Rectangle ABCD and A’B’C’D’ are similar.

a. What is the scale factor from ABCD to

A’B’C’D’?

b. What is the scale factor from A’B’C’D’ to

ABCD?

Answer:

C

Step-by-step explanation:

if the mean and median are the same, the graph will be symmetrical

18. C

(m / 2) - 6 = (m / 4) + 2

---Multiply everything by the LCM of the denominators

---LCM = 4

2m - 24 = m + 8

m - 24 = 8

m = 32

19. A

k / 12 = 25 / 100

---We can simplify 25/100

---We want to simplify enough to where the denominator of 25/100 is a multiple or factor of 12

k / 12 = 1 / 4

---4 x 3 = 12, 1 x 3 = 3

k = 3

20. A

9 / 5 = 3x / 100

---Cross multiply and solve algebraically

(5 * 3x) = (9 * 100)

15x = 900

x = 60

Hope this helps!

An estimate for the mean time spent watching television is approximately 24.375 hours.

Midpoint of 0 < h ≤ 10 = (0 + 10) / 2 = 5

Midpoint of 10 < h ≤ 20 = (10 + 20) / 2 = 15

Midpoint of 20 < h ≤ 30 = (20 + 30) / 2 = 25

Midpoint of 30 < h ≤ 40 = (30 + 40) / 2 = 35

Sum of observations = (5 × 5) + (15 × 15) + (25 × 50) + (35 × 10) = 125 + 225 + 1250 + 350 = 1950

Total number of observations = 5 + 15 + 50 + 10 = 80

Mean = Sum of observations / Total number of observations = 1950 / 80 = 24.375

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Answer:

Step-by-step explanation:

We are given a figure, consisting of a right triangle with a circle cut out of it.

To Calculate : area of the shaded region below,

First Let's find the Area of the right triangle:

Area(triangle) = 1/2 × Base × Height

= 1/2 × 56 × 56

= 1/2 × 3136

= 3136/2

= 1568 m².

Now, We are diameter of the circle = 20 m

Radius of the circle = 20/2 = 10 m

Acc. to question we have to use 3.14 as an

approximation for pi.

Area (circle) = 3.14 × 10 × 10

Area (circle) = 3.14 × 100

Area (circle) = 314 m²

Area of shaded region = Area of triangle - Area of circle =

= 1568 m² - 314 m²

= 1.254 square meters

Therefore, The approximate area of the shaded region is 1.254 square meters.

Option (a) is the required answer

Answer:

yes

Step-by-step explanation:

the factor theorem says if (x + a) is a factor of a polynomial f(x) then f(- a) = 0

given (x+ 5) then if f(- 5) = 0 , (x + 5) is a factor

4[tex](-5)^{4}[/tex] + 16(- 5)³ - 6(- 5)² + 73(- 5) + 15

= 4(625)+ 16(- 125) - 6(25) - 365 + 15

= 2500 - 2000  - 150 - 350

= 500 - 500

= 0

then (x + 5) is a factor of the given polynomial

Here is the data collected as a result of the experiment.
1) Numerical order is: 0.20, 0.20, 0.46, 0.46, 2.10, 2.10, 5.50, 11.80, 12.20, 12.20, 13.22, 14.65, 16.07, 17.49, 18.91, 20.33, 20.50, 20.50, 21.75, 23.18

2)  The data represents rainfall in millimeters over a 20-day period.

3) Mean = 11.69 mm.

4) The minimum rainfall is 0.20 mm ;

the maximum is 23.18 mm.

1) The data is listed as follows.

2) The data represents the amount of rainfall in millimeters over a 20 day period.

3) The mean rainfall over the 20 day period is 11.69 mm.

= (0.20 + 0.20 + 0.46 + 0.46 + 2.10 + 2.10 + 5.50 + 11.80 + 12.20 + 12.20 + 13.22 + 14.65 + 16.07 + 17.49 + 18.91 + 20.33 + 20.50 + 20.50 + 21.75 + 23.18) /20

Mean (x)  = 11.69 mm.

