Can you please help with ratios and rates

Mathematics

1 answer:

lisabon 2012 [21]3 years ago

3 0

Hey it's me my account messed up so I have to use this one, btw Im the one that helped you with your science (aka firstnamelastname

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a survey amony freshman at a certain university revealed that the number of hours spent studying the week before final exams was

Marat540 [252]

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

Let $\bar X$ = sample average time spent studying

The z-score probability distribution for sample mean is given by;

Z = $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ ~ N(0,1)

where, $\mu$ = population mean hours spent studying = 25 hours

$\sigma$ = standard deviation = 15 hours

n = sample of students = 36

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < $\bar X$ < 30 hours)

P(29 hours < $\bar X$ < 30 hours) = P( $\bar X$ < 30 hours) - P( $\bar X$ $\leq$ 29 hours)

P( $\bar X$ < 30 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ < $\frac{ 30-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z < 2) = 0.97725

P( $\bar X$ $\leq$ 29 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ $\leq$ $\frac{ 29-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z $\leq$ 1.60) = 0.94520

So, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.

Therefore, P(29 hours < $\bar X$ < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

7 0

2 years ago

An object in a simulation accelerates

notsponge [240]

Observe the graph below. This graph represents the scenario.

The question is ill formated, the complete question is

In a simulation, a moving object accelerates from rest to 4 meters per second in 2 seconds. For the following three seconds, it increases linearly until it reaches a speed of 10 meters per second. Following three seconds at that speed (acceleration = 0), the item progressively decelerates until it comes to rest two seconds later. Draw the graph of this scenario for 10 seconds?

I'll describe how the graph may show.

It will move diagonally upward from time 0 to 2 seconds until it reaches the y axis at a speed of 4 m/s.

Then, from 2 to 5, the position will move up diagonally until it reaches the y axis at a speed of 10 m/s.

The next 5 to 8 seconds will be horizontal.

After that, it will descend diagonally.

Observe the graph below. This graph represents the scenario.

Learn more about Acceleration here-

brainly.com/question/21509870

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