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Anna35 [415]
3 years ago
7

f the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular path were reduced, w

hat would happen?
Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
8 0
The frictional force between the tires and the road prevent the car from skidding off the road due to centripetal force.

If the frictional force is less than the centripetal force, the car will skid when it navigates a circular path.

The diagram below shows that when the car travels at tangential velocity, v, on a circular path with radius, r, the centripetal acceleration of v²/ r acts toward the center of the circle.

The resultant centripetal force is (mv²)/r, which should be balanced by the frictional force of μmg, where μ =  coefficient of kinetic friction., and mg is the normal reaction on a car with mass, m.

This principle is applied on racing tracks, where the road is inclined away from the circle to give the car an extra restoring force  to overcome the centripetal force.
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Suppose that a local TV station conducts a survey of a random sample of 120 registered voters in order to predict the winner of
forsale [732]

Answer:

a) The 99% CI for the true proportion of voters who prefer the Republican candidate is (0.3658, 0.6001). This means that we are 99% sure that the true population proportion of all voters who prefer the Republican candidate is (0.3658, 0.6001).

b) The upper bound of the confidence interval is above 0.5 = 50%, which meas that the candidate can be confidence of victory.

Step-by-step explanation:

Question a:

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 z-score that has a p-value of 1 - \frac{\alpha}{2}.

Sample of 120 registered voters in order to predict the winner of a local election. The Democrat candidate was favored by 62 of the respondents.

So 120 - 62 = 58 favored the Republican candidate, so:

n = 120, \pi = \frac{58}{120} = 0.4833

99% confidence level

So \alpha = 0.01, z is the value of Z that has a p-value of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.  

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4833 - 2.575\sqrt{\frac{0.4833*0.5167}{120}} = 0.3658

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4833 + 2.575\sqrt{\frac{0.4833*0.5167}{120}} = 0.6001

The 99% CI for the true proportion of voters who prefer the Republican candidate is (0.3658, 0.6001). This means that we are 99% sure that the true population proportion of all voters who prefer the Republican candidate is (0.3658, 0.6001).

b. If a candidate needs a simple majority of the votes to win the election, can the Republican candidate be confident of victory? Justify your response with an appropriate statistical argument.

The upper bound of the confidence interval is above 0.5 = 50%, which meas that the candidate can be confidence of victory.

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3 years ago
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