suppose baby kittens' weights are normally distributed with a mean of 12.3 and a standard deviation of 2.2. the z-score tells you how many units above the average (if z-score is positive) or below the average (if z-score is negative) any particular baby kitten's weight is. find the baby kitten weight that corresponds to the following z-scores given below. hint: use the formula x
X-μ Use the formula Z where is the mean, o is the standard deviation, and X is the baby kitten σ weight. a. Z= 1.98, X = b. Z-2.87, X =

Answer:

The time at which the football reaches its highest point is 0.5 seconds.

Step-by-step explanation:

The height of an object thrown into the air is determined by the formula h(t) = -16t^2 + vt + h, where t represents the time in seconds after the object is thrown, v is the initial velocity with which the object is thrown, and h is the initial height from which the object is thrown. In this case, the initial velocity of the object is 16 feet per second and the initial height is 0 feet.

To find the time at which the football reaches its highest point, we need to find the value of t that makes the value of h(t) maximum. This value of t can be found by taking the derivative of the equation h(t) and setting it equal to 0. The derivative of h(t) is given by h'(t) = -32t + 16. Setting h'(t) equal to 0, we get -32t + 16 = 0, which can be solved to get t = 0.5 seconds.

Therefore, the time at which the football reaches its highest point is 0.5 seconds.

Answer:

The correct answer is the 3rd option

H(t) = [tex]\left \{ {{0 if t < 0} \atop {1 if t \geq 0}} \right.[/tex]

As t→0 from the left, H(t) approaches 0, as t→0 from right H(t) approaches 1.

From Laplace transformation definition:

L{f(t)} = [tex]\int\limits^\alpha _0 {e^{-st} f(t)} \, dt[/tex]

Given, H(t) = [tex]\left \{ {{0 if t < 0} \atop {1 if t \geq 0}} \right.[/tex]

Then

L{H(t)} = [tex]\int\limits^1_0 {e^{-st} H(t) } \, dt+\int\limits^\alpha _1 {e^{-st} H(t) } \, dt[/tex]

[tex]=\int\limits^1_0 {e^{-st} (0) } \, dt+\int\limits^\alpha _1 {e^{-st} (1) } \, dt[/tex]

[tex]=[\frac{e^{-st} }{-s} ]^{\alpha }_{1}[/tex]

[tex]=\frac{e^{-st} }{s}[/tex]

L{H(t)} =[tex]\frac{e^{-st} }{s}[/tex]

As t→0 from the left, H(t) approaches 0, as t→0 from right H(t) approaches 1.

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Answer:

Your answer is A.

Step-by-step explanation:

There are $85.50 will they receive if the current exchange rate is 1 U.S. dollar to 112 Nepalese rupees.

What is convertibility of rupee?

Convertibility is the simplicity with which a national currency can be changed into gold or another asset through international exchanges. The rupee is a partially convertible currency in India; while smaller amounts can be exchanged at market rates, larger ones require permission.

Given:

Your friend just returned to the United States from their trip to Nepal and wants to exchange their 9,576 Nepalese rupees for U.S. dollars.

Since,

1 U.S dollar = 112 Nepalese rupees

Suppose x be the U.S dollars for 9576 Nepalese rupees.

So,

x = 9576/112 = 85.5

Hence, there are $85.50 will they receive if the current exchange rate is 1 U.S. dollar to 112 Nepalese rupees.

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The probability that a randomly selected bike is green given that the bike has road bike tires is 0.66

Given :

Nellie is shopping for a new bicycle. She is most interested in color and type of tires.

Probability :

Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event .

From the table given :

Total number of 6 road bikes = 4

Total road bikes = 2 + 4 = 6

Probability = 4 / 6

= 2 / 3

= 0.66

Hence the  probability that a randomly selected bike is green given that the bike has road bike tires is 0.66 .

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24 is represents the entry fee in  the amusement park.

A. The total amount of money Carlos will spend $24 if he goes on o rides.

B. The total amount of money Carlos will spend $26 if he goes on 1 ride

C. The change in the total amount of money he will spend $2 for every additional ride he goes on.

D.  The total amount of money Carlos will spend $224 if he goes on 100 rides.

What is a variable?

