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3 years ago

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the probabilities of having 0,1,2,3, or 4 people waiting in line at a grocery store are shown in the probability distribution be

low. what is the expected number of people waiting in line, rounded to the nearest tenth?

1 answer:

makvit [3.9K]3 years ago

8 0

Answer:

1.4

Step-by-step explanation:

Given the probability distribution :

X _____ 0 ____1 ____ 2 ____ 3 ____4

P(x) _ 0.16 _ 0.41 _ 0.32 __ 0.09 __ 0.02

Expected number of people waiting in line :

E(X) = Σ(x * p(x))

E(X) = (0*0.16) + (1*0.41) + (2*0.32) + (3*0.09) + (4*0.02)

E(X) = 1.4

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HELP ASAP (Geometry)

Andrei [34K]

1) Parallel line: $y=-2x-3$

2) Rectangle

3) Perpendicular line: $y = 0.5x + 2.5$

4) x-coordinate: 2.7

5) Distance: $d=\sqrt{(4-3)^2+(7-1)^2}$

6) 3/8

7) Perimeter: 12.4 units

8) Area: 8 square units

9) Two slopes of triangle ABC are opposite reciprocals

10) Perpendicular line: $y-5=-4(x-(-1))$

Step-by-step explanation:

1)

The equation of a line is in the form

$y=mx+q$

where m is the slope and q is the y-intercept.

Two lines are parallel to each other if they have same slope m.

The line given in this problem is

$y=-2x+7$

So its slope is $m=-2$ . Therefore, the only line parallel to this one is the line which have the same slope, which is:

$y=-2x-3$

Since it also has $m=-2$

2)

We can verify that this is a rectangle by checking that the two diagonals are congruent. We have:

- First diagonal: $d_1 = \sqrt{(-3-(-1))^2+(4-(-2))^2}=\sqrt{(-2)^2+(6)^2}=6.32$

- Second diagonal: $d_2 = \sqrt{(1-(-5))^2+(0-2)^2}=\sqrt{6^2+(-2)^2}=6.32$

The diagonals are congruent, so this is a rectangle.

3)

Given points A (0,1) and B (-2,5), the slope of the line is:

$m=\frac{5-1}{-2-0}=-2$

The slope of a line perpendicular to AB is equal to the inverse reciprocal of the slope of AB, so:

$m'=\frac{1}{2}$

And using the slope-intercept for,

$y-y_0 = m(x-x_0)$

Using the point $(x_0,y_0)=(7,1)$ we find:

$y-1=\frac{1}{2}(x-7)$

And re-arranging,

$y-1 = \frac{1}{2}x-\frac{7}{2}\\y=\frac{1}{2}x-\frac{5}{2}\\y=0.5x-2.5$

4)

The endpoints of the segment are X(1,2) and Y(6,7).

We have to divide the sgment into 1/3 and 2/3 parts from X to Y, so for the x-coordinate we get:

$x' = x_0 + \frac{1}{3}(x_1 - x_0) = 1+\frac{1}{3}(6-1)=2.7$

5)

The distance between two points $A(x_A,y_A)$ and $B(x_B,y_B)$ is given by

$d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$

In this problem, the two points are

E(3,1)

F(4,7)

So the distance is given by

$d=\sqrt{(4-3)^2+(7-1)^2}$

6)

We have:

A(3,4)

B(11,3)

Point C divides the segment into two parts with 3:5 ratio.

The distance between the x-coordinates of A and B is 8 units: this means that the x-coordinate of C falls 3 units to the right of the x-coordinate of A and 5 units to the left of the x-coordinate of B, so overall, the x-coordinate of C falls at

$\frac{3}{3+5}=\frac{3}{8}$

of the distance between A and B.

7)

To find the perimeter, we have to calculate the length of each side:

$d_{EF}=\sqrt{(x_E-x_F)^2+(y_E-y_F)^2}=\sqrt{(-1-2)^2+(6-4)^2}=3.6$

$d_{FG}=\sqrt{(x_G-x_F)^2+(y_G-y_F)^2}=\sqrt{(-1-2)^2+(3-4)^2}=3.2$

$d_{GH}=\sqrt{(x_G-x_H)^2+(y_G-y_H)^2}=\sqrt{(-1-(-3))^2+(3-3)^2}=2$

$d_{EH}=\sqrt{(x_E-x_H)^2+(y_E-y_H)^2}=\sqrt{(-1-(-3))^2+(6-3)^2}=3.6$

So the perimeter is

p = 3.6 + 3.2 + 2 + 3.6 = 12.4

8)