3) The minimum rainfall is 0.20 mm and the maximum is 23.18 mm.

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The value of theta in the given trigonometry function is 45⁰.

The value of theta in the given trigonometry function is calculated as follows;

9 sin²θ tanθ  -  9 sin²θ = 0

Solve the equation as follows;

9 sin²θ tanθ  =  9 sin²θ

divide both sides by  9 sin²θ;

(9 sin²θ tanθ)/9 sin²θ = 9 sin²θ /9 sin²θ

tanθ = 1

Now, solve for θ as follows;

θ = tan⁻¹ (1)

θ = 45⁰

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The number of kg of maize was left with Ali is 73350 Kg.

Given that, Ali had 3469 bags of maize each weighing 90 kg.

Ali sold 2654 of them.

Number of bags left = 3469-2654

= 815

Number of Kg's of maize left = 815×90

= 73350 Kg

Therefore, the number of kg of maize was left with Ali is 73350 Kg.

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The  equation accurately represents this statement is "Negative 3 less than 4.9 times a number, x, is the same as 12.8"

The statement is "Negative 3 less than 4.9 times a number, x, is the same as 12.8." We can break it down into two parts:

"4.9 times a number, x": This means we need to multiply a number, x, by 4.9. So the first part of the equation is 4.9x.

"Negative 3 less than 4.9 times a number, x, is the same as 12.8":

This means we need to subtract 3 from the result of multiplying x by 4.9. So the second part of the equation is -3.

Putting it all together, we get:

4.9x - 3 = 12.8

This equation accurately represents the given statement.

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The calculated value of the length RP is 107 units

From the question, we have the following parameters that can be used in our computation:

The circles and the tangent lines

Start by calculating the segment PQ using the following Pythagoras theorem

PQ² = (50 + 11)² + 11²

Evaluate

PQ² = 3842

So, we have

PQ = 62

The value of the length RP is then calculated as

RP = PQ + TU

So, we hav

RP = 62 + 45

Evaluate

RP = 107

Hence, the value of the length RP is 107 units

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The integral of  [tex]\int\limits^e_1 {x^{1/2}lnx } \, dx[/tex]  is (2/3)[tex]e^{3/2}[/tex] - (4/9).

To find the integral of   [tex]\int\limits^e_1 {x^{1/2}lnx } \, dx[/tex]  , we can use integration by parts. Let u = ln x and dv = [tex]x^{1/2}[/tex] dx. Then du/dx = 1/x and v = (2/3) [tex]x^{3/2}[/tex] .

Using the formula for integration by parts, we have:

∫[tex]x^{1/2}[/tex]lnx dx = uv - ∫v du/dx dx

= ln x * (2/3) [tex]x^{3/2}[/tex]  - ∫(2/3) [tex]x^{3/2}[/tex]  * (1/x) dx

= ln x * (2/3) [tex]x^{3/2}[/tex]  - (2/3) ∫ [tex]x^{1/2}[/tex]  dx

= ln x * (2/3) [tex]x^{3/2}[/tex]  - (4/9) [tex]x^{3/2}[/tex]  + C

where C is the constant of integration.

To evaluate the definite integral from 1 to e, we substitute e for x in the expression above and subtract the result when x = 1:

 [tex]\int\limits^e_1 {x^{1/2}lnx } \, dx[/tex]   = [(2/3) [tex]e^{3/2}[/tex] ln e - (4/9) [tex]e^{3/2}[/tex] ] - [(2/3)[tex]1^{3/2}[/tex]ln 1 - (4/9)[tex]1^{3/2}[/tex]]

= (2/3) [tex]e^{3/2}[/tex] - (4/9)

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Answer:

2(4x-3)

Step-by-step explanation:

Find common numbers between 8 and 6, which is 2. 2x4= 8 2x3=6

x is only on 8 so you can't put x on the outside of the bracket. so you put it with the 4 so it comes out correct

The given polynomial is a prime polynomial.

