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Musya8 [376]
3 years ago
11

Two friends race against each other in 100-meter and 800-meter races. Of the 200 races they ran this year, the wins, losses, and

ties are shown below.
100 meter (Event D) 800 meter (Event E)
A won the race (Event A) 30 50
B won the race (Event B)60 25
A and B tied (Event T)10 25
What percentage of all the races are ties? If someone wins the race today, is it more likely for A to win or B to win? How much more likely?
Mathematics
1 answer:
sveta [45]3 years ago
5 0
1. <span>What percentage of all the races are ties?
If</span>The chance for the race become tie is 10% for 100m and 25% for 800m. Assuming that the 200 race is divided equally, then there would be 100 race of 100-m and 100 race of 800-m.
The number of the event of ties would be:
(100*10% + 100*25%)/ 200= 35/200= 17.5%

2. If someone wins the race today, is it more likely for A to win or B to win? How much more likely?The event should be independent because the two events should not influence each other. That means the probability will not change and no one becomes more likely to win the next race.
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Answer:

<h3><u>Question 7</u></h3>

<u>Lateral Surface Area</u>

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\implies \sf L.A.=2(10 \times 6)+(3 \times 6)=138\:\:m^2

<u>Total Surface Area</u>

Area of the isosceles triangle:

\implies \sf A=\dfrac{1}{2}\times base \times height=\dfrac{1}{2}\cdot3 \cdot \sqrt{10^2-1.5^2}=\dfrac{3\sqrt{391}}{4}\:m^2

Total surface area:

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<u>Volume</u>

\sf \implies Vol.=area\:of\:base \times height=\left(\dfrac{3\sqrt{391}}{4}\right) \times 6=88.98\:\:m^3\:(2\:d.p.)

<h3><u>Question 8</u></h3>

<u>Lateral Surface Area</u>

The bases of a hexagonal prism are the pentagons.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the pentagons (bases).

\implies \sf L.A.=5(5 \times 6)=150\:\:cm^2

<u>Total Surface Area</u>

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\sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}a^2

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Therefore:

\implies \sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}\cdot 5^2=43.01\:\:cm^2\:(2\:d.p.)

Total surface area:

\sf \implies T.A.=2\:bases+L.A.=2(43.01)+150=236.02\:\:cm^2\:(2\:d.p.)

<u>Volume</u>

\sf \implies Vol.=area\:of\:base \times height=43.011193... \times 6=258.07\:\:cm^3\:(2\:d.p.)

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