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2 years ago

5

The standard Olympic marathon distance is 26 miles 385 yards. What is this distance in kilometres to 3 decimal places? (1 yard ?

0.9144 metre, 1 mile = 1760 yards)

1 answer:

Ivanshal [37]2 years ago

8 0

Answer:

50.465 km

Step-by-step explanation:

In order to find the distance in kilometers, you need to first convert the marathon distance into yards. Since the conversion measurement of 1 mile = 1760 yards is given, you can use this to find the total yards in 26 miles:

$\frac{1760yards}{1mile} =\frac{x}{26miles}$ or x = 45,760 yards

Adding this amount to the additional 385 in the marathon distance:

45760 + 385 = 46,145 yards

Since 1 yard = 0.9144 meters, we can find the number of meters the same way we found yards:

$\frac{1 yard}{0.9144 meters} =\frac{46145}{m}$ or m=50,465 meters

Since the question is asking for kilometers, we can need to divide the number of meters by 1000, or move the decimal point three places to the left:

50,465 ÷ 1000 = 50.465 km

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3 years ago

Probabilities with possible states of nature: s1, s2, and s3. Suppose that you are given a decision situation with three possibl

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Answer:

1. $P(s_1|I)=\frac{1}{11}$

2. $P(s_2|I)=\frac{8}{11}$

3. $P(s_3|I)=\frac{2}{11}$

Step-by-step explanation:

Given information:

$P(s_1)=0.1, P(s_2)=0.6, P(s_3)=0.3$

$P(I|s_1)=0.15,P(I|s_2)=0.2,P(I|s_3)=0.1$

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We need to find the value of P(s₁|I).

$P(s_1|I)=\frac{P(I|s_1)P(s_1)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

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$P(s_1|I)=\frac{0.015}{0.165}$

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Therefore the value of P(s₁|I) is $\frac{1}{11}$ .

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We need to find the value of P(s₂|I).

$P(s_2|I)=\frac{P(I|s_2)P(s_2)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_2|I)=\frac{(0.2)(0.6)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_2|I)=\frac{0.12}{0.015+0.12+0.03}$

$P(s_2|I)=\frac{0.12}{0.165}$

$P(s_2|I)=\frac{8}{11}$

Therefore the value of P(s₂|I) is $\frac{8}{11}$ .

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We need to find the value of P(s₃|I).

$P(s_3|I)=\frac{P(I|s_3)P(s_3)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_3|I)=\frac{(0.1)(0.3)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

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$P(s_3|I)=\frac{0.03}{0.165}$

$P(s_3|I)=\frac{2}{11}$

Therefore the value of P(s₃|I) is $\frac{2}{11}$ .

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