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3 years ago

5

In the game of backgammon, you get to double how much you move if you roll “doubles" on two dice (that is: 1 and 1, 2 and 2, 3 a

nd 3, etc.). The next three questions concern this situation. 23. What is the probability of rolling doubles on any one roll?
(a) 1/6

(b) 1/12

(c) 1/36

(d) 5/36

(e) 1/2

1 answer:

weeeeeb [17]3 years ago

3 0

*11* 12 13 14 15 16
21 *22* 23 24 25 26
31 32 *33* 34 35 36
41 42 43 *44* 45 46
51 52 53 54 *55* 56
61 62 63 64 65 *66*
(36 combos)

6/36 or 1/6 probability or rolling doubles.

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3 years ago

Please solve question 6, doing surds

Amiraneli [1.4K]

See the attached diagram, it has all the information you need.

(a) If the green radii are all 1, then the orange diameters are all 2 + √2, so that the orange radii are (2 + √2)/2 = 1 + √2/2.

This is because we can join the radii of two adjacent green circles to form the sides of a square with side length equal to twice the radius - i.e. the diameter - of the green circles. The diagonal of any square occurs in a ratio to the side length of √2 to 1. Then we get the diameter of an orange circle by summing this diagonal length and two green radii, and hence the radius by dividing this by 2.

(b) We get the blue diameter in the same way. It has length (2 + √2) (1 + √2/2) = 3 + 2√2, so that the blue radius is (3 + 2√2)/2 = 3/2 + √2.

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3 years ago

Consider the two functions:

koban [17]

Answer:

a) The x value of the point where the two equations intersect in terms of a is $x=\frac{40}{4+5a}$

b) The value of the functions at the point where they intersect is $\frac{10 (28 + 15 a)}{4 + 5 a}$

c) The partial derivative of f with respect to $x$ is $\frac{\partial f}{\partial x} = -5a$ and the partial derivative of f with respect to $a$ is $\frac{\partial f}{\partial x} = -5x$

d) The value of $\frac{\partial f}{\partial x}(3,2) = -10$ and $\frac{\partial f}{\partial a}(3,2) = -15$

e) $\upsilon_1=-\frac{3}{4} = -0.75$ and $\upsilon_2=-\frac{3}{4} = -0.75$

f) equation $\upsilon_1 = \frac{-5a\cdot x}{70-5ax}=\frac{ax}{ax-14}$ and $\upsilon_2 = \frac{-5a\cdot a}{70-5ax}=\frac{a^2}{ax-14}$

Step-by-step explanation:

a) In order to find the $x$ we just need to equal the equations and solve for $x$ :

$f(x,a)=g(x)\\70-5xa = 30+4x\\70-30 = 4x+5xa\\40 = x(4+5a)\\\boxed {x = \frac{40}{4+5a}}$

b) Since we need to find the value of the function in the intersection point we just need to substitute the result from a) in one of the functions. As a sanity check , I will do it in both and the value (in terms of $a$ ) must be the same.

$f(x,a)=70-5ax\\f(\frac{40}{4+5a}, a) = 70-5\cdot a \cdot \frac{40}{4+5a}\\f(\frac{40}{4+5a}, a) = 70 - \frac{200a}{4+5a}\\f(\frac{40}{4+5a}, a) = \frac{70(4+5a) -200a}{4+5a}\\f(\frac{40}{4+5a}, a) =\frac{280+350a-200a}{4+5a}\\\boxed{ f(\frac{40}{4+5a}, a) =\frac{10(28+15a)}{4+5a}}$

and for $g(x)$ :

$g(x)=30+4x\\g(\frac{40}{4+5a})=30+4\cdot \frac{40}{4+5a}\\g(\frac{40}{4+5a})=\frac{30(4+5a)+80}{4+5a}\\g(\frac{40}{4+5a})=\frac{120+150a+80}{4+5a}\\\boxed {g(\frac{40}{4+5a})=\frac{10(28+15a)}{4+5a}}$

c) $\frac{\partial f}{\partial x} = (70-5xa)^{'}=70^{'} - \frac{\partial (5xa)}{\partial x}=0-5a\\\frac{\partial f}{\partial x} =-5a$

$\frac{\partial f}{\partial a} = (70-5xa)^{'}=70^{'} - \frac{\partial (5xa)}{\partial a}=0-5x\\\frac{\partial f}{\partial a} =-5x$

d) Then evaluating:

$\frac{\partial f}{\partial x} =-5a\\\frac{\partial f}{\partial x} =-5\cdot 2=-10$

$\frac{\partial f}{\partial a} =-5x\\\frac{\partial f}{\partial a} =-5\cdot 3=-15$

e) Substituting the corresponding values:

$\upsilon_1 = \frac{\partial f(3,2)}{\partial x}\cdot \frac{3}{f(3,2)} \\\upsilon_1 = -10 \cdot \frac{3}{40} = -\frac{3}{4} = -0.75$

$\upsilon_2 = \frac{\partial f(3,2)}{\partial a}\cdot \frac{3}{f(3,2)} \\\upsilon_2 = -15 \cdot \frac{2}{40} = -\frac{3}{4} = -0.75$

f) Writing the equations:

$\upsilon_1=\frac{\partial f (x,a)}{\partial x}\cdot \frac{x}{f(x,a)}\\\upsilon_1=-5a\cdot \frac{x}{70-5xa}\\\upsilon_1=\frac{-5ax}{70-5ax}=\frac{-5ax}{-5(ax-14)}\\\boxed{\upsilon_1=\frac{ax}{ax-14} }$

$\upsilon_2=\frac{\partial f (x,a)}{\partial x}\cdot \frac{a}{f(x,a)}\\\upsilon_2=-5a\cdot \frac{a}{70-5xa}\\\upsilon_2=\frac{-5a^2}{70-5ax}=\frac{-5a^2}{-5(ax-14)}\\\boxed{\upsilon_2=\frac{a^2}{ax-14} }$

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4 years ago

Jaden claims that all quadratic equations have two real solutions. Is he correct?

kirza4 [7]

Answer:

No

Step-by-step explanation:

Quadratic equations usually has two solutions. However, in some cases, the other solution may not be real. When using quadratic formula to get solution, the squareroot generates a positive and negative number. Finally, some negative numbers may not be realistic to conclude as solutions for example when you find x as -5 but the expression needs number of items. In other cases, the squareroot of the number may be that the number is negative eg √-12 hence we don't get real numbers in this case. The solutions must be two numbers but not all the numbers will be real numbers.

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3 years ago

What does 3e-6 mean on the calculator?

Ostrovityanka [42]

it means 3 and then "e" means is up of the 3so its 3↖-6

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3 years ago