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rjkz [21]
3 years ago
8

A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 49.0 and 54.

0 minutes. Find the probability that a given class period runs between 50.75 and 51.25 minutes. Find the probability of selecting a class that runs between 50.75 and 51.25 minutes.
Mathematics
1 answer:
svlad2 [7]3 years ago
3 0

Answer:

The probability that a given class period runs between 50.75 and 51.25 minutes is 0.10.

Step-by-step explanation:

Let the random variable <em>X</em> represent the lengths of the classes.

The random variable <em>X</em> is uniformly distributed within the interval 49.0 and 54.0 minutes.

The probability density function of <em>X</em> is:

f_{X}(x)=\frac{1}{b-a};\ a

Compute the probability that a given class period runs between 50.75 and 51.25 minutes as follows:

P(50.75

Thus, the probability that a given class period runs between 50.75 and 51.25 minutes is 0.10.

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PLEASE HELPPPPPP. I NEED IT DONEEEEE. ​
agasfer [191]

<u><em>Answer:</em></u>

1. Couples tickets: $8

   Individual tickets: $4

2. Song download cost: $2

    Movie download cost: $15

<u><em>Step-by-step explanation:</em></u>

You can solve both of these questions by doing system of equations.

Let's start with question 1.

There are 2 variables in this equation:

- Couples tickets price: let's represent these with the letter C

- Individual tickets price: let's represent these with the letter K

Writing the two equations would look this these:

5c + 2k = 48

3c + 2k = 32

Next, isolate a variable. I am going to isolate K in the first equation.

2k = -5c + 48 \\k = -\frac{5}{2}c + 24

Now that we isolated a variable we can plug that back in to the second equation:

3c + 2(-\frac{5}{2}c + 24) = 32\\3c - \frac{10}{2}c + 48 = 32\\3c -5c + 48 = 32\\-2c + 48 = 32\\-2c = -16\\c = 8

We found that c is equal to 8 so we can put that back in to an equation to solve for k.

5(8) + 2k = 48

40 + 2k = 48

2k = 8

k = 4

Therefore, the price for couples tickets is $8 and the price for individual tickets is $4.

<u><em>Check #1:</em></u>

5(8) + 2(4) = 48

40 + 8 = 48

48 = 48

3(8) + 2(4) = 32

24 + 8 = 32

32 + 32

Now, let's go on to question 2.

There are 2 variables in this equation:

- Price of songs downloaded: let's represent these with S

- Price of movies downloaded: let's represent these with M

Writing the two equations would look like this:

15s + 11m = 195

15s + 8m = 150

There is a simple way to answer this system, however.

If you change the bottom equations signs to negative you can minus the second equation from the first equation like this:

15s + 11m = 195

-(15s +8m = 150)

Minus them to get this equation:

3m = 45

Solve

m = 15

We have found that each movie download costs $15, now let's plug this back into an equation:

15s + 11(15) = 195

15s + 165 = 195

15s = 30

s = 2

Each song costs $2 to download.

<em><u>Check #2:</u></em>

15(2) + 11(15) = 195

30 + 165 = 195

195 = 195

15(2) + 8(15) = 150

30 + 120 = 150

150 = 150

<em>I hope this helps!!</em>

<em>- Kay :)</em>

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Answer:

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