The given polynomial is

x² - x + 35

Since we know that discriminant of quadratic polynomial

p(x)= ax² + bx + c  is b² - 4ac

Therefore, discriminant of the given polynomial be,

⇒ (-1)² - 4x1x35

⇒ - 139

Since discriminant is negative number so it has no real roots.

And an irreducible or primal polynomial is a polynomial with integer coefficients that cannot be factored into polynomials of lower degree and also with integer coefficients.

Hence it is not factored completely.

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A. The slopes are the same but the y - intercepts are different.

In a graph, the slope is the steepness of a line, which is a measure of how quickly the y-coordinate changes with respect to the x-coordinate. It is represented by the ratio of the vertical change (rise) to the horizontal change (run) between any two points on the line.

The slope is often denoted by the letter m, and is calculated as:

m = (y₂ - y₁) / (x₂ - x₁)

where (x₁, y₁) and (x₂, y₂) are any two points on the line.

In a graph, the y-intercept is the point where a line or curve intersects the y-axis. It is the value of the dependent variable (usually denoted by y) when the independent variable (usually denoted by x) is equal to zero.

The y-intercept is often denoted by the letter b, and is expressed as a coordinate point (0, b), where 0 represents the x-coordinate and b represents the y-coordinate. For example, in the equation of a line y = mx + b, the y-intercept is the value of b.

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By making the financially irresponsible decision of maxing out your credit card limit, you might not have money for an emergency situation.

As, There is no foolproof technique to determine how much you can spend in excess of your credit limit, but you can go over it.

When determining the risk of accepting an over-the-limit transaction, card issuers may take into account a number of criteria, including your prior payment history.

Now, maxing out credit card leads to negative results.

So, either not have money for an emergency situation or not be able to buy the things you need.

But we can still spend after maxing out the limit.

Thus, not have money for an emergency situation is correct option.

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Answer:

No

Step-by-step explanation:

In order for a triangle to be a right angle, the sum of the squares of the two legs (a and b) must be equal the square of the longest side, known as the hypotenuse (c):

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

The square root of 155 is the longest side as it is approximately 12.45.  Thus, if the triangle is a right triangle, the square root of 155 must be the hypotenuse.

We can now check to see whether the sum of the squares of 7 and 9 equal the square of the square root of 155:

[tex]7^2+9^2=(\sqrt{155})^2\\ 49+81=155\\130=155[/tex]

The equation is not true since 130 is less than and not equal to 155.  Therefore, a triangle with side lengths 7, 9, and the square root of 155 cannot be a right triangle.

Sum

Mean = (4.1 + 3.2 + 2.8 + 2.6 + 3.7 + 3.1 + 9.4 + 2.5 + 3.5 + 3.8) / 10

Mean = 36.7 / 10

Mean = 3.67

To find the median, we need to put the values in order:

2.5, 2.6, 2.8, 3.1, 3.2, 3.5, 3.7, 3.8, 4.1, 9.4

The middle number is the median, which is 3.35 in this case (3.2+3.5)/2.

To find the mode, we look for the value that appears most often. In this case, there is no mode as no value appears more than once.

To find the first quartile (Q1), we need to find the value that separates the bottom 25% of the data from the top 75%. We can do this by finding the median of the lower half of the data:

2.5, 2.6, 2.8, 3.1, 3.2

The median of this lower half is 2.8, so Q1 = 2.8.

To find the third quartile (Q3), we need to find the value that separates the bottom 75% of the data from the top 25%. We can do this by finding the median of the upper half of the data:

3.7, 3.8, 4.1, 9.4

The median of this upper half is 3.95, so Q3 = 3.95.

