(m+n)¹⁰; The 6th term will contain m?n?

Suppose that an airline quotes a flight time of 128 minutes between two cities. Furthermore, suppose that historical flight reco

ANTONII [103]

Answer:

(a) The probability density function of X is:

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

(b) The value of P (129 ≤ X ≤ 146) is 0.3462.

(c) The probability that a randomly selected flight between the two cities will be at least 3 minutes late is 0.4327.

Step-by-step explanation:

The random variable X is defined as the flight time between the two cities.

Since the random variable X denotes time interval, the random variable X is continuous.

(a)

The random variable X is Uniformly distributed with parameters a = 10 minutes and b = 154 minutes.

The probability density function of X is:

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

(b)

Compute the value of P (129 ≤ X ≤ 146) as follows:

Apply continuity correction:

P (129 ≤ X ≤ 146) = P (129 - 0.50 < X < 146 + 0.50)

= P (128.50 < X < 146.50)

$=\int\limits^{146.50}_{128.50} {\frac{1}{154-102}} \, dx$

$=\frac{1}{52}\times \int\limits^{146.50}_{128.50} {1} \, dx$

$=\frac{1}{52}\times (146.50-128.50)$

$=0.3462$

Thus, the value of P (129 ≤ X ≤ 146) is 0.3462.

(c)

It is provided that a randomly selected flight between the two cities will be at least 3 minutes late, i.e. X ≥ 128 + 3 = 131.

Compute the value of P (X ≥ 131) as follows:

Apply continuity correction:

P (X ≥ 131) = P (X > 131 + 0.50)

= P (X > 131.50)

$=\int\limits^{154}_{131.50} {\frac{1}{154-102}} \, dx$

$=\frac{1}{52}\times \int\limits^{154}_{131.50} {1} \, dx$

$=\frac{1}{52}\times (154-131.50)$

$=0.4327$

Thus, the probability that a randomly selected flight between the two cities will be at least 3 minutes late is 0.4327.

6 0

3 years ago

1/2×4/5 can be written

Leno4ka [110]

I think it will be because if you multiply 1/2 times 4/5 it will be 0.4

6 0

3 years ago

Easy :) solve any ones you want

liberstina [14]

Answer:

3. D, Additive Property of Equality

4. x < 2

5. attached graph

Step-by-step explanation:

3. Additive Property of Equality: this means that you can add/subtract a number from one side, as long as you can do it to the other. as shown in steps 4 and 5, this is what they have done in order to get x by itself.

4. to solve -3x + 8 > 2x - 2, we want to get x by itself on one side. we can treat the inequality sign as an equal sign to solve

-3x + 8 > 2x - 2 < -- subtract 8 from both sides to get -3x alone

-8 -8

-3x > 2x - 10 < -- subtract 2x from both sides to get 2x with the -3x

-2x -2x

-5x > -10 <-- divide both sides by -5 to get x alone

when we divide by a negative number in an inequality, we flip the sign. so > becomes < once we divide by -5. this is important to ensure we get the right answer

-5x/-5 = x

-10/-5 = 2

x < 2

this inequality means that any value for x has to be less than 2 in order for the inequality to be true

on a number line it would look like the following:

<---------------2) - - - - - - - - >

the numbers to the left of 2 would be shaded in, as those are less than 2. the number 2 would have either an open circle on top of it or a parentheses facing to the left, either one works.

5. to graph y < 2/5x - 4, we want to get this equation into slope intercept form which is y = mx + b, where mx is the slope and b is the y-intercept

again, treat the inequality symbol as an = sign in this case and solve to get y by itself

y = 2/5x - 4 <--- this is already in slope intercept form, so we do not need to solve for y in this case.

to graph this, we would need to start at the y-intercept which is (0, -4). i will attach a graph to make this a bit easier visually

from there, you would use the slope to find another point on the graph. the slope is rise/run which also means y/x.

so you would need to go 2 units up from (0, -4), and 5 units right from where you landed

since this is an inequality, we will need to find any answers on the graph that satisfy the inequality by shading in the region where those points are solutions.

to do this we need to find a test point. a test point tells us whether or not that point and anything in its section is included in the solution. if the inequality is true with the test point, that answer and anything in that section is included in the answer. if its false, that point and everything in that section is not included.

you can always use the origin (0,0) if its not graphed on the line. in this case, we will use the origin as a test point. so plug in the test point into the original inequality

y < 2/5x - 4

0 < 2/5(0) - 4

0 < -4 <-- this is false and (0,0) and anything in its section is not included as a solution, so you would shade below the line as (0,0) was above the line, and anything above the line cannot be used as a solution to the inequality

graph the line connecting the two points using a dotted line as the inequality symbol is <, meaning points on that line are not solutions to the inequality

with the graph i attached, shade in the entire dark blue area. the graph is a bit rough but your graph should look similar to that

4 0

2 years ago

How do I find the critical points?

podryga [215]

Critical points is where the derivative (slope) is zero or does not exist. So to do this we have to find the derivative of our function:

$\frac{d}{dx}(x^{2} - 1)^{3}$

So we apply chain rule:

= $3(x^{2} - 1)^{2} * 2x$

Set our first derivative to zero and solve for x:

3(x^2 - 1) * 2x = 0

So we can see that (by plugging in) 0, -1 and 1 makes our solution true

So our critical value is x = 0, x = -1, x = 1

5 0

3 years ago

A statement of an employee’s biweekly earnings is given below. What is the employee’s gross pay?

Olin [163]

The employee's gross pay
$726.80

3 0

3 years ago

(m+n)¹⁰; The 6th term will contain m?n?​

(m+n)¹⁰; The 6th term will contain m?n?