Any qualities, quantity, or number that can be gauged or tallied qualifies as a variable. A data item is another name for a variable. Examples of variables include age, sex, company income and expenses, country of birth, capital expenditures, class grades, eye color, and vehicle kind.

Given equation is  y = 2x + 24

24 is represents the entry fee in  the amusement park.

where y represents the total amount of money, in dollars and cents, and a represents the number of rides Carlos goes on.

Putting x = 0 in the equation to find the total amount of money Carlos will spend if he goes on o rides:

y = (2 × 0) + 24

y = 24

Putting x = 1 in the equation to find the total amount of money Carlos will spend if he goes on 1 rides:

y = (2 × 1) + 24

y = 2 + 24

y = 26

The change in the total amount of money he will spend for every additional ride he goes on is $26 - 24 = $2

Putting x = 100 in the equation to find the total amount of money Carlos will spend if he goes on 100 rides:

y = (2 × 100) + 24

y = 200 + 24

y = 224

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The distribution of the time until the first sale of a book [tex]f(x)=\lambda p e^{\wedge_{-}} \lambda p x[/tex]

The probability that no books are sold during a particular hour [tex]\mathrm{P}(\mathrm{y}=0)=\mathrm{e}^{\wedge}-\lambda \mathrm{p}[/tex].

The expected number of customers who buy a book during a particular hour [tex]\mathrm{E}(\mathrm{y})=\lambda \mathrm{p}[/tex]

As per the details share in the above question are as follow,

Poisson process with a rate λ per hour

While customer buys a book with probability p

First we have to find the distribution of the time until the first sale of a book.

Second  to find the probability that no books are sold during a particular hour.

Third to find the expected number of customers who buy a book during a particular hour.

Giving the duration till a book sells its first copy

Consider,

Following are some instances of waiting times,

[tex]f(x)=\lambda p e^{\wedge_{-}} \lambda p x[/tex]

The distribution of the time until the first sale of a book [tex]f(x)=\lambda p e^{\wedge_{-}} \lambda p x[/tex]

Now, to calculate the likelihood that no books will be sold at a specific time.

[tex]\mathrm{P}(\mathrm{y}=0)=\mathrm{e}^{\wedge}-\lambda \mathrm{p}[/tex]

The probability that no books are sold during a particular hour [tex]\mathrm{P}(\mathrm{y}=0)=\mathrm{e}^{\wedge}-\lambda \mathrm{p}[/tex].

Consider the Poisson distribution to determine the anticipated number of book purchases during a specific hour.

[tex]\mathrm{E}(\mathrm{y})=\lambda \mathrm{p}[/tex]

The expected number of customers who buy a book during a particular hour [tex]\mathrm{E}(\mathrm{y})=\lambda \mathrm{p}[/tex].

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The Projectile Motion of vectors will be [tex]Proj_{v}(w)[/tex] = [tex]\left[\begin{array}{c}-1.5 \\0 \\-1.5\end{array}\right][/tex]

An item or particle that is projected in a gravitational field, such as from the surface of the Earth, and moves along a curved route while solely being affected by gravity is said to be in projectile motion.

According to the given information

Given, [tex]$\quad b_1=\left[\begin{array}{c}2 \\ 2 \\ -2\end{array}\right], \quad b_2=\left[\begin{array}{l}2 \\ 0 \\ 2\end{array}\right], \quad w=\left[\begin{array}{c}-2 \\ 1 \\ -1\end{array}\right]$[/tex]

We know,

[tex]Proj_{v}(w)[/tex] = [tex]Proj_{b1}(w) + Proj_{b2}[/tex]

[tex]& =\left(\frac{w \cdot b_1}{b_1 \cdot b_1}\right) b_1+\left(\frac{w \cdot b_2}{b_2 \cdot b_2}\right) b_2[/tex]

[tex]w \cdot b_1 &[/tex] =[tex]\left[\begin{array}{c}-2 \\1 \\-1\end{array}\right] \cdot\left[\begin{array}{c}2 \\2 \\-2\end{array}\right][/tex][tex]=-4+2+2=0 \\[/tex]

[tex]w \cdot b_2 &[/tex] =[tex]\left[\begin{array}{c}-2 \\1 \\-1\end{array}\right] \cdot\left[\begin{array}{l}2 \\0 \\2\end{array}\right][/tex] =[tex]-4+0-2=-6 \\[/tex]

