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laiz [17]
3 years ago
9

Amir ran the first 200-meter leg of the relay race with a speed of 8,260 millimeters per second. Ryder ran the last 200-meter le

g with an average speed of 930 centimeters per second. Who ran the race faster, and how much faster was that person’s speed to the nearest meter per second?
A. Amir ran approximately 1 meter per second faster than Ryder.
B. Ryder ran approximately 1 meter per second faster than Amir.
C. Amir ran approximately 104 meters per second faster than Ryder.
D. Ryder ran approximately 104 meters per second faster than Amir.
Mathematics
1 answer:
Maurinko [17]3 years ago
5 0

Answer: Amir is faster than Ryder by speed of 104 meter per second.



Step-by-step explanation:


Speed of Amir=8,260 mm per second

We know that 1m =1000mm

Convert into meter per second, divide speed by 1000

Speed of Amir

Speed of Ryder=930 centimeters per second

We know that 1m =100 cm

Convert into meter per second, divide speed by 100

Speed of Ryder

As distance covered by both of then is same in relay race thus we can compare their speeds

We can see 8.26<9.3 and difference=9.3-8.26=1.04

Amir is faster than Ryder by speed of 1.04 meter per second.

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Answer:

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error:

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

For this problem, we have that:

p = 0.1

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is

We need a sample size of at least n, in which n is found M = 0.04.

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

0.04 = 1.96\sqrt{\frac{0.1*0.9}{n}}

0.04\sqrt{n} = 0.588

\sqrt{n} = \frac{0.588}{0.04}

\sqrt{n} = 14.7

(\sqrt{n})^{2} = (14.7)^{2}

n = 216

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.

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Answer:

C).

Step-by-step explanation:

-3x+2 and[(-)(x^2+5x)

3x+2 and (-x^2+5x)

3x+2-x^2+5x

-x^2-8x+2

Hope this Helps :)

7 0
3 years ago
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