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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Answer:

No, they are not equivalent.

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Answer:

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j+j+j+j

remember multiplication is repeated addition!

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Answers

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:

You should plug in 2 for n in each equation, which, the correct one should give you 2. Then, to make sure, you plug in 3 in the equations that gave you 2.

Example:

an = 1/3 * 6 ^n-1

Plug in 2.

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an = 1/3 * 6 ^1

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Step 2.

an = 1/3 * 6 ^3-1

an = 1/3 * 6 ^2

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Hope this helps.

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Answer:

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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.

c) The graph of (C+D)(x) would show a steeper increase and higher values compared to the individual graphs of C(x) and D(x), indicating a higher concentration of the drugs when taken together.

a) The initial concentration of a drug in the bloodstream can be determined by evaluating the functions C(x) and D(x) at x = 0. Plugging in x = 0 into both functions, we find that C(0) = 5log(0 + 1) + 10 = 5log(1) + 10 = 5(0) + 10 = 10, and D(0) = 10log(0 + 1) + 5 = 10log(1) + 5 = 10(0) + 5 = 5. Therefore, 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) To find when C(x) = D(x), we set the two functions equal to each other and solve for x: 5log(x + 1) + 10 = 10log(x + 1) + 5. Subtracting 10log(x + 1) from both sides, we get 5log(x + 1) - 10log(x + 1) = 5. Simplifying, we have -5log(x + 1) = 5. Dividing by -5, we find log(x + 1) = -1. This equation can be rewritten in exponential form as 10^(-1) = x + 1, which simplifies to 0.1 = x + 1. Subtracting 1 from both sides, we get x = -0.9. However, since time cannot be negative, there is no solution to the equation. Therefore, there is no algebraic solution for when C(x) = D(x) in this situation.

c) The graph of (C+D)(x) would represent the combined concentration levels of both antibiotics when taken together. It would differ from the individual graphs of C(x) and D(x) by showing a higher overall concentration. Since (C+D)(x) is the sum of the two individual functions, the resulting graph would be the sum of their concentrations at each point in time. This means that at any given time, the concentration of the combined antibiotics would be the sum of the concentrations of levofloxacin and metronidazole at that time. The graph of (C+D)(x) would show a steeper increase and higher values compared to the individual graphs of C(x) and D(x), indicating a higher concentration of the drugs when taken together.

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If the family plans to move to their retirement home in 10 years, which pool would be more cost effective to install? How do

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Answer:

Step-by-step explanation:

but is their any answer choices

The first day, Brian saw 31 birds and twice as many squirrels as birds. The second day, Brian saw 20 birds and 42 squirrels. Which of the following is a good estimation of how many birds and squirrels Brian saw in the two days?

Answers

Answer:  A good estimation would be about 150.

Step-by-step explanation:

To estimate we can round instead of using exact numbers.

First day he saw about 30 birds and twice as many squirrels, meaning about 60 squirrels.  This is about 90 total when added together.

Second day he saw 20 birds and about 40 squirrels so about 60 when added together.

Add day one and day two totals for...

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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

2. m=y/x

m=0/1

m= 0

3. Sorry

PLSSSSSSSSSSSSSS HELPPP FOR REAL SOMEBODDYY REWARD BRAINLIEST!!!!!!!!!!!!!!!!!

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Answer:

1. x = 3

2. (1, 0)

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Step-by-step explanation:

Please Please help me

Answers

Answer:

1. 30

Step-by-step explanation:

First, place the numbers from least to greatest.

11,13,23,37,45,58

Then, find the middle number (or two middle numbers).

In this case, the middle numbers are 23 and 37.

Third, we find the mean of these numbers.

23+37=60

60/2=30

So your answer is 30.

Repeat this method for the other problems.

4. A Tesla dealership is having a sale on their cars. This week you can save 15% off
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Answers

Answer:

25,661.5

Step-by-step explanation:

30160x15%=4528.50

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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?

Answers

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?

Number of tires repaired per day = 24 tires

Number of time used to repair tire per day = 8 hours

What is Bernard's unit rate for the number of tires repaired in one hour?

Number of tires repaired in one hour = Number of tires repaired per day / Number of time used to repair tire per day

= 24 tires / 8 hours

= 3 tires per hour

The unit rate for the number of tires repaired in one hour is 3 tires per hour

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