[tex]b_2 \cdot b_2 &[/tex]  =[tex]\left[\begin{array}{c}2 \\0 \\2\end{array}\right] \cdot\left[\begin{array}{l}2 \\0 \\2\end{array}\right][/tex] =[tex]4+0+4=8[/tex]

Therefore

[tex]Proj_{v}(w)[/tex] = [tex]\left(\frac{0}{b_1 \cdot b_1}\right) b_1+\left(\frac{-6}{8}\right)\left[\begin{array}{l}2 \\0 \\2\end{array}\right] \\[/tex]

[tex]& =0+\frac{-3}{4}\left[\begin{array}{l}2 \\0 \\2\end{array}\right] \\[/tex]

[tex]& =\left[\begin{array}{c}\frac{-3}{2} \\0 \\-3 \\2\end{array}\right] \\[/tex]

[tex]Proj_{v}(w)[/tex] = [tex]\left[\begin{array}{c}-1.5 \\0 \\-1.5\end{array}\right][/tex]

The Projectile Motion of vectors will be [tex]Proj_{v}(w)[/tex] = [tex]\left[\begin{array}{c}-1.5 \\0 \\-1.5\end{array}\right][/tex]

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P(V|B)>P(V|A)

P(V|B)>P(V|A), so Bob is more likely to have virus, however is is still very small probability (only 15%), so in order to confirm illness, he should make one more test.

P(A/V) = 0.95

From the text we can also conclude,  that

P(A| V) = 0.1

P(B|V) = 0.9

P(B| V) = 0.5

P(V)= 0.1

What we need to calculate and compare is P( V |A) and P(V|B)

P(V∩A) = P(A) . P(V|A) => P(V|A)

P(V∩A) / P(A)

P(V∩A) means, that Joe has a virus and its detected so.,

P(V∩A) = P(A) . P(V|A) = 0.01 × 0.95

= 0.0095

P(A) is sum of two options: "Joe has virus and it is detected" and "Joe has no virus, but it was mistakenly detected", therefore:

P(A) = P(A) . P(V|A) + P(≈V)

P(A|≈V) = 0.01 . 0.95 + 0.99 . 0.1

= 0.1085

Dividing those 2 numbers we obtain

P(V|A) = 0.0095 / 0.1085

= 0.0875576036

Analogically,

P(V|B) = P(V∩B) / P(B)

= 0.01 . 0.9 / 0.01 . 0.9 + 0.99 . 0.1

= 0.1538461we observe that P(V|B) > P(V|A) ,  

so Bob is more likely to have virus, however is is still very small probability (only 15%), so in order to confirm illness, he should make one more test.

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The perimeter of the state of Wyoming, given the rectangle ABCD and the vertices, is 1, 282 miles

The perimeter is the distance around the shape so to find the perimeter of the shape, you need to find the distance between the sides of the rectangle and then add them up.

The distance between vertices can be found by the formula :

= ((x2 – x1)² + (y2 – y1)²)

The distance between AB is therefore 3.65 miles.

The distance between BC is 2.76 miles

The distance between CD is 3.65 miles.

The distance between AD is 2.76 miles

The perimeter of Wyoming is therefore :

= ( 3.65 x 2) + ( 2.76 x 2 )

= 12.82 miles x 100

= 1, 282 miles

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The general solutions for tanx−3cotx=0 is [tex]\frac{\pi}{3} \text { and } \frac{2 \pi}{3}[/tex]

[tex]\begin{aligned}& \tan x-\frac{3}{\tan x}=0 \\& \tan ^2 x-3=0 \text { (condition } \tan \times \text { not zero) } \\& \tan ^2 x=3 \rightarrow \tan x=\pm \sqrt{3}\end{aligned}[/tex][tex]a. $\tan x=\sqrt{3} \rightarrow x=\frac{\pi}{3}$b. $\tan x=-\sqrt{3} \rightarrow x=\frac{2 \pi}{3}$Extended answers:$$\begin{aligned}& x=\frac{\pi}{3}+k \pi \\& x=\left(2 \frac{\pi}{3}\right)+k \pi\end{aligned}[/tex]

What is the general equation for tan's solution?

