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kow [346]
3 years ago
11

4(x-2)+2x = 10

Mathematics
1 answer:
sleet_krkn [62]3 years ago
8 0

Answer:

x = 3

Step-by-step explanation:

Given:

4(x-2)+2x=10

Step 1: 4x-8+2x=10

Step 2:4x+2x-8=10

Step 3: 4x+2x-8+8=10+8

Now Solving for next steps we get

Step 4: 6x=18

dividing both side by 6 we get

Step 5:\frac{6x}{6} =\frac{18}{6}

And So the Final Answer is:

x = 3

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Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive
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Answer:

(a) 0.343

(b) 0.657

(c) 0.189

(d) 0.216

(e) 0.353

Step-by-step explanation:

Let P(a vehicle passing the test) = p

                        p = \frac{70}{100} = 0.7  

Let P(a vehicle not passing the test) = q

                         q = 1 - p

                         q = 1 - 0.7 = 0.3

(a) P(all of the next three vehicles inspected pass) = P(ppp)

                           = 0.7 × 0.7 × 0.7

                           = 0.343

(b) P(at least one of the next three inspected fails) = P(qpp or qqp or pqp or pqq or ppq or qpq or qqq)

      = (0.3 × 0.7 × 0.7) + (0.3 × 0.3 × 0.7) + (0.7 × 0.3 × 0.7) + (0.7 × 0.3 × 0.3) + (0.7 × 0.7 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.3)

      = 0.147 + 0.063 + 0.147 + 0.063 + 0.147 + 0.063 + 0.027

      = 0.657

(c) P(exactly one of the next three inspected passes) = P(pqq or qpq or qqp)

                 =  (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

                 = 0.063 + 0.063 + 0.063

                 = 0.189

(d) P(at most one of the next three vehicles inspected passes) = P(pqq or qpq or qqp or qqq)

                 =  (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7) + (0.3 × 0.3 × 0.3)

                 = 0.063 + 0.063 + 0.063 + 0.027

                 = 0.216

(e) Given that at least one of the next 3 vehicles passes inspection, what is the probability that all 3 pass (a conditional probability)?

P(at least one of the next three vehicles inspected passes) = P(ppp or ppq or pqp or qpp or pqq or qpq or qqp)

=  (0.7 × 0.7 × 0.7) + (0.7 × 0.7 × 0.3) + (0.7 × 0.3 × 0.7) + (0.3 × 0.7 × 0.7) + (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

= 0.343 + 0.147 + 0.147 + 0.147 + 0.063 + 0.063 + 0.063

                  = 0.973  

With the condition that at least one of the next 3 vehicles passes inspection, the probability that all 3 pass is,

                         = \frac{P(all\ of\ the\ next\ three\ vehicles\ inspected\ pass)}{P(at\ least\ one\ of\ the\ next\ three\ vehicles\ inspected\ passes)}

                         = \frac{0.343}{0.973}

                         = 0.353

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