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Paraphin [41]
3 years ago
12

Five countries enter a race. The first three racers in order, Jamaica finished before Barbados, but behind Trinidad. Haiti finis

hed before Guyana, but behind Barbados. Who won the Bronze (finished in 3rd place)?
Mathematics
1 answer:
AveGali [126]3 years ago
8 0

Answer:

Barbados won the Bronze.

Step-by-step explanation:

The information indicates that Jamaica finished before Barbados, but behind Trinidad which means that Trinidad was the first one, then Jamaica and the third one was Barbados:

1. Trinidad

2. Jamaica

3. Barbados

Then, it indicates that Haiti finished before Guyana, but behind Barbados which indicates that Haiti was in 4th place after Barbados and  the last one was Guyana:

1. Trinidad

2. Jamaica

3. Barbados

4. Haiti

5.Guyana

According to this, the answer is that Barbados won the Bronze.

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denis23 [38]

Answer:

(1) The possible outcomes are: X = {0, 1, 2, 3}.

(2) The number of times should Hartley spin a difference of 1 is 36.

(3) The number of times should Hartley spin a difference of 0 is 24.

Step-by-step explanation:

The number of sections on the spinner is 4 labelled as {1, 2, 3, 4}.

The total number of spins for each of the spinner is, <em>n</em> = 96.

(1)

The sample space of spinning both the spinners together are:

S = {(1, 1), (1, 2), (1, 3), (1, 4)

      (2, 1), (2, 2), (2, 3), (2, 4)

      (3, 1), (3, 2), (3, 3), (3, 4)

      (4, 1), (4, 2), (4, 3), (4, 4)}

Total = 16.

The possible outcomes are:

X = {0, 1, 2, 3}.

(2)

The sample space with the difference 1 are:

S₁ = {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3)}

n (S₁) = 6

The probability of the difference 1 is:

P(\text{Diff}=1)=\frac{n(S_{1})}{N}=\frac{6}{16}=\frac{3}{8}

The spinners were spinner 96 times.

The expected number of times would Hartley spin a difference of 1 is:

E(\text{Diff}=1)=P(\text{Diff}=1)\times n\\\\=\frac{3}{8}\times 96\\\\=36

Thus, the number of times should Hartley spin a difference of 1 is 36.

(3)

The sample space with the difference 0 are:

S₂ = {(1, 1), (2, 2), (3, 3), (4, 4)}

n (S₂) = 4

The probability of the difference 0 is:

P(\text{Diff}=0)=\frac{n(S_{2})}{N}=\frac{4}{16}=\frac{1}{4}

The spinners were spinner 96 times.

The expected number of times would Hartley spin a difference of 0 is:

E(\text{Diff}=0)=P(\text{Diff}=0)\times n\\\\=\frac{1}{4}\times 96\\\\=24

Thus, the number of times should Hartley spin a difference of 0 is 24.

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