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agasfer [191]
2 years ago
5

The random variable X, representing the number of accidents in a certain intersection in a week, has the following probability d

istribution: x 0 1 2 3 4 5 P(X = x) 0.20 0.30 0.20 0.15 0.10 0.05 What is the probability that in a given week there will be at most 3 accidents? 0.70 0.85 0.35 0.15 1.00
Mathematics
1 answer:
ollegr [7]2 years ago
7 0

Answer: 0.70

Step-by-step explanation:

Given : The random variable X, representing the number of accidents in a certain intersection in a week, has the following probability distribution:

       x       0       1         2        3       4       5

P(X = x) 0.20   0.30   0.20   0.15   0.10  0.05

Using the above probability distribution , the  the probability that in a given week there will be at most 3 accidents is given by :_

P(\leq3)=P(0)+P(1)+P(2)+P(3)\\\\=0.20+0.30+0.20=0.70

Hence, the required probability = 0.70

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2 years ago
∠1 ​ and ∠2 are supplementary. ∠1=124°∠2=(2x+4)° Select from the drop down menu to correctly answer the question.
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Remark
Supplementary angles add up to 180o. 

Givens
<1 = 124
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I imagine you are looking for either x or <2. 

Equation
<1 + <2 = 180          Substitute the givens
124 + 2x + 4 = 180  Collect like terms on the left.
128 + 2x = 180         Subtract 128 from both sides
2x = 180 - 128          Collect like terms on the right   
2x = 52                     divide by 2
x = 52/2
x = 26 <<<<<<<<< answer

<1 = 124
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<2 = 52 + 4
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We need choices if you want an exact answer.

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