To find the outlier, we can use the rule that any value more than 1.5 times the interquartile range (IQR) away from the nearest quartile is considered an outlier. The IQR is the difference between Q3 and Q1:

IQR = Q3 - Q1

IQR = 3.95 - 2.8

IQR = 1.15

1.5 times the IQR is 1.5 * 1.15 = 1.725.

The only value that is more than 1.725 away from either Q1 or Q3 is 9.4. Therefore, 9.4 is the outlier in this data set.

Answer:

Step-by-step explanation:

Mean is the average of the data vales : 38.7 (sum of all values) divided by 10 (the number of values).  Mean = 38.7/10 = 3.87

Median is the "middle" number" = put the date in order and find the middle value:

2.5, 2.6, 2.8, 3.1, 3.2, 3.5, 3.7, 3.8, 4.1, 9.4, since there is no middle data value, find the average of the 2 in the middle.

3.2 + 3.5/2 = 6.7 ÷ 2 = 3.35 - median

All values appear once so there is no mode.

First quartile is the middle of the lower set of data = 2.8

Third quartile is the middle of the upper set = 3.8

The outlier is 9.4.

How to calculate the outlier:

First you need the IQR which is the diffence of Q3 - Q1,

so the IQR is 3.8- 2.8 = 1

Outliers are the quartile + or - (1.5)(IQR)

Q1 -(1.5)(1) = 2.8 - 1.5 = 1.3

Q3 - (1.5)(1)= 3.8 + 1.5 =4.5

So anything below 1.3 or above 4.5 is an outlier.

There is one 9.4

1. It would be better to use the median to describe the data set.

2. It would be better to use the median to describe the data set.

3. It would be better to use the mean to describe the data set as the average salary of the workers is to be calculated.

4. It would be better to use the mean to describe the data set as the average test grade of the class is to be calculated.

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The coordinates of Y' are (6, 4) and the coordinates of Z' are (5, -1).

To find the coordinates of Y' and Z', we need to apply the same translation that maps X to X'.

We know that the vector that connects X to X' is (3 - (-1), 1 - 0) = (4, 1).

So, to translate Y to Y', we add the vector (4, 1) to the coordinates of Y:

Y' = Y + (4, 1)

= (2, 3) + (4, 1)

= (6, 4)

Similarly, to translate Z to Z', we add the vector (4, 1) to the coordinates of Z:

Z' = Z + (4, 1)

= (1, -2) + (4, 1)

= (5, -1)

Therefore, the coordinates of Y' are (6, 4) and the coordinates of Z' are (5, -1).

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Answer: D

Step-by-step explanation: 60x9=540 and 540x80=43,200

Kyle has $306 in his account so, right answer is Option D.

Let Kyle's money = x

Hayden's money = y

Jeremy's money = z

Given that Kyle has three times as much money in his savings account as  Hayden.

x = 3y   .....Eqn 1

Hayden has twice as much money in his savings account  as Jeremy.

z = 2y  ..... Eqn 2

Now Combined Kyle, Hayden, and Jeremy have a total of  $612.

x+y+z=612   .....Eqn 3

Putting value of x and z from Eqn 1 and 2 in Eqn 3

6y = 612

y = 106

Putting in Eqn 1;

x = 306

Thus, Kyle has $306 in his account.

The right answer is Option D.

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The correct option regarding which bus has the least spread among the travel times is given as follows:

Bus 14, with an IQR of 6.

To obtain measures of spread, we can use the dot plot to visualize the frequency of each observation in the data-set, and then calculate the interquartile range (IQR).

For a group of 15 students, we can find the quartiles as follows:

The first quartile (Q1) is the median of the first half of the data-set, which consists of the first seven students.

Looking at the dot plot, the fourth dot represents the median of the first half. For Bus 14, Q1 = 12. For Bus 18, Q1 = 9.

The third quartile (Q3) is the median of the second half of the data-set, which consists of the last seven students. Looking at the dot plot, the eleventh dot represents the median of the second half. For Bus 14, Q3 = 18.