Tan + Tan Equals Tan

n Z (i.e., n = 0, 1, 2, 3, etc.), so n = n + n. Therefore, n Z (i.e., n = 0, 1, 2, 3, etc.) is the value for which the general solution of tan = tan is given as = n +. Because cot = 1/tan and cot = 1/tan, the expression cot = cot is identical to tan = tant

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The angle opposite to the two congruent sides of an isosceles triangle are congruent.

To prove that the angles opposite the two congruent sides of an isosceles triangle are congruent, we can use the following steps:

This completes the proof of the theorem

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all can be the answer I guess

The minimum value, the first quartile (Q1), the median, the third quartile (Q3), and the maximum value must all be calculated before we can create a box plot for the rating points data.

The median of the lower and upper halves of the data must be determined in order to determine the first and third quartiles, respectively. The numbers from the minimum to the median make up the lower half of the data, while the values from the median to the highest make up the upper half.

We must first organize the numbers in numerical order: 93.5, 96.3, 97.9, 98.9, 99.3, 101.1, 103.9, 104.8, 105.1, 109.5. This will help us determine the median of the lower half of the data. The lower half's mean, which is (99.3 + 101.1) / 2 = 100.2, is then the average of the fifth and sixth values.

Again, the numbers must be arranged in numerical order to determine the median of the upper half of the data: 93.5, 96.3, 97.9, 98.9, 99.3, 101.1, 103.9, 104.8, 105.1, and 109.5. The average of the sixth and seventh numbers, or (101.1 + 103.9) / 2, equals 102.5, which is the median of the top half.

Following that, Q1 = 100 indicates the first and third quartiles.

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Answer: (i) 10% (ii) 60% (iii) 40%

Step-by-step explanation:

(i)  Newar language = 65%

  Gurung language = 45%

Total covers = 110 but to sum it up to base that is 100% then 110-100 is 10.

so, 10% of people speak both the language.

(ii) people who speak only Newari only would be 60% as 5% among them speak Gurung language.

(iii) people who speak only Gurung only would be 40% as 5% among them speak Newari language.

x=19

According to question,

3x-12=2x+7    

Subtracting 7 from both sides

3x-12-7=2x+7-7

⇒3x-19=2x

Subtracting 3x from both sides

3x-19-3x=2x-3x

⇒-19=-x

⇒x=19

∴The value of x is 19.

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The complete question is Find the value of x in the equation 3x-12=2x+7 when 7 and 3x is subtracted from both sides.

An inequality is a mathematical statement that compares two expressions using an inequality symbol. In this case, we have two inequalities that involve the variables \(x\) and \(y\): \(y > -\frac{1}{2} x 2\) and \(2x y \le 8\).

To find the points that satisfy these two inequalities, we must first solve for both \(x\) and \(y\).

To find the points that satisfy the first inequality, we can solve for \(y\) and substitute it into the second inequality:

\[y > -\frac{1}{2} x 2 \implies y = -\frac{1}{2} x 2 + k \quad \text{where} \quad k > 0\]

\[2x (-\frac{1}{2} x 2 + k) \le 8 \implies x^2 - 4x + 8 \le 0\]

Solving for \(x\) yields two solutions: \(x = 2 \pm \sqrt{2}\). To find the points that satisfy both inequalities, we must test both of these solutions in the original inequalities. For \(x = 2 + \sqrt{2}\), we have:

\[y > -\frac{1}{2} \cdot (2 + \sqrt{2}) \cdot 2 \implies y > 4 - 4\sqrt{2}\]

\[2 \cdot (2 + \sqrt{2}) \cdot y \le 8 \implies 8 + 8\sqrt{2} \le 8 \quad \text{which is true}\]

Therefore, the point \((2 + \sqrt{2}, 4 - 4\sqrt{2})\) satisfies both inequalities. For \(x = 2 - \sqrt{2}\), we have:

\[y > -\frac{1}{2} \cdot (2 - \sqrt{2}) \cdot 2 \implies y > 4 + 4\sqrt{2}\]

\[2 \cdot (2 - \sqrt{2}) \cdot y \le 8 \implies 8 - 8\sqrt{2} \le 8 \quad \text{which is true}\]

Therefore, the point \((2 - \sqrt{2}, 4 + 4\sqrt{2})\) also satisfies both inequalities.

