A set of eight cards were labeled as M, U, L, T, I, P, L, Y. What is the sample space for choosing one card?

S = {I, U, Y}
S = {M, L, T, P}
S = {I, L, M, P, T, U, Y}
S = {I, L, L, M, P, T, U, Y}
Question 2(Multiple Choice Worth 2 points)
(Sample Spaces LC)

A paper bag has five colored marbles. The marbles are red, green, blue, yellow, and orange. List the sample space when choosing one marble.

S = {1, 2, 3, 4, 5}
S = {red, blue, green, yellow}
S = (g, r, b, y, o}
S = (green, blue, yellow, orange}
Question 3(Multiple Choice Worth 2 points)
(Sample Spaces LC)

A four-part spinner is spun once.

spinner with four parts labeled W, X, Y, and Z

List the sample space for the experiment.

S = {X, Y, Z}
S = {W, Z, Z}
S = {W, Y, Y, Z}
S = {W, X, Y, Z}
Question 4(Multiple Choice Worth 2 points)
(Sample Spaces LC)

Which event will have a sample space of S = {h, t}?

Flipping a fair, two-sided coin
Rolling a six-sided die
Spinning a spinner with three sections
Choosing a tile from a pair of tiles, one with the letter A and one with the letter B
Question 5(Multiple Choice Worth 2 points)
(Sample Spaces MC)

Julia had a bag filled with gumballs. There were 2 lemon-lime, 3 watermelon, and 5 grape gumballs. What is the correct sample space for the gumballs in her bag?

Sample space = 2, 3, 5
Sample space = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Sample space = lemon-lime, watermelon, grape
Sample space = lemon-lime, lemon-lime, watermelon, watermelon, watermelon, grape, grape, grape, grape, grape
Question 6 (Essay Worth 4 points)
(Sample Spaces HC)

Part A: Create your own experiment with 5 or more possible outcomes. (2 points)
Part B: Create a sample space for a single experiment and explain how you determined the sample space. (2 points)

Answer:

C

Step-by-step explanation:

[tex]-15x+60\leq 105 \\ \\ -15x \leq 45 \\ \\ x \geq -3[/tex]

[tex]14x+11 \leq -31 \\ \\ 14x \leq -42 \\ \\ x \leq -3[/tex]

The intersection is x = 3.

the volume of the hexagonal prism is 184. 75 cm^3

It is important to note that the volume of a hexagonal prism is given as;

V = (2/√3) × ap2 × h

Where

From the information given, we have the values of the following parameters;

height = 10cm

apothem = 4cm

base edge = 6cm

Now, let's  substitute the value of the parameters

Volume = (2/√3) × ap^2 × h

Volume = 2/√3 × 4^2 × 10

Multiply through

Volume = 1. 1547 × 16 × 10

Volume = 184. 75 cm^3

Thus, the volume of the hexagonal prism is 184. 75 cm^3

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With convolution theorem the equation is proved.

According to the statement

we have given that the equation and we have to evaluate with the convolution theorem.

Then for this purpose, we know that the

A convolution integral is an integral that expresses the amount of overlap of one function as it is shifted over another function.

And the given equation is solved with this given integral.

So, According to this theorem the equation becomes the

[tex]\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{ \mathscr{L} (e^{-\tau} \cos \tau ) }{s} \\\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{\frac{s+1}{(s+1)^2+1}}{s} \\\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{1}{s}\left (\frac{s+1}{(s+1)^2+1} \right).[/tex]

Then after solving, it become and with theorem it says that the

[tex]\mathscr{L} \left( \int_{0}^{t} f(\tau) d\tau \right) = \frac{\mathscr{L} ( f(\tau))}{s} .[/tex]

Hence by this way the given equation with convolution theorem is proved.

So, With convolution theorem the equation is proved.

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The sponge function in cryptography takes a string of any length as input and returns a string of any requested variable length.

According to the statement

we have to explain about the function in which cryptography takes a string of any length as input and returns a string of any requested variable length.

