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Fudgin [204]
2 years ago
15

Given the speeds of each runner below, determine who runs the fastest.

Mathematics
2 answers:
malfutka [58]2 years ago
7 0

Answer:

brooke 14 feet in one sec

Step-by-step explanation:

stephanie 12 feet 1 sec

liz 8 feet in 1 sec

will approx 6 feet in 1 sec

brooke 14 feet in one sec

alexandr402 [8]2 years ago
5 0

Answer:

Brooke

Step-by-step explanation:

We can solve this by figuring out how far each person runs a second

Stephanie runs 12 feet a second so nothing needs to be done here

Liz runs 306 feet in 38 seconds. To solve for the distance ran in 1 second we can divide (306/38)= around 8 feet

Will runes 1 mile in 491 seconds. A quick search will tell use that 1 mile is equal to 5280 feet and 5280/491 = 10.75

Brooke runs 851 feet in 1 minute or 60 seconds. 851/60 =14.18

Brooke obviously runs the furthest in one second and therefore she's the fastest runner

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Answer:

(1) A Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

Step-by-step explanation:

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If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

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p₁ = 0.10

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Check the Normal approximation conditions for population 1, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.10=1

Thus, a Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

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Check the Normal approximation conditions for population 2, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3)

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n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

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