} \approx \$20.44[/tex]

The value of the six trigonometric functions as ( -3 ,-5 ) is the point on the terminal side of an angle is given by:

sin α = -5 /√34                       cosec α =  - √34 / 5

cos α = -3 / √34                    sec α = -√34 / 3

tan α = 5 / 3                             cot α = 3 /5

As given in the question,

In standard position ( -3 ,-5 ) represents the terminal side of an angle.

Here in right angled triangle,

Base = -3

Height = -5

Using Pythagoras theorem ,

Hypotenuse = √ ( -3 )² + ( -5 )²

                    = √9 + 25

                    = √34

Value of the six trigonometric functions are given by :

sin α = -5 /√34

cos α = -3 / √34

tan α = 5 / 3

cosec α =  - √34 / 5

sec α = -√34 / 3

cot α = 3 /5

Therefore, the value of the six trigonometric functions as per the given terminal point of an angle is equal to :

sin α = -5 /√34                       cosec α =  - √34 / 5

cos α = -3 / √34                    sec α = -√34 / 3

tan α = 5 / 3                             cot α = 3 /5

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It has been proven that there exists a positive integer n < 10^6 such that 5^n has six consecutive zeros in its decimal representation. This is done by induction, where the base case is when n = 6, and the inductive step is when n is increased by 1.

We can prove this by induction.

When n = 6, 5^6 = 15,625 and it has six consecutive zeros in its decimal representation.

Suppose that there exists a positive integer n < 10^6 such that 5^n has six consecutive zeros in its decimal representation.

Let k = n + 1. Then 5^k = 5^n * 5 and it has six consecutive zeros in its decimal representation.

Thus, by induction, we can conclude that there exists a positive integer n < 10^6 such that 5^n has six consecutive zeros in its decimal representation.

It has been proven that there exists a positive integer n < 10^6 such that 5^n has six consecutive zeros in its decimal representation. This is done by induction, where the base case is when n = 6, and the inductive step is when n is increased by 1.

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Countless pointy corners are called cusps. A cusp is a place where a vertical tangent exists, but where one side's derivative is positive and the other side's derivative is negative.

The above example of a paradigm is y=x23. The derivative's upper limit as you move left toward zero is. As you move toward zero from the right, the derivative's limit is +. A vertical tangent exists at x=0.

Cusps have vertical tangents (when describing functions). A vertical tangent, however, is not a cusp by itself. The y=x13 curve is smooth, showing no behavior like a pointed needle, although it does have a vertical tangent at the origin.

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The formula for the bill total for the week by summing the daily bill amounts is  (=sum(B25:H25)).

Numerous common mathematical operations are carried out by the Excel Math Functions, such as fundamental arithmetic, conditional sums and products, exponents and logarithms, and trigonometric ratios. Please take note that the Excel Statistical Operations and Excel Engineering Functions categories also offer other math-related Excel functions.

Given finding the  bill total for the week in cell B26

To sum the daily bill amounts, use Excel's Sum function, which is a pretty simple mathematical procedure. You can manually input the sum function by typing (=sum) in the field designated for the sum, after which you provide the parameters (=sum(B25:H25)). When you hit "enter," the outcome is presented immediately. By selecting the Sum Function from the Ribbon or by hitting Alt + =, this can also be done automatically.

Hence the formula is  (=sum(B25:H25)).

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Answer:

Step-by-step explanation:

250

Answer:250

Step-by-step explanation: just add 10 till you get 250

The total cost of production of six dining tables is $1300.

Production is the act of combining a variety of material and immaterial inputs to create an output. In a perfect world, this output would be a good or valuable good or service for people to use. The three main production factors of land, labor, and capital are known as primary producers of goods or services.

These crucial inputs are not significantly altered by the output process or the end product, nor are they made a crucial component of it. In classical economics, materials and energy are regarded as secondary factors because they are byproducts of land, labor, and capital. Technology and entrepreneurship are sometimes seen as evolved variables in production in addition to the conventional components of production.

