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WARRIOR [948]
3 years ago
6

Let x be the amount of time (in minutes) that a particular San Francisco commuter must wait for a BART train. Suppose that the d

ensity curve is as pictured below (a uniform distribution): A horizontal line segment is graphed on the coordinate plane. The horizontal x axis is labeled "Minutes" and has two tick marks at 0 and 20. The vertical axis is labeled "Density" and has one tick mark at 0.05. The line enters the viewing window at (0, 0.05) and stops at (20, 0.05). (a) What is the probability that x is less than 8 min? more than 14 min? P (x is less than 8 minutes) = P (x is more than 14 minutes) = (b) What is the probability that x is between 7 and 11 min? P (x is between 7 and 11 minutes) = (c) Find the value c for which P(x < c) = .9. c = mins
Mathematics
1 answer:
larisa [96]3 years ago
3 0

Answer:

a) P(X

P(X>14) = 1-P(X

b) P(7< X

c) We want to find a value c who satisfy this condition:

P(x

And using the cumulative distribution function we have this:

P(x

And solving for c we got:

c = 20*0.9 = 18

Step-by-step explanation:

For this case we define the random variable X as he amount of time (in minutes) that a particular San Francisco commuter must wait for a BART train, and we know that the distribution for X is given by:

X \sim Unif (a=0, b =20)

Part a

We want this probability:

P(X

And for this case we can use the cumulative distribution function given by:

F(x) = \frac{x-a}{b-a} = \frac{x-0}{20-0}= \frac{x}{20}

And using the cumulative distribution function we got:

P(X

For the probability P(X>14) if we use the cumulative distribution function and the complement rule we got:

P(X>14) = 1-P(X

Part b

We want this probability:

P(7< X

And using the cdf we got:

P(7< X

Part c

We want to find a value c who satisfy this condition:

P(x

And using the cumulative distribution function we have this:

P(x

And solving for c we got:

c = 20*0.9 = 18

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an airplane lands at an airport 80 miles west and 30 miles north of where it took off. How far apart are the airports.
MArishka [77]

Answer:

85.44 miles apart

Step-by-step explanation:

According to the question;

  • The airplane flew 80 miles west and 30 miles north

We are supposed to determine how far apart are the airports.

  • The shortest distance will be the displacement from the initial point.
  • We are going to use the Pythagoras theorem;
  • a² + b² = c² , where a and b are the legs of a right angled triangle and c is the hypotenuse.
  • Therefore, in this case; our legs will be 30 miles and 80 miles, so we need to determine the hypotenuse;

That is;

c² = 80² + 30²

    = 6400 + 900

    = 7300

c = √7300

  = 85.44

Thus, the airports are 85.44 miles apart

   

7 0
3 years ago
Let u = &lt;9,4&gt;, v = &lt;-2,5&gt; Find u + v.
love history [14]

Answer:

<u><em>u +v = < 7,9 ></em></u>

Step-by-step explanation:

You are given u = <9,4> and v = <-2,5>. You are asked to find u+v. all you need to do is to add them with their respective position. u+v = <9-2, 4+5>, u+v = <7,9>. This is the correct answer.

3 0
2 years ago
-. Find the value of x. Round to the nearest degree.
Mice21 [21]
X is 44.4 degrees and round it off to the nearest degree is 44
7 0
2 years ago
Triangle Congruence:
Serga [27]
Where are the triangles so I can compare to find the triangle congruence?
8 0
3 years ago
Read 2 more answers
P(x)=(x-2)(x-3)(x+7) find the constant term
hram777 [196]
Short method: multiply constants
= ( - 2) \times ( - 3) \times (7) \\  = 6 \times 7 \\  = 42

Long mothod: expand it
= ( {x}^{2} - 5x + 6)(x + 7) \\  = ... \\  =  ... + 42

Answer: 42
5 0
3 years ago
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