So, For this purpose,

we know that the

A sponge function or sponge construction is any of a class of algorithms with finite internal state that take an input bit stream of any length and produce an output bit stream of any desired length.

So from definition and its working process it is clear that for this purpose the sponge function is used.

this function returns the string of any variable length.

So, The sponge function in cryptography takes a string of any length as input and returns a string of any requested variable length.

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Answer:

x =  -1,  7,  3 + i,  3 - i.

Step-by-step explanation:

(x^2-6x+9)^2-15(x^2-6x+10)=1

(x^2 - 6x + 9)^2  - 15(x^2 - 6x + 9) - 15*1 = 1

(x^2 - 6x + 9)^2  - 15(x^2 - 6x + 9) - 16 = 0

Let Z = x^2 - 6x + 9, then we have:

Z^2 - 15Z - 16 = 0

(Z - 16)(Z + 1) = 0

Z = 16 or Z = -1

so  x^2 - 6x + 9 = -1 or x^2 - 6x + 9 = 16

x^2 - 6x + 9 = -1

---> x^2 - 6x + 10 = 0

Using the Quadratic Formula:

---> x = [6 +/- √((-6)^2 - 4* 1* 10) / 2

---> x = 6/2 +/- √-4/2

---> x = 3 + i , 3 - i.

x^2 - 6x + 9 = 16

---> x^2 - 6x - 7 = 0

---> (x - 7)(x + 1) = 0

---> x = 7, -1.

Answer:

7

Step-by-step explanation:

The start this question by looking at two important things, how much money we have and how much money it costs per person. The friends have a total of $284.25 and we don't know how much it costs per person. To find this we must set up an equation and solve it!

Because each person must pay for parking and a ticket, we can find the cost for one person by adding the parking and ticket cost together.

$9.25 + $27.50 = $36.75

Now that we have solved this equation, we know that it costs $36.75 for one person. To find how many total people can go we dived the total amount of money we have by how much it costs per person. Let's call the number of people that can go 'p'.

p = [tex]\frac{284.25}{36.75}[/tex]

Once we simplify/solve this equation we get 7 [tex]\frac{36}{49}[/tex] so essentially, we get the whole number 7 and a long decimal, but the only important part is the 7. We take the whole number from our answer which is 7.

We now know the answer: 7.

Let's check our work!

7 * 36.75 = 257.75

284.25 - 257.75 = 26.5

26.5 < 36.75 so we are correct!

The final answer is 7!

Have an amazing day!

The speed of the wind is 50 miles per hour.

The term speed is defined as the ratio of the distance to the time taken. Now we can see that the movement of the plane and the wind were once in the same direction and then in opposite direction. This could be used to obtain a pair of simultaneous equations that could be used to solve the problem.

Hence;

300 = (250+s)* t = 250t + st ----- (1)

200 = (250-s)* t = 250t   - st  ------- (2)

Adding equations (1) and (2)

500 = 500t

t = 1 hour

To obtain the speed of the wind;

300 =250t + st

300 = 250(1) + (s * 1)

300 = 250 + s

300 - 250 = s

s = 50 miles per hour

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Missing parts;

A plane has a cruising speed of 250 miles per hour when there is no wind. At this speed, the plane flew 300 miles with the wind in the same amount of time it flew 200 miles against the wind. Find the speed of the wind.

The standard form polynomial representing the volume of this shipping container is determined as: 24x^5 + 78x^4 - 147x^3 - 624x^2 - 360x.

A standard form polynomial is a polynomial expression written whereby the term with the highest degree or power on a variable is written first in the expression, followed by the least, then the constant of the polynomial comes last.

What is the Volume of a Rectangular Prism?

The Volume of a rectangular prism = (length)(width)(height).