Each dining table for $1,000

The second it's $1,100.

The marginal cost of each table increases by $200

The total cost of his production is

Fixed cost + Variable cost

= $1000 + 300= $1300

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Answer:

Option B

Step-by-step explanation:

If you replace the X in the P(X) it becomes -1 and then you'll change every x in the equation to -1

2(-1)³ + 2(-1)² -3(-1) -3

which = -2 +2 +3 -3

= 0

Since the answer is 0 it means x + 1 is a factor of 2x³ + 2x² - 3x -3

Answer:

-----------------------

Factorize the given polynomial:

As we see the polynomial is divisible by x + 1.

Correct choice is B.

Therefore, additivity and homogeneity of T are satisfied, T is linear.

T must be linear for both b and c to be true. T is linear, hence additivity is true for every p, q, and P. (R). In order to make our computations as straightforward as feasible, it would be a good idea for us to use straightforward polynomials in P(R). p, q ∈ P(R), where p(x) = [tex]\frac{\pi }{2}[/tex] and q(x) =  [tex]\frac{\pi }{2}[/tex] for all x ∈ R and so we have

T(p + q) =(3(p + q)(4) +5(p + q)'(6)+b(p + q)(1)(p + q)(2) , [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](p + q)(x) d(x) +               c sin((p + q)(0))

           = (3(p(4)+q(4)) + 5(p'(6)+q'(6))+b(p(1)+q(1))(p(2)+q(2)) , [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](p(x)+q(x))d(x)+ c sin(p(0)+q(0)))

           = (3([tex]\frac{\pi }{2}[/tex] +[tex]\frac{\pi }{2}[/tex])+5(0+0)+b([tex]\frac{\pi }{2}[/tex]+[tex]\frac{\pi }{2}[/tex])([tex]\frac{\pi }{2}[/tex]+[tex]\frac{\pi }{2}[/tex]),  [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex]([tex]\frac{\pi }{2}[/tex]+[tex]\frac{\pi }{2}[/tex])d(x)+c sin([tex]\frac{\pi }{2}[/tex]+[tex]\frac{\pi }{2}[/tex]))

           = (3([tex]\pi[/tex])+b[tex]\pi ^{2}[/tex],[tex]\frac{15\pi }{4}[/tex])

and

Tp + Tq = (3p(4) +5(p)'(6)+bp(1)(p)(2) , [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](p)(x) d(x) + c sin((p)(0)) +(3(q)(4) +5(q)'(6)+b(q)(1)(q)(2) , [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](q)(x) d(x) +c sin(q(0)))

           = (3([tex]\frac{\pi }{2}[/tex])+5(0)+b([tex]\frac{\pi }{2}[/tex])([tex]\frac{\pi }{2}[/tex]),  [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex]([tex]\frac{\pi }{2}[/tex])d(x)+c sin([tex]\frac{\pi }{2}[/tex])) +  (3([tex]\frac{\pi }{2}[/tex])+5(0)+b([tex]\frac{\pi }{2}[/tex])([tex]\frac{\pi }{2}[/tex]),  [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex]([tex]\frac{\pi }{2}[/tex])d(x)+c sin([tex]\frac{\pi }{2}[/tex]))

           =(3([tex]\frac{\pi }{2}[/tex])+[tex]\frac{\pi b}{4}[/tex],[tex]\frac{\pi }{2}[/tex]  [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex]d(x) + c) +(3([tex]\frac{\pi }{2}[/tex])+[tex]\frac{\pi b}{4}[/tex],[tex]\frac{\pi }{2}[/tex]  [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex]d(x) + c)

           =(3[tex]\pi[/tex]+ [tex]\frac{\pi b}{2}[/tex],[tex]\frac{15\pi }{4}[/tex]+2c)

Since T is linear, additivity of T holds and implies that we have

              (3([tex]\pi[/tex])+b[tex]\pi ^{2}[/tex],[tex]\frac{15\pi }{4}[/tex]) = T(p+q)

                                     =Tp+Tq

                                     =(3[tex]\pi[/tex]+ [tex]\frac{\pi b}{2}[/tex],[tex]\frac{15\pi }{4}[/tex]+2c)

from which we can equate the coordinates to obtain the equations 3π + πb/2= 3π +πb/2 and 15π/4 =15π/4+2c, which imply b = 0 and c = 0, respectively.

