The formula for __________________________ is x = t e or raw score (x) equals the true score (t) plus error (e).

A given linear equation has infinitely many solutions.

Given: −3y 3y 4 = 4

            3y+4=3y+4

The given linear equation satisfies any value of 'y'.

So, A given linear equation has infinitely many solutions.

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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.

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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Here is the five-number summary, range and interquartile range:

Minimum: 9

Lower quartile:   11

Median: 24

Upper quartile:  34

Maximum: 45

Interquartile range: 23

The minimum number is the smallest number in the data. The  minimum number is 9. The maximum number is the biggest number in the dataset. The maximum number is 45.

The first quartile, third quartile and the interquartile range are used to measure dispersion.

The first quartile : 1/4(n + 1)

1/4 x (12) = 3rd term = 11

The third quartile : 3/4 x (n + 1)

3/4 x (12) = 9th term = 34

Interquartile range = third quartile - first quartile

34 - 11 = 23

Median can be described as the number that is at the center of the dataset when the number is arranged in either ascending or descending order.

Median = 25

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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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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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The circumference of the given circle is 25.1 cm.

According to the statement

we have given that the the radius of circle which is 4cm.

And we have to find the circumference of the circle.

So, For this purpose

The circumference is the length of any great circle, the intersection of the sphere with any plane passing through its center.To calculate the circumference of a circle, multiply the diameter of the circle with π (pi).

The formula to calculate it is

C = 2(pi)r

So, substitute the values in it then

Here use pi value is 3.14

C = 2 *3.14*4

C = 25.12.

The circumference of the given circle is according to one decimal place is 25.12 cm.

So, The circumference of the given circle is 25.1 cm.

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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)

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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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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Answer:

11.7647%

Hope this helps!

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.

The given expression is equivalent to the expression (B) [tex](x-1)^{2} -3[/tex].

To complete the square to transform the expression:

Therefore, the given expression is equivalent to the expression (B) [tex](x-1)^{2} -3[/tex].

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The correct question is shown below:

Complete the square to transform the expression x^2− 2x − 2 into the form a(x − h)^2+ k.

(A) (x − 1)2 + 3

(B) (x − 1)2 − 3

(C) (x − 2)2 − 3

(D) (x − 2)2 + 3

The critical values of the function are x = 0 and x = 64

The critical values of a function f(x) are the values of x for which f'(x) = 0.

Given,

f(x) = [tex]\frac{x^{4} }{5(x-8)^{2} }[/tex]

The derivative is found as follows, applying the quotient rule:

f'(x) = [tex]\frac{[5(x^{4} )(x-8)^{2}-5x^{4} [(x-8)^{2} ]' }{[5(x-8)^{2} ]^{2} }[/tex]

     = [tex]\frac{2x^{3}(x-64) }{5(x-8)^{3} }[/tex]

The zeros of the function are the zeros of the numerator, thus:

[tex]2x^{3} (x-64) =0\\2x^{3} =0-------- > x =0\\[/tex]

x - 64 = 0------------->x = 64

The critical values are x = 0 and x = 64

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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!

Answer:

answer is 3 ft

Step-by-step explanation:

area of circle= 3*(1)^2

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°

Given the point U and point V, the midpoint between of the line segment UV is ( 0, -5/2 ).

The midpoint formula is used to find the midpoint of a line segment.

It is expressed as;

[ (x₁ + x₂ )/2 , (y₁ + y₂ )/2 ]

Given the data in the graph,

Point U = ( 3, -1 ) and Point V = ( -3, -4 )

We substitute into the formular above.

[ (x₁ + x₂ )/2 , (y₁ + y₂ )/2 ]

[ (3 + (-3) )/2 , ( (-1) + (-4) )/2 ]

[ 0/2, (-1-4)/2 ]

[ 0/2, -5 /2 ]

( 0, -5/2 )

Given the point U and point V, the midpoint between of the line segment UV is ( 0, -5/2 ).

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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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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 "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:

4112

Step-by-step explanation:

When you plug in 3 for x you get 6.

Then you just do 4^6 which is 4096.

Then add 16 which gets you 4112.

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 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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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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The amount of space (volume) that is in the larger spherical ball than the smaller = 1,084.8 in.³

To find the volume of a sphere, which is the amount of space the solid contains, the formula used is given as: 4/3 πr³, where r is the radius of the sphere, which is half the measure of the diameter. Volume of sphere (V) = 4/3 πr³.

Diameter of the smaller spherical ball = 5 inches

Radius of the smaller spherical ball = 5/2 = 2.5 inches

Volume of the smaller spherical ball = 4/3 πr³ = 4/3 × π × 2.5³

Volume of the smaller spherical ball ≈ 65.5 in.³

Diameter of the larger spherical ball = 13 inches

Radius of the larger spherical ball = 13/2 = 6.5 inches

Volume of the larger spherical ball = 4/3 πr³ = 4/3 × π × 6.5³

Volume of the larger spherical ball ≈ 1,150.3 in.³

The amount of space (volume) that is in the larger spherical ball than the smaller = 1,150.3  - 65.5

The amount of space (volume) that is in the larger spherical ball than the smaller = 1,084.8 in.³

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

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