The shipping container is a rectangular prism with the following dimensions:

Length of container = 4x² + 3x

Width of container = x² - 8

Height = 6x + 15

Plug in the values

Volume of container = (4x² + 3x)(x² - 8 )(6x + 15)

Expand

Volume of container = 24x^5 + 78x^4 - 147x^3 - 624x^2 - 360x

Thus, using the formula for the volume of a rectangular prism, the standard form polynomial representing the volume of this shipping container is determined as: 24x^5 + 78x^4 - 147x^3 - 624x^2 - 360x.

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Answer: [tex]\Large\boxed{B.~\displaystyle 75~m^3}[/tex]

Step-by-step explanation:

Given information

Height = 2 m

Base = 5 m

Length = 15 m

Given the formula for Triangular Prism Volume

[tex]V~=~\displaystyle \frac{1}{2}\times ~b~\times~h~\times~l[/tex]

[tex]\to V=Volume\\\to b=base\\\to h = height\\\to l = length[/tex]

Substitute values into the formula

[tex]V~=~\displaystyle \frac{1}{2}\times ~(5)~\times~(2)~\times~(15)[/tex]

Simplify by multiplication

[tex]V~=~\displaystyle \frac{5}{2}~\times ~30[/tex]

[tex]\Large\boxed{V~=~\displaystyle 75~m^3}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

The total surface area of the solid object is 2660.925 square cm the answer is 2660.925 square cm.

In geometry, it is defined as the three-dimensional shape having two circular shapes at a distance called the height of the cylinder.

We know the volume of the cylinder is given by:

[tex]\rm V = \pi r^2 h[/tex]

The surface area of the

= surface area of half sphere + surface area of the cylinder - surface area

  of one circular base

The radius r = (64-50)/2 = 7 cm

= (1/2)[4π(7)²] + 2π(7)(50) + 2π(7)² - π(7)²

= (1/2)[615.75] + 2506.99 - 153.94

= 307.875 + 2506.99 - 153.94

= 2660.925 square cm

Thus, the total surface area of the solid object is 2660.925 square cm the answer is 2660.925 square cm.

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Answer:

a) horizontal compression with a factor of 0.5 and a horizontal reflection over the y-axis.

b) Pick one the two correct answers:

translation of 2 units right

translation of 2 units down

Step-by-step explanation:

If function f(x) is transformed into f(ax) then it is stretched or compressed horizontally.

If |a| > 1 it is compressed horizontally.

If 0 < |a| < 1, it is stretched horizontally.

If a is negative, then it is reflected over the y-axis.

a) Compare y = -2x with y = x.

The change is in that x became -2x.

Here, a = -2.

Since |-2| = 2, and 2 > 1, it has a compression of a factor of 2 horizontally.

Also, since -2 is a negative number, it is reflected over the y-axis.

Answer: horizontal compression with a factor of 0.5 and a horizontal reflection over the y-axis.

If function f(x) is transformed into f(x) + b then it is translated vertically b units. If b > 0, the translation is b units up. If b < 0, the translation is b units down.

b) y = x - 2

This can be thought of the function f(x) becoming f(x) - 2.

It is a translation of 2 units down.

Interestingly, in this case, this can also be thought of x being replaced by x - 2 which is a translation of 2 units to the right.

Answer:

There are two correct answers (use only one of the two below):

translation of 2 units right

translation of 2 units down

The SEC routinely require public companies to include their review reports in their 10-Q filings because it is said to be made up of fewer details and the financial statements inclusive that are known to be unaudited.

Form 10-Q is known to be a type of a report  that is often needed by the Securities and Exchange Commission (SEC)  and it is one that all public company need to file on quarterly basis.

Note also that The SEC is known to have made and need the form in accord to the pursuant of  the Securities Exchange Act of 1934  and this form helps them to keep investors well  informed about the state of their financial health and all that is taking place at  the companies they have invest in or will choose to invest in.

Hence,  Form 10-Q is asked for because it is often used to make comparison of a company’s previous financial quarter to that of its current financial quarter.

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Answer:

463833

expand

Medium

Solution

verified

Verified by Toppr

In △ABD and △ACD, we have

DB=DC ∣ Given

∠ADB=∠ADC ∣ since AD⊥BC

AD=AD ∣ Common

∴ by SAS criterion of congruence, we have.

