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KatRina [158]
4 years ago
13

3.95m + 8.95b = 47.65

Mathematics
1 answer:
nataly862011 [7]4 years ago
6 0
3.95m + 8.95b = 47.65  Equation 
 
                      \fbox{Solution for variable m}

3.95m=47.65-8.95b
  \text{Subtract 8.95b from both sides} 

m= \dfrac{47.65-8.95b}{3.95}  \text{Divide both sides by 3.95} 

                      \fbox{Solution for variable b}        
 
 3.95m + 8.95b = 47.65  Equation    

8.95b=47.65-3.95m
     \text{Subtract 3.95m from both sides}

b= \dfrac{47.65-3.95m}{8.95} \text{Divide both sides by 8.95} 
         
                                  \text{Hence Solved}
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Determine the intercepts of the line that correspond to the following table of values.
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y = mx + b
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Find three consecutive intergers such that three times the third is 102 more than the sum of the first and second.
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Step-by-step explanation:

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A pharmacist receives a shipment of 21 bottles of a drug and has three of the bottles tested. If five of the 21 bottles are cont
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Answer:

The probability is 0.8722

Step-by-step explanation:

There are 21 bottles

5 of them are contaminated.

there are 21 - 5 = 16 non-contaminated bottles.

So, if we grab a bottle at random, the probability that this bottle is contaminated will be equal to the quotient between the number of contaminated bottles and the total number of bottles, this is:

p = 5/21

Now we want to find the probability that, for 3 tested bottles, that less than two (0 or 1 ) are contaminated.

Let's see each case on its own.

0 bottles:

The probability of getting a non-contaminated bottle in the first try is equal to the quotient between the number of non-contaminated bottles and the total number of bottles, this is:

p₁ = 16/21

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p₂ = 15/20

and similarly, for the third bottle the probability is:

p₃ = 14/19

The joint probability is the product of the individual probabilities, we get:

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Let's assume that the first one is contaminated.

The probability of getting a contaminated bottle in the first draw is equal to the quotient between the number of contaminated bottles and the total number of bottles, so:

p₁ = 5/21

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p₂ = 16/20

and for the third bottle we have the probability:

p₃ = 15/19

The joint probability is:

p = p₁*p₂*p₃ = (5/21)*(16/20)*(15/19)

Also notice that we only looked at the case where the first bottle is contaminated, we also have the case where the second one is contaminated and the case where the third one is  contaminated, so there are 3 permutations, then the probability of having one contaminated bottle is:

Q = 3*p = 3*(5/21)*(16/20)*(15/19) = 0.4511

Then the probability of having less than two contaminated bottles is:

probability = P + Q = 0.4211 + 0.4511 = 0.8722

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3 years ago
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