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kobusy [5.1K]
2 years ago
11

PLEASE HELP FAST!! WILL MARK AS BRAINLIEST IF GIVEN GOID ANSWER....​

Mathematics
1 answer:
alexandr1967 [171]2 years ago
6 0

Answer:

Step-by-step explanation:

I am quite not sure what do you mean by corresponding.

like, what side should RS in ΔRQS would be the side in ΔRSP

if so, then the answer would be PR

we know this because RS is the hypotenuse of ΔRQS

and PR is the hypotenuse of ΔRPS

But if you meant by the question that which side would be equal to the side in the bigger triangle, then it would be RS itself, because it's a common side in both triangles

hope this helps!

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Neko [114]
So, for number two, you fill in the numbers that theyve given you
x. y
1. 5
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8 0
3 years ago
A system contains n atoms, each of which can only have zero or one quanta of energy. How many ways can you arrange r quanta of e
My name is Ann [436]

Answer:

\mathbf{a)} 2\\ \\ \mathbf{b)} 184 \; 756 \\ \\\mathbf{c)}  \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}

Step-by-step explanation:

If the system contains n atoms, we can arrange r quanta of energy in

                         \binom{n}{r} = \dfrac{n!}{r!(n-r)!}

ways.

\mathbf{a)}

In this case,

                                n  = 2, r=1.

Therefore,

                    \binom{n}{r} = \binom{2}{1} = \dfrac{2!}{1!(2-1)!} = \frac{2 \cdot 1}{1 \cdot 1} = 2

which means that we can arrange 1 quanta of energy in 2 ways.

\mathbf{b)}

In this case,

                                n  = 20, r=10.

Therefore,

                    \binom{n}{r} = \binom{20}{10} = \dfrac{20!}{10!(20-10)!} = \frac{10! \cdot 11 \cdot 12 \cdot \ldots \cdot 20}{10!10!} = \frac{11 \cdot 12 \cdot \ldots \cdot 20}{10 \cdot 9 \cdot \ldots \cdot 1} = 184 \; 756

which means that we can arrange 10 quanta of energy in 184 756 ways.

\mathbf{c)}

In this case,

                                n = 2 \times 10^{23}, r = 10^{23}.

Therefore, we obtain that the number of ways is

                    \binom{n}{r} = \binom{2\times 10^{23}}{10^{23}} = \dfrac{(2\times 10^{23})!}{(10^{23})!(2\times 10^{23} - 10^{23})!} = \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}

3 0
3 years ago
9 is 9% of what number?
hjlf

Answer:

are you sure this is high school level stuff? its 9% of 100

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
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Brut [27]

Answer:

Step-by-step explanation:

area of top=7×6=42 yd²

area of bottom=3×6=18 yd²

area of two rectangles=2[4×6]=48 yd²

area of two trapezoids=2[1/2(7+3)×3.5]=35 yd²

total surface area=42+18+48+35=143 yd²

7 0
3 years ago
We have two fair three-sided dice, indexed by i = 1, 2. Each die has sides labeled 1, 2, and 3. We roll the two dice independent
Bogdan [553]

Answer:

(a) P(X = 0) = 1/3

(b) P(X = 1) = 2/9

(c) P(X = −2) = 1/9

(d) P(X = 3) = 0

(a) P(Y = 0) = 0

(b) P(Y = 1) = 1/3

(c) P(Y = 2) = 1/3

Step-by-step explanation:

Given:

- Two 3-sided fair die.

- Random Variable X_1 denotes the number you get for rolling 1st die.

- Random Variable X_2 denotes the number you get for rolling 2nd die.

- Random Variable X = X_2 - X_1.

Solution:

- First we will develop a probability distribution of X such that it is defined by the difference of second and first roll of die.

- Possible outcomes of X : { - 2 , -1 , 0 ,1 , 2 }

- The corresponding probabilities for each outcome are:

                  ( X = -2 ):  { X_2 = 1 , X_1 = 3 }

                  P ( X = -2 ):  P ( X_2 = 1 ) * P ( X_1 = 3 )

                                 :  ( 1 / 3 ) * ( 1 / 3 )

                                 : ( 1 / 9 )

   

                  ( X = -1 ):  { X_2 = 1 , X_1 = 2 } + { X_2 = 2 , X_1 = 3 }

                 P ( X = -1 ):  P ( X_2 = 1 ) * P ( X_1 = 3 ) + P ( X_2 = 2 ) * P ( X_1 = 3)

                                 :  ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

                                 : ( 2 / 9 )

         

       ( X = 0 ):  { X_2 = 1 , X_1 = 1 } + { X_2 = 2 , X_1 = 2 } +  { X_2 = 3 , X_1 = 3 }

       P ( X = -1 ):P ( X_2 = 1 )*P ( X_1 = 1 )+P( X_2 = 2 )*P ( X_1 = 2)+P( X_2 = 3 )*P ( X_1 = 3)

                                 :  ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

                                 : ( 3 / 9 ) = ( 1 / 3 )

       

                    ( X = 1 ):  { X_2 = 2 , X_1 = 1 } + { X_2 = 3 , X_1 = 2 }

                 P ( X = 1 ):  P ( X_2 = 2 ) * P ( X_1 = 1 ) + P ( X_2 = 3 ) * P ( X_1 = 2)

                                 :  ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

                                 : ( 2 / 9 )

                    ( X = 2 ):  { X_2 = 1 , X_1 = 3 }

                  P ( X = 2 ):  P ( X_2 = 3 ) * P ( X_1 = 1 )

                                    :  ( 1 / 3 ) * ( 1 / 3 )

                                    : ( 1 / 9 )                  

- The distribution Y = X_2,

                          P(Y=0) = 0

                          P(Y=1) =  1/3

                          P(Y=2) = 1/ 3

- The probability for each number of 3 sided die is same = 1 / 3.

7 0
3 years ago
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