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Novay_Z [31]
3 years ago
10

If the maximum acceleration that is tolerable for passengers in a subway train is 1.74 m/s2 and subway stations are located 930

m apart, what is the maximum speed a subway train can attain between stations?
Mathematics
1 answer:
Free_Kalibri [48]3 years ago
6 0
Given that the subway stations are 930 m apart, the train have to be accerelated for half the distance and then decerelated for the rest of the distance.

Recall that the distance travelled by an object with an initial velocity, u, for a period of time, t, at an accereration, a, is given by
s=ut+ \frac{1}{2} at^2

But, we assume that the train accelerates from rest, thus
s=\frac{1}{2} at^2 \\  \\ \Rightarrow465=\frac{1}{2}(1.74)t^2 \\  \\ \Rightarrow t^2=534.48 \\  \\  \Rightarrow t=\sqrt{534.48}=23.12

The maximum speed is attained at half the center of the distance between subway stations (i.e. at distance = 465 m).

Thus, maximum speed = distance / time = 465 / 23.12 = 20.11 m/s.
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The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first deter
JulsSmile [24]

Answer:

1) n=114

2) n=59

3) On this case no, because if we survey just the adults of the nearest college that would be a convenience sample. And when we use "convenience sample" we have some problems associated to bias. This methodology it's not appropiate in order to have a good estimation of the parameter of interest. It's better use a random, cluster or stratified sampling.

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

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

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 80% of confidence, our significance level would be given by \alpha=1-0.80=0.2 and \alpha/2 =0.1. And the critical value would be given by:

z_{\alpha/2}=-1.28, z_{1-\alpha/2}=1.28

Part 1

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.06 or 6% points, and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

Since we don't have a prior estimate of \het p we can use 0.5 as a good estimate, replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.06}{1.28})^2}=113.77  

And rounded up we have that n=114

Part 2

On this case we have a prior estimate for the population proportion and is \hat p =0.85 so replacing the values into equation (b) we got:

n=\frac{0.85(1-0.85)}{(\frac{0.06}{1.28})^2}=58.027

And rounded up we have that n=59

Part 3

On this case no, because if we survey just the adults of the nearest college that would be a convenience sample. And when we use "convenience sample" we have some problems associated to bias. This methodology it's not appropiate in order to have a good estimation of the parameter of interest. It's better use a random, cluster or stratified sampling.

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A 20-gallon container is filled halfway with a mixture that is $90\%$ vinegar and $10\%$ water. How many gallons of water must b
Yuki888 [10]

Answer:

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10 gallons to start with

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amount of vinegar before and after water addition is constant

.9(10) = .6(10+x)  

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Step-by-step explanation:

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The divide.

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