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The following composition of two functions ( f o g ) ( 0 )  = 1 and ( g o f ) ( 1 ) = 1 .

Given :

The functions f and g are defined by the following tables:

Tables :

x       -2       -1      0       1

f ( x )   1        2     1        0

x          1       -3      0       -4

g ( x )   -2      0      1         5

( f o g ) ( 0 ) = ( f ( g ( 0 ) )

from the table g ( 0 ) = 1

= ( f ( 0 ) )

=  1

( g o f ) ( 1 ) = ( g ( f ( 1 ) )

= ( g ( 0 ) )

= 1

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Answer: To find the mean of the number of students who left school early, we must add all of the numbers together and then divide the sum by the number of days.

48 + 62 + 75 + 43 + 32 + 52 + 70 + 63 + 81 + 40 + 38 + 67 = 671

The mean of the number of students who left school early is 56.75 (671 / 12).

This mean represents the average number of students who left school early each day over the 12-day period. It tells us that on average, about 56.75 students left school early each day.  

Step-by-step explanation:  2. The mean of the group of teachers' ages is currently (25+32+33+35+41+43+45+48+52+55+60+62)/12=45.5. If another teacher with an age of 45 is added to the data, the mean will be (25+32+33+35+41+43+45+45+48+52+55+60+62)/13=44.5. The mean will be slightly lower, as the additional value of 45 is being added to the sum, but the number of values being averaged is also increasing by 1.

When x = 9, the inequality - 4 + x ≤ 9 has the necessary value.

In mathematics, an inequality displays the connection between two values in an unbalanced algebraic statement. One of the two variables on the two sides of the inequality may be greater than, greater than or equal to, less than, or less than or equal to another value, according to inequality signals.

Mathematics that contains at least two terms with variables or integers that are not equal is referred to as being unequal.

Inequality, two integers or algebraic expressions with a declared order link (greater than, greater than, or equal to, less than, or less than or equal to).

As per option (B),

When substituting the values the equation -4 + 9 is less than or equal to 5.

-4 + 9 is 5 so this equation is true

The required value of inequality - 4 + x ≤ 9  is true when x = 9,

Therefore, the correct answer is option (B) -4 + x ≤ 5.

The complete question is:

Which inequality is true when x = 9?

A) 4 - x ≥ 5

B) -4 + x ≤  5

C) 4 + x ≤ 5  

D) -4 - x ≥ 5

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Spotify charges $0.98 for individual song and $10 for entire album

According to the question,

Spotify charges "x" dollars for individual songs and "y" for entire album

Also, Person A pays $14.9 to download 5 individual songs and 1 album

Which can be written as

=> 5x + y = 14.9 --------------(1)

Person B pays $22.95 to download 3 individual songs and 2 albums

=> 3x + 2y = 22.95 ---------(2)

Solving these two linear equations by elimination method

Multiplying equation (1) by 2

=> 10x + 2y = 29.8

Now, subtracting these equation

10x + 2y - 3x - 2y = 29.8 - 22.95

=> 7x = 6.85

=> x = 0.98

Replacing value of x in equation (1),

=> 5(0.98) + y = 14.9

=> 4.9 + y = 14.9

=> y = 10

Hence , Spotify charges $0.98 for individual song and $10 for entire album

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(a) The probability that O1 gets the Volvo is  3/10 .

(b) The probability that O2 or O5 will get the Volvo is 1/2  .

In the question ,

it is given that ,

the five officials : O1, O2, O3 , O4 , O5 are to be assigned five different city cars: a Escort,  Lexus,  Nissan, the Taurus, and the Volvo .

From the given data , the table formed is shown below .

From the table , we can observe that ,

O1 will not drive the Escort or a Nissan . So, he has chance of getting one car among Lexus, Taurus & Volvo = 3 possibilities

O2 will not drive the Taurus​ . So, he has chance of getting one car among Escort , Nissan ,Lexus, & Volvo = 4 possibilities

O3 will not drive the Lexus or Volvo​ . So, he has chance of getting one car among Escort , Nissan ,Taurus = 3 possibilities

O4 will not drive the Lexus so he has chance of getting one car among Escort , Nissan ,Taurus, Volvo = 4 possibilities

O5 will not drive the Escort or a Nissan . So, he has chance of getting one car among Lexus, Taurus & Volvo = 3 possibilities .

