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sladkih [1.3K]
2 years ago
12

The event coordinator asks you to determine how many students participated in the track-and-field day. The total number of stude

nts in seventh and eighth grade combined is 584; 5/8 of them are seventh graders and 3/8 of them are eighth graders. If 4/5 of the seventh graders participated in track-and-field day and 7/8 of the eighth graders participated, about how many total students participated? Describe the process you used to find your answer.
Mathematics
1 answer:
kakasveta [241]2 years ago
8 0
584 is the total number of students in the school.

5/8 are in seventh grade and 3/8 are in eighth grade.

4/5 of the seventh graders participated in the track meet, so there were (4/5 · 5/8) · 584 students in the seventh grade participating in the track meet.

7/8 of the eighth graders participated, so there were (7/8 · 3/8) · 584 students in the eighth grade participating in the track meet.

So, all together, there were

(4/5 · 5/8) · 584 + (7/8 · 3/8) · 584 students from the school in the track meet.

Let's simplify as you asked:

(4/5 · 5/8) · 584 + (7/8 · 3/8) · 584 = [(4/5 · 5/8) + (7/8 · 3/8)] · 584 (distributive property - factoring)

= [20/40 + 21/64] · 584 (multiply fractions)

= (1/2 + 21/64) 584 (reduce the first fraction to lowest terms)

= (32/64 + 21/64) 584 (getting a common denominator)

= (53/64) 584 (combine/add the two fractions)

= 483.625 (multiply together)

All together, there were: 483.625 students in the meet.
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Eighty percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive
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Answer:

(a) P(all of the next three vehicles inspected pass) = 0.512 .

(b) P(at least one of the next three inspected fails) = 0.488 .

(c) P(exactly one of the next three inspected passes) = 0.096 .

(d) P(at most one of the next three vehicles inspected passes)<em> = </em>0.104 .

(e) Probability that all three pass given that at least one of the next three vehicles passes inspection = 0.516 .

Step-by-step explanation:

We are given the Probability of all vehicles examined at a certain emissions inspection station pass the inspection to be 80%.

So, Probability that the next vehicle examined fails the inspection is 20%.

Also, it is given that successive vehicles pass or fail independently of one another.

(a) P(all of the next three vehicles inspected pass) = Probability that first vehicle, second vehicle and third vehicle also pass the inspection

              = 0.8 * 0.8 * 0.8 = 0.512

(b) P(at least one of the next three inspected fails) =

       <em> 1 - P(none of the next three inspected fails) = 1 - P(all next three passes)</em>

    = 1 - (0.8 * 0.8 * 0.8) = 1 - 0.512 = 0.488 .

(c) P(exactly one of the next three inspected passes) is given by ;

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  • <em>Second vehicle pass the inspection, first and third vehicle doesn't pass</em>
  • <em>Third vehicle pass the inspection, first and second vehicle doesn't pass</em>

Hence, P(exactly one of the next three inspected passes) = Add all above cases ;

             (0.8 * 0.2 * 0.2) + (0.2 * 0.8 * 0.2) + (0.2 * 0.2 * 0.8) = 0.096 .

(d) <em>P(at most one of the next three vehicles inspected passes) = P(that none</em>

<em>      of the next three vehicle passes) + P(Only one of the next three passes)</em>

We have calculated the P(Only one of the next three passes) in the above part of this question;

 Hence, P(at most one of the next three vehicles inspected passes) =

              (0.2 * 0.2 * 0.2) + (0.8 * 0.2 * 0.2) + (0.2 * 0.8 * 0.2) + (0.2 * 0.2 * 0.8)

             = 0.008 + 0.096 = 0.104 .

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<em>P(All three passes/At least one of the next three vehicles passes inspection)</em>

= P( All next three passes \bigcap At least one of next three passes) /

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=  P( All next three passes ) / P(At least one of next three passes)

= P( All next three passes ) / 1 - P(none of the next three passes)

 =  \frac{0.8*0.8*0.8}{1-(0.2*0.2*0.2)} = 0.516 .

Therefore, Probability that all three pass given that at least one of the next three vehicles passes inspection = 0.516 .

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