Backward direction: If b = 0 and c = 0, then T is linear. Suppose b = 0 and c = 0. Then the map T : R ^3 → R^2 becomes

Tp = (3p(4) +5(p)'(6) , [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](p)(x) d(x) )

we need to prove that T is linear

• Additivity: For all p, q ∈ P(R), we have

 T(p+q) = (3(p + q)(4) +5(p + q)'(6) , [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](p + q)(x) d(x))

             =(3(p(4)+q(4)) + 5(p'(6)+q'(6)) , [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](p(x)+q(x))d(x))

             =  (3p(4) +5(p)'(6)+3q(4)+5q'(6) , [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](p)(x) d(x) + [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex]q(x) d(x))

              = (3p(4) +5(p)'(6) , [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](p)(x) d(x)) +(3(q)(4) +5(q)'(6), [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](q)(x) d(x))

             =Tp+Tq

• Homogeneity: For all λ ∈ F and for all (x, y, z) ∈ R^3, we have

                 T(λp) = (3(λp)(4) + 5(λp)'(6), [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](λp)(x) d(x))

                           =(3λp(4) + 5λp'(6), [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex]λ(p)(x) d(x))

                          =(λ(3p(4) + 5p'(6)),λ [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex]p)(x) d(x))

                          =λ(3p(4) + 5p'(6), [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex]p)(x) d(x))

                         =λTp

 Therefore, additivity and homogeneity of T are satisfied, T is linear.

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Answer:

The original number is 62.

Step-by-step explanation:

Solution :

Let,

Units digit = x

Tens digit = 8 - x

★ Original number :

10 (8 - x) + x

80 - 10x + x

80 - 9x

When the digits of the number are interchanged :

10 (x) + 8 - x

10 x + 8 - x

9x + 8

(80 - 9x) - 36 = 9x + 8

80 - 36 - 9x = 9x + 8

44 - 9x = 9x + 8

44 - 8 = 9x + 9x

36 = 18x

x = 36/18

x = 2

Units digit = 2

Tens digit = 8 - x

8 - 2 = 6

Therefore, the original number is 62.

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Answer:

68 oz.

Step-by-step explanation:

The bag weighs 4.25lbs

1lb=16oz.

4.25*16=68oz.

Commercial paper with a maturity of 270 number of days or less is a type of short-term debt instrument that can be used for a variety of purposes, including funding operating expenses and managing cash flow.

It is typically issued by large corporations with high credit ratings. Commercial paper typically has a lower interest rate than other types of debt instruments, making it an attractive option for businesses seeking to raise capital quickly.

Commercial paper is a type of short-term debt instrument with a maturity of 270 number of days or less. It is typically issued by large corporations with high credit ratings and has a lower interest rate than other types of debt instruments, making it an attractive option for businesses seeking to raise capital quickly.

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Answer: [tex]V=-62000x+15200000[/tex]

Step-by-step explanation:

The slope of the line is [tex]\frac{900000-1520000}{100-90}=-62000[/tex].

Using point-slope form, the equation is:

[tex]V-900000=-62000(x-100)\\\\V-900000=-62000x+620000\\\\V=-62000x+15200000[/tex]

It is possible to find the value of S20 by applying arithmetic progression. similar by applying the sum formula.

Formula is used to determine S20.

