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defon
3 years ago
13

What conclusions can you draw about the two track teams from the data shown?

Mathematics
2 answers:
IceJOKER [234]3 years ago
8 0
Where is the data that is shown?
kumpel [21]3 years ago
3 0
Which data shown exactly?
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6 stools and 4chairs cost$58 but 5 stools and 2 chairs cost$ 35 find the cost of each stool and each chair
Ivahew [28]
Stool = s, chair = c
6s+4c=58 ---(1)
5s+2c=35 ---(2)

(2)*2; 10s+4c=70 ---(3)
(3)-(1); 4s=12
so s=3
from (2)
5s+2c= 35
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A job fair was held at the Student Union. 25% of the students who attended received job offers. Of all of the students at the jo
vovangra [49]

Answer:

A) Both events are not independent.

B) Both events are not mutually exclusive

C) 8.33%

D) 80%

Step-by-step explanation:

A) Both events are not independent. This is because, If B occurs it means that it is very likely that J will occur as well.

B) Both events are not mutually exclusive. This is because it is possible for both events J and B to occur at the same time.

C) we want to find the probability that Joe who is not a business student will receive the job offer.

This is;

P(J|Not B) = P(J & Not B)/P(Not B)

Now,

P(J & Not B) = P(J) – (P(B) × P(J | B))

25% of the students who attended received job offers. Thus; P(J) = 0.25

40% were from the College of Business. Thus;

P(B) = 0.4

Among the business students, 50% received job offers. Thus;

P(J|B) = 0.5

Thus;

P(J & Not B) = 0.25 - (0.4 × 0.5)

P(J & Not B) = 0.25 - 0.2

P(J & Not B) = 0.05

Since P(B) = 0.4

Then, P(Not B) = 1 - 0.4 = 0.6

Thus;

P(J|Not B) = 0.05/0.6

P(J|Not B) = 0.0833 = 8.33%

D) This probability is represented by;

P(B | J) = P(B & J)/P(J)

P(B & J) = (P(B) × P(J | B)) = (0.4 × 0.5) = 0.2

P(B | J) = 0.2/0.25

P(B | J) = 0.8 = 80%

5 0
3 years ago
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