The number of minutes that it takes students to fill out an online survey has an approximately normal distribution with mean 11 minutes and standard deviation 2.5 minutes.
a. What percent of students take more than 12 minutes to fill out the survey?
b. What percent of student take between 9 and 14 minutes to fill out the survey?
c. 75% of students fill the survey in less than how many minutes?
d. 80% of students will be within how many standard deviations of the mean?

Answers

Answer 1

Given: The number of minutes that it takes students to fill out an online survey has an approximately normal distribution with mean 11 minutes and standard deviation 2.5 minutes.

a. About 34.46% of students take more than 12 minutes to fill out the survey.

b. About 17.3% of students take between 9 and 14 minutes to fill out the survey.

c. 75% of students fill out the survey in less than 12.675 minutes.

d. 80% of students will be within 1.28 standard deviations of the mean.

a. In this problem, we have μ=11 and σ=2.5.

We need to find out the percent of students who take more than 12 minutes to fill out the survey.

Using z-score formula, we get

z=(x−μ)/σ

=(12−11)/2.5

=0.4

Now we can use a standard normal distribution table to find the percentage of students taking more than 12 minutes. Looking up the z-score of 0.4, we get the probability of 0.3446 or 34.46% approximately.

Therefore, about 34.46% of students take more than 12 minutes to fill out the survey.

b. Now we need to find out the percentage of students who take between 9 and 14 minutes to fill out the survey.

Using z-score formula for the lower and upper limits, we get

z_(lower)=(9−11)/2.5

=−0.8

z_(upper)=(14−11)/2.5

=1.2

Now we can use a standard normal distribution table to find the percentage of students taking between 9 and 14 minutes. Looking up the z-score of -0.8 and 1.2, we get the probabilities of 0.2119 and 0.3849 respectively.

The difference between these probabilities gives us the answer:0.3849−0.2119=0.173.

Therefore, about 17.3% of students take between 9 and 14 minutes to fill out the survey.

c. Now we need to find out the time taken by 75% of students to fill out the survey.

Using a standard normal distribution table, we can find the z-score that corresponds to the probability of 0.75.

This is approximately 0.67. Using the z-score formula, we can find out the time taken by 75% of students.

z=0.67

=(x−11)/2.5

Solving for x, we get x=12.675.

Therefore, 75% of students fill out the survey in less than 12.675 minutes.

d. Finally, we need to find out how many standard deviations away from the mean do we have to go to capture 80% of the students.

Using a standard normal distribution table, we can find the z-score that corresponds to the probability of 0.9. This is approximately 1.28.

Using the z-score formula, we can find out the deviation from the mean that corresponds to this z-score.

1.28=(x−11)/2.5

Solving for x, we get x=14.2.

Therefore, the deviation from the mean is 14.2−11=3.2 minutes.

Since 80% of the students lie within this deviation, we can say that 80% of students will be within 3.2/2.5=1.28 standard deviations of the mean.

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Ik this is late, but people who are looking for the answer can see this.

Answer:

an =1/3*6^n-1

Step-by-step explanation:

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Both antibiotics were prescribed in high dosage slow release capsules. The function C(x)=5log(x+1)+10 models the concentration of levofloxacin in mol/L over a time x, in hours. The function D(x) = 10log(x+1)+5 models the concentration of metronidazole in mol/L over a time x in hours. a) Which of the two drugs has a higher initial concentration in the blood stream? Justify your answer with an explanation. (2K, A) b) Determine when C(x) = D(x) algebraically and state what this represents in this situation. (6K, A, T) c) If Mathews is instructed to take both antibiotics at once the concentration levels could be modeled by the function(C+D)(x). How would the graph of (C+D)(x) differ from the individual graphs of C(x) and D(x)? Explain. (2T)

Answers

a) The initial concentration of levofloxacin (represented by C(x)) in the bloodstream is higher than the initial concentration of metronidazole (represented by D(x))

b) There is no algebraic solution for when C(x) = D(x) in this situation.

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Please help me will mark brainliet

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Answer:

The degree of the term is 5

Don't forget to mark me as brainliest

Plz help I’ll give brainliest

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1. y= -5

2. m=0

3. I dont know

1. 3x+4y=-20

x=0

3×0+4y=-20

0+4y=-20

y=-5

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m= 0

3. Sorry

Bernard works for a business that sells and repairs tires. He can repair tires in an hour work day. What is Bernard's unit rate for the number of tires repaired in one hour?

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Answer:

3 tires per hour

Step-by-step explanation:

Bernard works for a business that sells and repairs tires. He can repair 24 tires in an 8 hour work day. What is Bernard's unit rate for the number of tires repaired in one hour?

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