△ABD≅△ACD

⇒AB=AC ∣ Since corresponding parts of congruent triangles are equal

Hence, △ ABC is isosceles.

Answer:

A.

Step-by-step explanation:

With the given information, triangles ABD and ACD can be proved congruent, and by CPCTC, segments AB and AC are congruent.

The initial-value is  x(t)=F0/2w^2 (sin wt-wt cos wt)

An initial-value hassle is a differential equation wherein is an open set of, together with a point within the domain called the initial situation. A technique to an initial cost hassle is a function that is a technique to the differential equation and satisfies.

Inside the discipline of differential equations, preliminary cost trouble (additionally called Cauchy trouble by using a few authors) is a normal differential equation collectively with a detailed fee, called the preliminary situation, of the unknown function at a given factor in the domain of the solution.

In multivariable calculus, an initial-value hassle is a normal differential equation collectively with a preliminary circumstance that specifies the fee of the unknown characteristic at a given factor in the area. Modeling a system in physics or different sciences often amounts to fixing an initial cost problem.

Consider a differential equation x+x=F, sin wt, x(0) = 0,X(0) = 0

Apply Laplace transformation on both sides, and we get

(s²L{x(t)}-sy(0)-y'(0)) + w²L {x(t)} =·

Fow

Fow

(s² + w² ) { x(t)} = 3 + w²

L{x(t)}=

Fow

(5²+w²)²

x(t)=FwL

[(s² + w²)²]

Fow

x(t)=(sin wt-wt coswt)

2w

x(t)=

Fo

wtcos

2w (sin wt-wt cos wt),

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The minimum value of data is 24,lower quartile is 29,median is 41, upper quartile is 50 and maximum value is 56 and the interquartile range is 21.

Given a data about ages of 13 history teacher as under:

24, 27, 29, 29, 35, 39, 43, 45, 46, 49, 51, 51, 56.

We are required to find the minimum value, lower quartile,median,upper quartile,maximum value, interquartile range.

The minmum value is 24.

Lower quartile=(n+1)/4 th term

=(13+1)/4

=7/2

=3.5

Lower quartile=(29+29)/2

=29

Median=(n/2)th term

=13/2 th term

=6.5 th term

Median=(39+43)/2

=82/2

=41

Upper quartile=3(n+1)/4   th term

=3(13+1)/4

=3*14/4

=10.5 th term

Upper quartile=(49+51)/2=100/2=50

Inter quartile range=Upper quartile- lower quartile

=50-29

=21

Hence the minimum value of data is 24,lower quartile is 29,median is 41, upper quartile is 50 and maximum value is 56 and the interquartile range is 21.

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The terms of this series may be arranged without changing the value of the series. The sum of the reciprocals of the squares of the odd positive integers is [tex]\pi ^{2} /8[/tex].

In mathematics, a sequence is the cumulative sum of a given collection of terms. Usually, those phrases are actual or complicated numbers, but plenty of extra generalities are feasible.

A series is described as an arrangement of numbers in a specific order. then again, a chain is described as the sum of the factors of a sequence.

In mathematics, a series is, more or less speaking, a description of the operation of including infinitely many quantities, one after the alternative, to a given beginning quantity. The look at of series is a primary part of calculus and its generalization, mathematical analysis.

k=1

1/(1)2+1/(2)2+1/(3)2+1/(4)2+1/(5)2+1/(6)2+1/(7)2+.

