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Anika [276]
3 years ago
6

(a+b-c)-(b-a-c) I NEED IT RN FOR A TEST OR ELSE I AM DEAD FOR LIFE

Mathematics
2 answers:
Zepler [3.9K]3 years ago
5 0

Answer:

<h2><em><u>2a</u></em></h2>

Step-by-step explanation:

(a+b-c)-(b-a-c)

= a + b - c - b + a + c

= a + a + b - b - c + c

= <em><u>2a (Ans)</u></em>

Marat540 [252]3 years ago
3 0
Here is the answer-I just took a screenshot off of an app called Symbolab which is really good for algebra

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1/3 divided by 8??????????
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Answer: 1/24

Step-by-step explanation: if you take 8 and change it to 1/8, you can multiply it by 1/3 and 1x1 is 1 and 8x3 is 24 so it’s 1 over 24 or 1/24

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What is 84:7 in its lowest form
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Divide both the numerator and denominator by the GCD so it’s 12/1
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Consider an experiment where two 6-sided dice are rolled. We can describe the ordered sample space as below where the first coor
agasfer [191]

Answer:

  • E = { (4,1) , (3,2) , (2,3) , (1,4) }
  • P(E)=\frac{1}{9}
  • P(F|E)=\frac{1}{4}

Step-by-step explanation:

Let's start writing the sample space for this experiment :

S= { (1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6) , (2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6) , (3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6) , (4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6) , (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) , (6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6) }

Let's also define the event E ⇒

E : '' The sum of the two dice is 5 ''

We can describe the event by listing all the favorables cases from S ⇒

E = { (4,1) , (3,2) , (2,3) , (1,4) }

In order to calculate P(E) we are going to divide all the cases favorables to E over the total cases from S. We can do this because all 36 of these possible outcomes from S are equally likely. ⇒

P(E)=\frac{4}{36}=\frac{1}{9} ⇒

P(E)=\frac{1}{9}

Finally we are going to define the event F ⇒

F : '' The number of the first die is exactly 1 more than the number on the second die ''

⇒

F = { (2,1) , (3,2) , (4,3) , (5,4) , (6,5) }

Now given two events A and B ⇒

P ( A ∩ B ) = P(A,B)

We define the conditional probability as

P(A|B)=\frac{P(A,B)}{P(B)} with P(B)>0

We need to find P(F|E) therefore we can apply the conditional probability equation :

P(F|E)=\frac{P(F,E)}{P(E)}   (I)

We calculate P(E)=\frac{1}{9} at the beginning of the question. We only need P(F,E).

Looking at the sets E and F we find that (3,2) is the unique result which is in both sets. Therefore is 1 result over the 36 possible results. ⇒

P(F,E)=\frac{1}{36}

Replacing both probabilities calculated in (I) :

P(F|E)=\frac{P(F,E)}{P(E)}=\frac{\frac{1}{36}}{\frac{1}{9}}=\frac{1}{4}=0.25

We find out that P(F|E)=\frac{1}{4}=0.25

6 0
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Step-by-step explanation:

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Step-by-step explanation:

can u please help me

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