First term (a) here equals 32, and common difference (d) is equal to 2.9

Sn = n/2 × ( 2a + (n-1) × d)

S20 = 20/2 × ( 2×32 + ( 20 - 1 ) × 2.9 )

S20 = 20/2 × ( 64 + (19 × 2.9) )

S20 = 20/2 × ( 64 + 55.1 )

S20 = 20/2 × 119.1

S20 = 1191

Therefore, value of S20 is 1191

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Answer:

Step-by-step explanation:

[tex]s_{n}=\frac{n}{2} [2a+(n-1)d]\\s_{20}=\frac{20}{2} [2 \times 32+(20-1)\times 2.9]\\=10[64+19\times 2.9]\\=640+551\\=1191[/tex]

Answer: -24

Step-by-step explanation:

⇒4(-8)+8

in such questions, first solve the bracket. In this question there is only a single term within the bracket, -8. Brackets usually mean multiplication. So we multiply 8 by 4 to get 32. Now since we need to give 32 a sign as well, so in case of multiplication we multiply the sign of 8 and 4 together i.e negative and positive respectively. Negative positive multiplied give us negative. Therefore the sign of 32 will be negative.

⇒- 32 + 8

now again we are stuck with the sign issue. Notice that we have both a negative and a positive sign as well. How do we know if we are supposed to add or subtract? In order to decide which operation we apply on the two numbers, we first multiply the signs. Negative positive multiplied give negative, therfore we will do subtraction. 32 subtracted from 8 gives 24. Now again we are supposed to give 24 a sign. Unlike in multiplication, when we do addition or subtraction, instead of multiplying the signs to give the result a sign, we look for the sign of the greater operand, i.e 32, because 32 is greater than 8 and has a negative sign, so we give 24 a negative sign. H

⇒-24 answer

⇒4(-8)+8

⇒- 32 + 8

⇒-24 answer. hope the helps ...

1447.5 is 75% of all years have an FTES at or below .

What does FTEs mean?

An employee's scheduled hours are divided by the employer's hours for a full-time workweek to determine their full-time equivalent (FTE). Employees who are scheduled to work 40 hours per week for an employer are considered 1.0 FTEs.

                              A worker's scheduled hours are divided by the company's work hours on a weekly full-time basis to determine their full-time equivalent (FTE). A 40-hour workweek means that there will be 1.0 FTEs of employees working for that company throughout that time.

We know that,

75% of all the data are at or below the value of third quartile.

Hence, required correct answer is,

75% of all years have an FTES at or below 1447.5 .

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There amount are 504 different groups of 3 side dishes that a customer can choose from.

There are 9 possible side dishes, and the customer can choose any 3 of them. The number of different groups of 3 side dishes can be determined using the formula:

nCr = n!/(r!(n-r)!).

In this case,

n = 9 and r = 3,

so the number of possible groups is

9!/(3!(9-3)!) = 504.

When a customer buys a family-sized meal at a certain restaurant, they get to choose 3 side dishes from 9 options. The number of different groups of 3 side dishes possible for the customer to choose from can be determined using the formula for combination: nCr = n!/(r!(n-r)!). In this case, n = 9 (the number of possible side dishes) and r = 3 (the number of side dishes chosen). Applying the formula, the number of possible groups of 3 side dishes is 9!/(3!(9-3)!) = 504. This means that there are 504 different groups of 3 side dishes that a customer can choose from.

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What is the exponential function?

An exponential function is a type of mathematical function in which the output (the y-value) is a constant multiplied by a fixed number (the base) raised to the power of the input (the x-value). The general form of an exponential function is y = ab^x, where a and b are constants and x is the input value.

1 .To estimate f(1/3) using the graph points (0, 12) and (1, 0.75), we can use the fact that an exponential function has the form y = ab^x, where a and b are constants and x is the input value. By looking at the graph, we can see that the y-intercept (when x = 0) is 12, which means that a = 12. We can also see that the function passes through the point (1, 0.75), which means that f(1) = 0.75. Using this information, we can write the equation for the function as:

y = 12b^x

f(1/3) = 12b^(1/3)

Since we don't know the value of b, it's impossible to know the exact value of f(1/3) from the given information. However, it tells us that the amount of medicine in the bloodstream decreases exponentially and how fast it decreases depends on the value of b.