up to ∞ terms = 2/6

[1/(1)2+1/(3)2+1/(5)2+1/(7)2+]+[1/(2)²+1/(4)²+1/(6)²+

..∞0] = T²/6

→ [1/(1)² + 1/(3)² + 1/(5)2+1/(7)2+......00] + [1/4 (1)² + 1/4(2)²+

1/4(3)²+....0] =²/6

[1/(1)²+1/(3)²+1/(5)2+1/(7)2+.......)] + 1/4[1/(1)² + 1/(2)²+

1/(3)²+....x] = 2/6

⇒ [1/(1)² + 1/(3)² + 1/(5)²+1/(7)²+..] + 1/4 [π²/6] = 2/6

⇒ [1/(1)² + 1/(3)² + 1/(5)²+1/(7)²+] = (1-1/4)/6

⇒ [1/(1)²+1/(3)2+1/(5)2+1/(7)2+..∞ = 3/4 x π²/6

=

↑

[1/(1)2+1/(3)2+1/(5)2+1/(7)²+] = 2/8

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[tex](k \circ p)(x)=k(p(x))=k(x-3) \\ \\ =2(x-3)^2-5 \\ \\ =2(x^2 - 6x+9)-5 \\ \\ =\boxed{2x^2 - 12x+13}[/tex]

For instance, f[g (x)] exists the composite function of f(x) and g(x). The composite function f[g (x)] exists read as “f of g of x”.

The composite function exists [tex]$k(p(x))=2 x^{2}-12 x+13$[/tex]

Therefore, the correct answer is option b. [tex]$k(p(x))=2 x^{2}-12 x+13$[/tex].

A composite function exists generally as a function that exists written inside another function. The composition of a function exists done by replacing one function with another function.

Given function exists, [tex]$k(x)=2 x^{2}-5[/tex] and p(x) = (x - 3)

To find composite function k(p(x)).

k(p(x)) = k(x-3)

[tex]$&k(p(x))=2(x-3)^{2}-5 \\[/tex]

simplifying the above equation, we get

[tex]$&k(p(x))=2\left(x^{2}+9-6 x\right)-5 \\[/tex]

[tex]$&k(p(x))=2 x^{2}+18-12 x-5 \\[/tex]

[tex]$&k(p(x))=2 x^{2}-12 x+13[/tex]

The composite function exists [tex]$k(p(x))=2 x^{2}-12 x+13$[/tex]

Therefore, the correct answer is option b. [tex]$k(p(x))=2 x^{2}-12 x+13$[/tex].

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Answer:

B. 68

Step-by-step explanation:

x is a raw score to be standardized;

μ is the mean of the population;

σ is the standard deviation of the population.

Therefore the mean is zero. Seventy is -1z, or -1 standard deviation.

Step 2: 34.13% of the cases fall between -1 standard deviation and the mean. Thus there is a 34.13% chance that the score will fall between 70 and 75. This, of course, assumes a normal curve.

Step 3: An 80 is +1z or +1 standard deviation assuming a normal curve.

Step 4: Thirty four percent of the cases fall between +1 standard deviation and the mean. Thus there is a 34.13% chance that the score will fall between 75 and 80. This, of course, assumes a normal curve.

Step 5: Score between +1z and -1z, or +1 and -1 standard deviation account for 68.26% of the cases.

[tex]y > 2x + 1 \\ 3 > 2(1) + 1 \\ 3 > 3 \\ false[/tex]

[tex]y < - 3x \\ 3 < - 3(1) \\3 < - 3 \\ also \: false[/tex]

The "given series" should be for [tex]\cos(x)[/tex], not [tex]x[/tex], so that

[tex]\cos(x) = 1 - \dfrac{x^2}2 + \dfrac{x^4}{24} - \dfrac{x^6}{720} + \cdots[/tex]

In the limit (which should say [tex]x\to\infty[/tex], not [tex]n[/tex]), we have

[tex]\displaystyle \lim_{x\to\infty} \frac{\frac{x^2}{1+\cos(x)}}{x^4} = \lim_{x\to\infty} \frac{1}{x^2\left(2 - \frac{x^2}2 + \frac{x^4}{24} - \cdots\right)} = \boxed{0}[/tex]

Answer:

Using the usual notations and formulas,

Using the usual notations and formulas,mean, mu = 3550

Using the usual notations and formulas,mean, mu = 3550standard deviation, sigma = 870