The area of a triangle is

$A=\frac{1}{2}(base)(height)$

For this triangle,

Base = XW

Height = YZ

We calculate the length of the base and of the height:

$Base =XW=\sqrt{(x_X-x_W)^2+(y_X-y_W)^2}=\sqrt{(6-2)^2+(3-(-1))^2}=5.7$

$Height =YZ=\sqrt{(x_Y-x_Z)^2+(y_Y-y_Z)^2}=\sqrt{(7-5)^2+(0-2)^2}=2.8$

So the area is

$A=\frac{1}{2}(XW)(YZ)=\frac{1}{2}(5.7)(2.8)=8$

9)

A triangle is a right triangle when there is one right angle. This means that two sides of the triangle are perpendicular to each other: however, two lines are perpendicular when their slopes are opposite reciprocals. Therefore, this means that the true statement is

"Two slopes of triangle ABC are opposite reciprocals"

10)

The initial line is

$y=\frac{1}{4}x-6$

A line perpendicular to this one must have a slope which is the opposite reciprocal, so

$m'=-4$

Using the slope-intercept form,

$y-y_0 = m'(x-x_0)$

And using the point

$(x_0,y_0)=(-1,5)$

we find:

$y-5=-4(x-(-1))$

Learn more about parallel and perpendicular lines:

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8 0

3 years ago

How many times greater is 46 than .46

anyanavicka [17]

100 times bigger .46x100=46

3 0

3 years ago

Read 2 more answers

Solve the following, round to 2 decimal places. <br>36.67 ÷ 12

Lubov Fominskaja [6]

So 36.67\12= 3.055833 and the 3 repeats so round up the 5 to get 3.06
Hope I’m correct!

5 0

3 years ago

Read 2 more answers

How many solutions exist for the system of equations graphed below?

Anestetic [448]

Answer:

One

General Formulas and Concepts:

<u>Algebra I</u>

Reading a coordinate plane
Solving systems of equations by graphing

Step-by-step explanation:

If 2 lines are parallel, they will have no solution.

If 2 lines are the same, they will have infinite amount of solutions.

We see from the graph that the 2 lines intersect at one point, near x = 1.5.

∴ our systems has 1 solution.

4 0

3 years ago

Read 2 more answers

Let $f(x) = 2x^2 + 3x - 9,$ $g(x) = 5x + 11,$ and $h(x) = -3x^2 + 1.$ Find $f(x) - g(x) + h(x).$

Viefleur [7K]

QUESTION 1

Given that:

$f(x)=2x^2+3x-9$ ,

$g(x)=5x+11$ ,

and

$h(x)=-3x^2+1$

Then;

$f(x)-g(x)+h(x)=2x^2+3x-9-(5x+11)+(-3x^2+1)$

$f(x)-g(x)+h(x)=2x^2+3x-9-5x-11-3x^2+1$

Group similar terms;

$f(x)-g(x)+h(x)=2x^2-3x^2+3x-5x-11-9+1$

Simplify;

$f(x)-g(x)+h(x)=-x^2-2x-19$

QUESTION 2

Given that;

$f(x)=4x-7$ .

$g(x)=(x+1)^2$

and

$s(x)=f(x)+g(x)$

Substitute the functions;

$s(x)=4x-7+(x+1)^2$

Substitute x=3

$s(3)=4(3)-7+(3+1)^2$

$s(3)=12-7+(4)^2$

$s(3)=5+16$

$s(3)=21$

QUESTION 3

Given:

$f(x)=3x+2$

$g(x)=x^2-5x-1$

$f(g(x))=f(x^2-5x-1)$

This implies that;

$f(g(x))=3(x^2-5x-1)+2$

Expand the parenthesis;

$f(g(x))=3x^2-15x-3+2$

$f(g(x))=3x^2-15x-1$

QUESTION 4

The given function is;

$f(x)=3(x-6)^2+1$

Let

$y=3(x-6)^2+1$

$\Rightarrow y-1=3(x-6)^2$

$\Rightarrow \frac{y-1}{3}=(x-6)^2$

$\Rightarrow \sqrt{\frac{y-1}{3}}=x-6$

$\Rightarrow x=6+\sqrt{\frac{y-1}{3}}$

The range is:

$\frac{y-1}{3}\ge0$

$y-1\ge0$

$y\ge1$

The interval notation is;

$[1,+\infty)$

6 0

4 years ago