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Ksivusya [100]
3 years ago
5

5(1+s)=9s+6 ----------------

Mathematics
2 answers:
liq [111]3 years ago
8 0

Answer:

Solution given

5(1+s)=9s+6

distribute

5+5s=9s+6

subtract both side by 5s

5+5s-5s=9s-5s+6

solve like terms

5=4s+6

subtract both side by 6

5-6=4s+6-6

-1=4s

divide both side by 4

-1/4=s

emmainna [20.7K]3 years ago
5 0

Answer:

s=-\frac{1}{4}

s=-0.25

Step-by-step explanation:

5\left(1+s\right)=9s+6

Apply distributive law:-

5\left(1+s\right)

=5+5s

5+5s=9s+6

Now, Subtract 5 from both sides :-

5+5s-5=9s+6-5

5s=9s+1

Subtract 9s from both sides:-

5s-9s=9s+1-9s

-4s=1

Divide both sides by -4:-

\frac{-4s}{-4}=\frac{1}{-4}

s=-\frac{1}{4}

<u>OAmalOHopeO</u>

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Answer:

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

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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

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= 6x² + 8x - 9x - 12 ← collect like terms

= 6x² - x - 12 ← in expanded form


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