For Bus 18, Q3 = 16.

We can then calculate the IQR as the difference between Q3 and Q1:

For Bus 14, IQR = Q3 - Q1 = 18 - 12 = 6.

For Bus 18, IQR = Q3 - Q1 = 16 - 9 = 7.

Based on the IQR values, we can conclude that Bus 14 is more consistent than Bus 18, as it has a lower IQR.

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Answer:

3.36%

Step-by-step explanation:

(10.14-9.81)/(9.81)(100)=3.36% error

There are 2000 milligrams in 2 grams of a substance, and it constitutes 40% of a 5-gram sample.

Convert grams to milligrams:

2 grams = 2 * 1000 milligrams

2 grams = 2000 milligrams

Calculate the percentage in a 5-gram sample:

Percentage = (2000 milligrams / 5000 milligrams) * 100

Percentage = (2 / 5) * 100

Percentage = 0.4 * 100

Percentage = 40%

Thus, there are 2000 milligrams in 2 grams of a substance, and it constitutes 40% of a 5-gram sample.

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​2 grams = ​ milligrams

How many milligrams are there in 2 grams of a substance, and what is the percentage of this amount in a 5-gram sample?

The minimum grade you need to get a B in the course is given as follows:

x = 80.625.

The mean of a data-set is given by the sum of all observations in the data-set divided by the number of observations, which is also called the cardinality of the data-set.

The grades and their percentages are given as follows:

You want a grade of 80, hence the minimum test average is obtained as follows:

0.15 x 73 + 0.2 x 79 + 0.25 x 84 + 0.4x = 80

47.75 + 0.4x = 80

x = (80 - 47.75)/0.4

x = 80.625.

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The graph of the piecewise function is given by the image presented at the end of the answer.

A piece-wise function is a function that has different definitions, depending on the input of the function.

All the five definitions of the function in this problem are constant, hence the graph will be composed by horizontal lines.

The definitions are given as follows:

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Answer:

False.

Step-by-step explanation:

A integer in math is defined as a whole number. Therefore, the presence of fractions (which, in this case, is 1/2) and decimals, would mean that the term is in fact, not a integer.

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Answer:

[tex]sec(-\theta)=-3.1[/tex]

[tex]sin(-\theta)=-0.62[/tex]

Step-by-step explanation:

Part a.

[tex]sec(\theta)[/tex] is related to [tex]cos(\theta)[/tex] through a reciprocal relationship [tex]sec(\theta)=\frac{1}{cos(\theta)}[/tex].

Since [tex]cos(\theta)[/tex] is an even function (reflecting left-right across the y-axis doesn't change the graph), then [tex]sec(\theta)[/tex] is also an even function.

For any input in the domain, even functions produce the same output if the opposite of the input is used.  In other words, if "f" is an even function, for all x in the domain of f, [tex]f(x)=f(-x)[/tex].

Thus, for the secant function, if a mystery value "x" is used as an input, and -3.1 is obtained as an output, then if the opposite of x, or -x, is input into the secant function, the output will also be -3.1.

[tex]sec(-\theta)=-3.1[/tex]

Part b.

The [tex]sin(\theta)[/tex] function does not reflect left-right across the y-axis to produce the same graph, so the sine function is not even.

However, the sine function can be rotated 180 degree about the origin, sometimes thought of as reflecting through the origin, to produce the same graph.  Visually, this is an "odd" function.

For any input in the domain, if the opposite of the input is used, odd functions produce the opposite of the original output.  In other words, if "g" is an odd function, for all x in the domain of g, [tex]-g(x)=g(-x)[/tex].

Thus, for the sine function, if a mystery value "x" is used as an input, and 0.62 is obtained as an output, then if the opposite of x, or -x, is input into the sine function, the output will be the opposite of 0.62, meaning -0.62.

[tex]sin(-\theta)=-0.62[/tex]