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klemol [59]
3 years ago
11

nikita ran a 5-kilometer race in 39 minutes (0.65 of an hour) without training beforehand. in the first part of the race, her av

erage speed was 8.75 kilometers per hour. for the second part of the race, she started to get tired, so her average speed dropped to 6 kilometers per hour. which expression represents nikita’s distance for the second part of the race?
Mathematics
1 answer:
Nataliya [291]3 years ago
5 0
Total distance 5 km; at 5km / 0.65 h =

Second part distance: x; at 6 km/h, during t2

First part distance: 5 - x; at 8.75 km/h, during t1

V = d/t ⇒ t = d/V

t2 = x/6
t1=[5-x]/8.75

t2 + t1 = 0.65

x/6 + [5-x]/8.75 = 0.65

x/6 + 5/8.75 - x/8.75 = 13/20

 x/6 - x/8.75 = 13/20 - 5/8.75

x/6 - 4x/35 =13/20 - 20/35

35x - 24x = (35*6)(35*13 - 20*20)/(20*35)

11 x = 16.5
x = 16.5/11 = 1.5 km












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A programmer plans to develop a new software system. In planning for the operating system that he will​ use, he needs to estimat
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Using the z-distribution, we have that:

a) A sample of 601 is needed.

b) A sample of 93 is needed.

c) A.  ​Yes, using the additional survey information from part​ (b) dramatically reduces the sample size.

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which z is the z-score that has a p-value of \frac{1+\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so z = 1.96.

For this problem, we consider that we want it to be within 4%.

Item a:

  • The sample size is <u>n for which M = 0.04.</u>
  • There is no estimate, hence \pi = 0.5

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}

0.04\sqrt{n} = 1.96\sqrt{0.5(0.5)}

\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.04}

(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.04}\right)^2

n = 600.25

Rounding up:

A sample of 601 is needed.

Item b:

The estimate is \pi = 0.96, hence:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.96(0.04)}{n}}

0.04\sqrt{n} = 1.96\sqrt{0.96(0.04)}

\sqrt{n} = \frac{1.96\sqrt{0.96(0.04)}}{0.04}

(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.96(0.04)}}{0.04}\right)^2

n = 92.2

Rounding up:

A sample of 93 is needed.

Item c:

The closer the estimate is to \pi = 0.5, the larger the sample size needed, hence, the correct option is A.

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