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nekit [7.7K]
3 years ago
13

Samantha’s younger brother just turned 3 years old. Fred’s brother is now 30 months old. Who brother is older? Explain

Mathematics
1 answer:
Alina [70]3 years ago
6 0
1. You have that:
 
 - <span>Samantha’s younger brother  turned 3 years old.
 - </span><span>Fred’s brother is 30 months old.
 
 2. one (1) year has twelve (12) months. </span>Keeping this on mind, you have to follow the proccedure below:<span>
 
 3 yearsx12 months=36 months
  
 3. As you can see, </span><span>Samantha’s younger brother just turned 36 months old</span>  and Fred’s brother is 30 months old. Therefore, you can conclude that Samantha’s younger brother is older than Fred’s brother.


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gavmur [86]

Answer:

Step-by-step explanation:

6x + 3y = -18

Taking 3 common from left side

so, 3(2x+y) = -18

2x+y = -18/3

2x+y = -6 (equation 1)

and, 7x + 7y = 0

Taking 7 as commom from left side

7(x + y) = 0

x + y = 0/7

x + y = 0 (equation 2)

now , by using elimination method

subtracting equation2 from equation1

2x + y - (x + y)= -6 - 0

2x + y - x - y = -6

x = -6

so, x = -6

substituting value of x in equation 1

2(-6) + y = -6

-12 + y = -6

y = -6 +12

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7 0
3 years ago
Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive
NeX [460]

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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