Worked a total of

Elena-2011 [213]

Domain: (infinity, 4]
range: [-6, infinity)

4 0

2 years ago

A game of chance involves rolling an unevenly balanced 4-sided die. The probability that a roll comes up 1 is 0.21, the probabil

WITCHER [35]

Answer:

2.65 dollars

Step-by-step explanation:

The expected value for a discrete variable is calculated as:

E(x)=x1P(x1) + x2P(x2) + ... + xnP(xn)

Where x1, x2, ... , xn are the possibles values of the variable and P(x1), P(x2), ... , P(xn) are the probabilities of x1, x2, ... , xn respectively.

In this case, the roll can comes up 1, 2, 3 or 4 and you can win 1, 2, 3 or 4 dollars respectively. So, taking into account that they are mutually exclusive events, the probability that the player win 1, 2, 3 or 4 dollars is:

P(1) = 0.21

P(2) = P(1∪2) - P(1) = 0.42 - 0.21 = 0.21

P(3) = P(2∪3) - P(2) = 0.51 - 0.21 = 0.3

P(4) = 1 - P(1) - P(2) - P(3) = 1 - 0.21 - 0.21 - 0.3 = 0.28

Therefore, If you win the amount that appears on the die the expected winning are:

E(x) = $1P(1) + $2P(2) + $3P(3) + $4P(4)

E(x) = $1(0.21) + $2(0.21) + $3(0.3) + $4(0.28)

E(x) = $2.65

4 0

3 years ago

Quanto é 60 x 11 + 346²

Bond [772]

Answer:

120376

Step-by-step explanation:

Primeiro peguei 60x11 e os multipliquei e obtive 660. Em seguida, descobri que 346 ^ 2 é 119716. Somei meus dois números e obtive 120376.

Espero que isso ajude você! :)

8 0

3 years ago

ASAP (only the answer pls)

atroni [7]

Answer:

B..........................................

3 0

2 years ago

Can someone please help me with this

Mama L [17]

Answer:

27.

Equation of 2 tangent lines at the given curve going through point P is:

y = -7x + 1

y = x + 1

28.

Part 1: 400 feet

Part 2: Velocity is +96 feet/second (or 96 feet/second UPWARD) & Speed is 96 feet per second

Part 3: acceleration at any time t is -32 feet/second squared

Part 4: t = 10 seconds

29.

Part 1: Average Rate of Change = -15

Part 2: The instantaneous rate of change at x = 2 is -8 & at x = 3 is -23

Step-by-step explanation:

27.

First of all, the equation of tangent line is given by:

$y-y_1=m(x-x_1)$

Where m is the slope, or the derivative of the function

Now,

If we take a point x, the corresponding y point would be $x^2-3x+5$ , so the point would be $(x,x^2-3x+5)$

Also, the derivative is:

$f(x)=x^2-3x+5\\f'(x)=2x-3$

Hence, we can equate the DERIVATIVE (slope) and the slope expression through the point given (0,1) and the point we found $(x,x^2-3x+5)$

The slope is $\frac{y_2-y_1}{x_2-x_1}$

So we have:

$\frac{x^2-3x+5-1}{x-0}\\=\frac{x^2-3x+4}{x}$

Now, we equate:

$2x-3=\frac{x^2-3x+4}{x}$

We need to solve this for x. Shown below:

$2x-3=\frac{x^2-3x+4}{x}\\x(2x-3)=x^2-3x+4\\2x^2-3x=x^2-3x+4\\x^2=4\\x=-2,2$

So, this is the x values of the point of tangency. We evaluate the derivative at these 2 points, respectively.

$f'(x)=2x-3\\f'(-2)=2(-2)-3=-7\\f'(2)=2(2)-3=1$

Now, we find 2 equations of tangent lines through the point (0,1) and with respective slopes of -7 and 1. Shown below:

$y-y_1=m(x-x_1)\\y-1=-7(x-0)\\y-1=-7x\\y=-7x+1$

and

$y-y_1=m(x-x_1)\\y-1=1(x-0)\\y-1=x\\y=x+1$

So equation of 2 tangent lines at the given curve going through point P is:

y = -7x + 1

y = x + 1

28.

Part 1:

The highest point is basically the maximum value of the position function. To get maximum, we differentiate and set it equal to 0. Let's do this:

$s(t)=160t-16t^2\\s'(t)=160-32t\\s'(t)=0\\160-32t=0\\32t=160\\t=\frac{160}{32}\\t=5$

So, at t = 5, it reaches max height. We plug in t = 5 into position equation to find max height:

$s(t)=160t-16t^2\\s(5)=160(5)-16(5^2)\\=400$

max height = 400 feet

Part 2:

Velocity is speed, but with direction.

We also know the position function differentiated, is the velocity function.

Let's first find time(s) when position is at 256 feet. So we set position function to 256 and find t:

$s(t)=160t-16t^2\\256=160t-16t^2\\16t^2-160t+256=0\\t^2-10t+16=0\\(t-2)(t-8)=0\\t=2,8$

At t = 2, the velocity is:

$s'(t)=v(t)=160-32t\\v(2)=160-32(2)\\v(2)=96$

It is going UPWARD at this point, so the velocity is +96 feet/second or 96 feet/second going UPWARD

The corresponding speed (without +, -, direction) is simply 96 feet/second

Part 3:

We know the acceleration is the differentiation of the velocity function. let's find it:

$v(t)=160-32t\\v'(t)=a(t)=-32$

hence, the acceleration at any time t is -32 feet/second squared

Part 4:

The rock hits the ground when the position is 0 (at ground). So we equate the position function, s(t), to 0 and find time when it hits the ground. Shown below:

$s(t)=160t-16t^2\\0=160t-16t^2\\16t^2-160t=0\\16t(t-10)=0\\t=0,10$