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Olin [163]
3 years ago
10

A man drove 11 mi directly east from his home, made a left turn at an intersection, and then traveled 9 mi north to his place of

work. If a road was made directly from his home to his place of work, what would its distance be to the nearest tenth of a mile?

Mathematics
1 answer:
ludmilkaskok [199]3 years ago
3 0

Answer: 14 miles

Step-by-step explanation:

In the attached figure we can see the two segments the man drove, where the d is the distance of a road that goes from the man's home directly to his work.

It should be noted that dis also the hypotenuse of the righ triangle formed. Hence, we can use the Pithagorean theorem to find this distance:

d=\sqrt{(11 mi)^{2} + (9 mi)^{2}}

Then:

d=14.21 mi \approx 14 mi

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3 years ago
In a certain population of mussels (Mytilus edulis), 80% of the individuals are infected with an intestinal parasite. A marine b
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Answer:

The probability that 85% or more of the sampled mussels will be infected is 0.1057.

Step-by-step explanation:

Let <em>X</em> = number of mussels infected with an intestinal parasite.

The probability that a random selected mussel is infected with an intestinal parasite is, <em>p</em> = 0.80.

A random sample of <em>n</em> = 100 mussels from the population are examined by a marine biologist.

The random variable <em>X</em> follows a Binomial distribution with parameters n = 100 and p = 0.80.

But the sample selected is too large, i.e. <em>n</em> = 100 > 30.

So a Normal approximation to binomial can be applied to approximate the distribution of \hat p<em>, </em>the sample proportion of mussels infected with an intestinal parasite, if the following conditions are satisfied:

  1. np ≥ 10
  2. n(1 - p) ≥ 10

Check the conditions as follows:

n × p = 100 × 0.80 = 80 > 10

n × (1 - p) = 100 × (1 - 0.80) = 20 > 10

Thus, a Normal approximation to binomial can be applied.

So,  the distribution of \hat p is:

\hat p\sim N(p, \frac{p(1-p)}{n}  )

Compute the probability that 85% or more of the sampled mussels will be infected as follows:

Apply continuity correction:

P(\hat p\geq 0.85)=P(\hat p>0.85+0.50)

                   =P(\hat p>0.90)\\

                   =P(\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}>\frac{0.85-0.80}{\sqrt{\frac{0.80(1-0.80)}{100}}})

                   =P(Z>1.25)\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that 85% or more of the sampled mussels will be infected is 0.1057.

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