Using the usual notations and formulas,mean, mu = 3550standard deviation, sigma = 870Observed value, X = 3000

Using the usual notations and formulas,mean, mu = 3550standard deviation, sigma = 870Observed value, X = 3000We calculate

Using the usual notations and formulas,mean, mu = 3550standard deviation, sigma = 870Observed value, X = 3000We calculateZ= (X-mu)/sigma = (3000-3550)/870 = -0.6321839

Using the usual notations and formulas,mean, mu = 3550standard deviation, sigma = 870Observed value, X = 3000We calculateZ= (X-mu)/sigma = (3000-3550)/870 = -0.6321839Probability of weight below 3000 lbs

Using the usual notations and formulas,mean, mu = 3550standard deviation, sigma = 870Observed value, X = 3000We calculateZ= (X-mu)/sigma = (3000-3550)/870 = -0.6321839Probability of weight below 3000 lbs=P(X < 3000) = P(z < Z) = P(z < - 0.6321839) =

Using the usual notations and formulas,mean, mu = 3550standard deviation, sigma = 870Observed value, X = 3000We calculateZ= (X-mu)/sigma = (3000-3550)/870 = -0.6321839Probability of weight below 3000 lbs=P(X < 3000) = P(z < Z) = P(z < - 0.6321839) =0.2636334

Using the usual notations and formulas,mean, mu = 3550standard deviation, sigma = 870Observed value, X = 3000We calculateZ= (X-mu)/sigma = (3000-3550)/870 = -0.6321839Probability of weight below 3000 lbs=P(X < 3000) = P(z < Z) = P(z < - 0.6321839) =0.2636334Answer:

Using the usual notations and formulas,mean, mu = 3550standard deviation, sigma = 870Observed value, X = 3000We calculateZ= (X-mu)/sigma = (3000-3550)/870 = -0.6321839Probability of weight below 3000 lbs=P(X < 3000) = P(z < Z) = P(z < - 0.6321839) =0.2636334Answer:Probability that a car randomly selected is less than 3000

Using the usual notations and formulas,mean, mu = 3550standard deviation, sigma = 870Observed value, X = 3000We calculateZ= (X-mu)/sigma = (3000-3550)/870 = -0.6321839Probability of weight below 3000 lbs=P(X < 3000) = P(z < Z) = P(z < - 0.6321839) =0.2636334Answer:Probability that a car randomly selected is less than 3000=P(X < 3000) = 0.2636 (to 4 decimals)

Using the usual notations and formulas,mean, mu = 3550standard deviation, sigma = 870Observed value, X = 3000We calculateZ= (X-mu)/sigma = (3000-3550)/870 = -0.6321839Probability of weight below 3000 lbs=P(X < 3000) = P(z < Z) = P(z < - 0.6321839) =0.2636334Answer:Probability that a car randomly selected is less than 3000=P(X < 3000) = 0.2636 (to 4 decimals)Probability that a car randomly selected is greater than 3000

Using the usual notations and formulas,mean, mu = 3550standard deviation, sigma = 870Observed value, X = 3000We calculateZ= (X-mu)/sigma = (3000-3550)/870 = -0.6321839Probability of weight below 3000 lbs=P(X < 3000) = P(z < Z) = P(z < - 0.6321839) =0.2636334Answer:Probability that a car randomly selected is less than 3000=P(X < 3000) = 0.2636 (to 4 decimals)Probability that a car randomly selected is greater than 3000=1 - P(X < 3000) = 1 - 0.2636 (to 4 decimals) =0.7364 (to 4 decimals)

Using a linear function, the total charge from parking in the lot for 72 hours is of:

b. $146.

A linear function is modeled by:

y = mx + b

In which:

Both the basic fee, which is the y-intercept, and the hourly fee, which is the slope, are of $2, hence the cost of parking x hours is given by:

C(x) = 2x + 2

Hence the cost for parking 72 hours is:

C(72) = 2 x 72 + 2 = 2 x 73 = $146.

