24. If one person will be doing all of the driving, __________ hours driving a day should be considered the limit.

Advanced Placement (AP)

2 answers:

g100num [7]3 years ago

7 0

Twelve hours a day

Yay

marissa [1.9K]3 years ago

6 0

This question is incomplete because it is missing the options, here is the complete question

If one person will be doing all of the driving, __________ hours driving a day should be considered the limit.

a. two to three

b. four or five

c. six to eight

d. ten

According to this, the correct answer is C. Six to eight hours

Explanation:

Driving requires complete concentration and attention, besides driving implies staying in only one position, while still moving. Due to this, as you drive the physical and mental effort causes you feel tired and therefore, your concentration and precision reduce. Because of this, it is recommended to drive only six to eight hours a day if only one person will be driving, because this is the amount of time a person can do the same activity without feeling excessively tired. This is important because in this way you avoid accidents that can be a result of the driver being too tired, and not paying enough attention while driving.

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If the wait time is less than 30 minutes for 75% of all patients in the emergency room, what is the probability that the

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The probability that the proportion of patients who wait less than 30 minutes is 0.582 or less is 0.0020

<h3>What is probability? </h3>

Probability can be defined as the likelihood of an event to occur. In statistics, the mean of the sample distribution typically shows the probability of the population.

From the parameters given:

The sample size (n) = 55 patients
Let's assume that the mean (x) = 32 (i.e. 58.2%) of the patients

The sample proportion $\mathbf{\hat p}$ can be computed by using the expression:

$\mathbf{\hat p = \dfrac{x}{n}}$

$\mathbf{\hat p = \dfrac{32}{55}}$

$\mathbf{\hat p =0.582}$

If the percentage of the probability of all patients in the emergency room = 0.75

The probability that the proportion of patients who wait less than 30 minutes is 0.582 or less can be computed as:

$\mathbf{( \hat P \leq 0.582) = P \Big( \dfrac{\hat p - p}{\sqrt{\dfrac{p(1-p)}{n}}} \leq \dfrac{0.582 - 0.75}{\sqrt{\dfrac{0.75(1-0.75)}{55}}} \Big )}$