2. To find the equation that defines f, we can use the point (1, 0.75). We know that f(1) = 0.75, so we can substitute this into the equation we found earlier:

0.75 = 12b^1

0.75 = 12b

b = 0.0625

So, the equation that defines f is:

y = 12 * 0.0625^x

or

y = 0.75 * 0.0625^x

This equation tells us that the amount of medicine in the bloodstream decreases exponentially with time, with a rate determined by the value of b = 0.0625.

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h(x)=f(x)/g(x) separating a fraction like this, the quotient rule is applicable.

The numerator is the first function.

The denominator is the second function.

In essence, it is [(derivative of first function)* second function - (derivative of second function)* first function] divided by the square of the second function. So the functions would the quotient rule be considered the best method for finding the derivative-

When separating a fraction like this, the quotient rule is applicable. f(x)/g(x)

This is differentiated by squaring the denominator and using the product rule to the numerator:

h(x)=f(x)/g(x)

therefore:

h′(x)=(f′(x)g(x)−f(x)g′(x))/g2(x)

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a. The sample mean time to get to work for public transportation is 31.3 minutes and the sample mean time to get to work for the automobile is 32.1 minutes.

b. The sample standard deviation for public transportation is 4.2 minutes and the sample standard deviation for the automobile is 1.7 minutes.

c. Based on the results from parts (a) and (b), public transportation should be preferred. The sample mean time for public transportation is slightly lower than the sample mean time for the automobile, and the sample standard deviation for public transportation is much lower than the sample standard deviation for the automobile.

This indicates that public transportation is more consistent and reliable than the automobile, and therefore should be preferred as a method of transportation to get to work each day.

Complete question:

Public transportation and the automobile are two methods an employee can use to get to work.

each day. Samples of times recorded for each method are shown. Times are in minutes.

Public Transportation: 28 29 32 37 37 33 25 29 32 41 34

Automobile: 29 31 33 32 34 30 31 32 35 33

a. Compute the sample mean time to get to work for each method.

b. Compute the sample standard deviation for each method.

c. On the basis of your results from parts (a) and (b), which method of transportation should be preferred? Explain.

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Answer: 3x + 5

x = x + 2

3 (x+2) -1

3x + 6 - 1

3x+5

Adults make up 0.02100 of the probability for whom the test would be positive.

The proportion of adults for which the test would be positive can be calculated by multiplying the probability of a randomly selected adult having the rare disease (0.001) by the probability that the test will be positive for a person with the disease (0.99) and adding it to the probability that the test will be positive for a person without the disease (0.02).

0.001 x 0.99 = 0.00099

0.00099 + 0.02 = 0.02099

0.02099 = 0.02100 (rounded)

Therefore, the proportion of adults for which the test would be positive is 0.02100.The proportion of adults for which the test would be positive is 0.02100, which is calculated by taking into account the probability of a randomly selected adult having the rare disease (0.001) and the probability that the test will be positive for a person with the disease (0.99) and a person without the disease (0.02). This reflects how the test is not perfect and can sometimes give false positives or false negatives. The result of 0.02100 indicates that out of every 1000 adults, 2 of them will have a positive test result.

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The point-slope form of the equation that passes through the points  [tex](2, 4)[/tex]and  [tex](-3, -6)[/tex]  is:   [tex]y - 4 = -5/5 (x - 2)[/tex].

To find the equation in point-slope form, you can use the point-slope formula:   [tex]y - y1 = m(x - x1)[/tex]

Where  [tex](x1, y1)[/tex]  is a point on the line, and m is the slope of the line, which can be found by using the two points given:

[tex]m = (y2 - y1) / (x2 - x1)[/tex]

[tex]m = (-6 - 4) / (-3 - 2) = -10/5 = -2[/tex]

Then you can substitute the values into the point-slope formula:

[tex]y - 4 = -2(x - 2)[/tex]

[tex]y = -2x + 8[/tex]

[tex]y - 4 = -5/5 (x - 2)[/tex]

This is the final equation in point-slope form that passes through the points  [tex](2, 4)[/tex]  and  [tex](-3, -6).[/tex]

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