The  20 possible arrangements for E,L,N,T,V are :

{ (O2, O1, O3, O4, O5),  (O2, O1, O3, O5, O4),  (O2, O1, O4, O3, O5),  

(O2, O5, O3, O1, O4),  (O2, O5, O3, O4, O1),(O2, O5, O4, O3, O1),  

(O3, O1, O2, O4, O5),  (O3, O1, O2, O5, O4),  (O3, O1, O4, O5, O2),  

(O3, O2, O4, O1, O5), (O3, O2, O4, O5, O1),  (O3, O5, O2, O1, O4),  

(O3, O5, O2, O4, O1),  (O3, O5, O4, O1, O2),  (O4, O1, O2, O3, O5),

(O4, O1, O3, O5, O2),  (O4, O2, O3, O1, O5),  (O4, O2, O3, O5, O1),  

(O4, O5, O2, O3, O1),  (O4, O5, O3, O1, O2)  }

Part(a)  : The probability that O1 gets the Volvo is = 6/20 = 3/10 .

and

Part(b) :

Probability of O2 or O5 gets the Volvo = (Probability of O2 getting Volvo
) + (Probability of O5 getting Volvo) .

= 4/20 + 6/20

= 10/20 = 1/2 .

Therefore , the required probability for (a)  is 3/10 and for (b) is 1/2  .

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Answer: 806.25

Step-by-step explanation: First, we need to find how much the sales tax is worth in dollars. Recall the formula:

Worth = Wole number x part percent / 100

Now, plug in the numbers:

Worth = 750 x 7.5 / 100

So, 750 x 7.5 = 5625

So, 5625 / 10 = 56.25

But remember, $56.25 is the amount of the SALES TAX. To find the total, we need to add 56.25 to 750. So,

750 + 56.25 = 806.25

Therefore, $806.25 is the TOTAL AMOUNT she needs to pay.

The solution of the logarithmic equation:

ln(x - 7) = 3

Is x = 27.09

Here we have the following logarithmic function:

ln(x - 7) = 3

And we want to solve this for x.

Remember that the logarithmic function is the inverse of the exponential function, so now we can use the exponential function in both sides so we get:

exp( ln(x - 7) ) = exp(3)

x - 7 = exp(3)

Now we can add 7 in both sides so we isolate x, we will get:

x = exp(3) + 7

x = 27.09

That is the solution of the logarithmic equation.

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Ratio is the relation between two amounts that shows the number of times one value contains within the other.

What is Ratio?

A ratio in mathematics illustrates how many times one number contains another. For example, if a dish of vegetables contains eight tomato and six carrots, the tomato-to-carrot ratio is eight to six (that is, 8:6, which is equivalent to the ratio 4:3). Similarly, the proportion of carrots to tomato is 6:8 (or 3:4), while the proportion of tomato to overall fruit is 8:14. (or 4:7). A ratio's numbers can be any quantity, such as a count of persons or things, or measures of lengths, weights, time, and so forth. In most situations, both numbers must be positive.

Ratio is the relation between two amounts that shows the number of times one value contains within the other.

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Answer:

False

Step-by-step explanation:

Lines that intersect at 90 degrees are called perpendicular lines.

Skew lines do not intercept.

Answer:

a = -3

Step-by-step explanation:

The explanation is in the image

NP finished. Only two jobs share resources with the reduction provided in section A.

to put in the same order or rank. : to bring into a common action, movement, or condition : harmonize. coordinate schedules. She'll be coordinating the relief effort.

The overall issue is NP-complete. From Independent Set, this is a decrease. We substitute jobs for the graph's vertex points. If there is an edge connecting vertices v and w, we establish a resource called r v,w that v and w can share.

If k = 2, we are simply checking to see if there are any jobs that are resource-exclusive. By checking every pair of jobs, we can accomplish this using sheer force.

NP finished. Only two jobs share resources with the reduction provided in section A.

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