Which means that option b is correct.

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3 ways of computing the sum on the number line are:

Here we have the sum:

8 + (-3) + (-2)

The first way of doing the sum in a number line is starting on the first number, which is 8.

Now, we also can think the sum as:

(-3) + 8 + (-2)

So now we can start at the value -3, then move 8 units to the right, and then 2 units to the left.

Or think the sum as:

(-2) + (-3) + 8

So we start at -2, then we move 3 units to the left, and finally 8 units to the right.

So the different ways of computing the sum on a number line depends on where we start.

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It costs $0.50 to mail a postcard to Canada, cost of mailing a one ounce letter to Canada is $0.60 and Ahmad wrote to 21 friends and spent $12.00 for postage. This means that he wrote 15 letters and 6 postcards.

Given Information:

Cost of mailing a postcard = $0.50

Cost of mailing a letter = $0.60

Total number of friends Ahmad wrote to = 21

Total cost of postage = $12.00

Let the number of postcards written by Ahmad be x and that of letters be y.

Then, x + y = 21 ............... (1)

0.50x + 0.60y = 12 ................ (2)

Multiplying equation (1) by 0.50, we get,

0.50x + 0.50y = 10.5 ................. (3)

Subtracting equation (3) from (2), we get,

0.1y = 1.5

⇒ y = 15

From equation (1), x = 21 - y

x = 21 - 15

x = 6

Thus, Ahmad wrote 6 postcards at a mailing cost of $0.50 each and 15 letters at a mailing cost of $0.60 each.

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Answer:

answer is 3 ft

Step-by-step explanation:

area of circle= 3*(1)^2

The value of P² + Q² + P · Q including elimination of radical denominators is equal to 13.

Herein we have two irrational terms whose denominators are radical expressions, which can be treated by algebraic handling and properties for radical expressions:

P = (√2 + 1) / (√2 - 1)

P = [(√2 + 1) / (√2 - 1)] · [(√2 + 1) / (√2 + 1)]

P = (√2 + 1)² / (2 - 1)

P = (√2 + 1)²

P = 2 + 2√2 + 1

P = 3 + 2√2

Q = (√2 - 1) / (√2 + 1)

Q = [(√2 - 1) / (√2 + 1)] · [(√2 - 1) / (√2 - 1)]

Q = (√2 - 1)²

Q = 2 - 2√2 + 1

Q = 3 - 2√2

Then, the value of P² + Q² + P · Q is:

M = (3 + 2√2)² + (3 - 2√2)² + (3 + 2√2) · (3 - 2√2)

M = 9 + 12√2 + 8 + 9 - 12√2 - 8 + 3 - 8

M = 9 + 8 + 9 - 8 + 3 - 8

M = 21 - 8

M = 13

The value of P² + Q² + P · Q including elimination of radical denominators is equal to 13.

The statement is poorly formatted and reports typing mistakes, correct form is presented below:

If P = (√2 + 1) / (√2 - 1) and Q = (√2 - 1) / (√2 + 1), then find P² + Q² + P · Q.

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Given the number of espresso sold at Chestnut Hill Coffee Cafe as either a single-shot or a double-shot, the percentage of double-shot espressos sold is 43.48%.

Percentage is simply number or ratio expressed as a fraction of 100.

It is expressed as;

Percentage = ( Part / Whole ) × 100%

Given the data in the question;

Percentage = ( Part / Whole ) × 100%

Percentage = ( Part double / Whole ) × 100%

Percentage = ( 20 / 46 ) × 100%

Percentage = 0.43 × 100%

Percentage = 43.48%

Given the number of espresso sold at Chestnut Hill Coffee Cafe as either a single-shot or a double-shot, the percentage of double-shot espressos sold is 43.48%.

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Answer:

x = 30° and y = 52°

Step-by-step explanation:

angles on the circle subtended by the same arc are congruent

x and 30° are on the same arc, then x = 30°

y and 52° are on the same arc, then y = 52°