All categories

3 years ago

3. Two fair dice are thrown once, find the probability of getting a total score greater than 5

Mathematics

2 answers:

zmey [24]3 years ago

8 0

Answer:

26/36

Step-by-step explanation:

The probability of something = the amount of favourable outcomes/total possible outcomes.

The total number of dice combinations is 36 (6x6).

In this case there are more outcomes more than 5 than less than 5 so for the sake of simplicity we are going to find the number of outcomes less than 5 and then find the inverse.

There are only 10 possible dice combos which add to totals 5 or under:

1,1 1,2 2,1 1,3 3,1 1,4 4,1 2,2 2,3 and 3,2

That means there is a 10/36 probability of getting a total score less than 5 so to get the probability of getting more than 5, we just do 36/36 - 10/36 (because all probabilities add to 1)

Therefore there is a 26/36 probability that the toatals will add to greater than 5.

Alternatively you could also draw up a possibilities table and count the favourable outcomes.

Hope this helped!

IceJOKER [234]3 years ago

4 0

Answer:

13/18.

Step-by-step explanation:

The favourable outcomes are (1,5) (1,6) (2,4) (2,5) (2,6) (3,3) (3,4) (3,5) (3,6) (4,2)(4,3) (4,4) 4,5) (4,6) 5,1) (5,2) 5,3) (5,4) (5,5) (5,6 )(6,1) to (6,6).

= total of 26.

As the total of all possible outcomes is 36 the answer is 26/36

= 13/18.

You might be interested in

A second-order linear differential equation for y(t) is said to be homogeneous if... every term involve either y or its derivati

swat32

Answer: Hello!

A second order differential equation has the next shape:

$y''(t) + p(t)y'(t) + q(t)y(t) = g(t)$

where p(t), q(t) and g(t) are functions of t, that can be constant numbers for example.

And is called homogeneus when g(t) = 0, so you have:

$y''(t) + p(t)y'(t) + q(t)y(t) = 0$

Then a second order differential equation is homogeneus ef every term involve either y or the derivatives of y.

4 0

3 years ago

Which graph represents the function f(x)=-|x|-2?

skad [1K]

Answer:

ind the absolute value vertex. In this case, the vertex for y=−|x|−2 is (0,−2).

(0,−2)

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

(−∞,∞)

Set-Builder Notation: {x|x ∈ R}

For each x value, there is one y value. Select few x values from the domain. It would be more useful to select the values so that they are around the x value of the absolute value vertex.

x y

−2 −4

−1 −3

0 −2

1 −3

2 −4

Step-by-step explanation:

5 0

3 years ago

2x^3-x^2-3x=210<br> the answer is 5 but I want to know why.

mel-nik [20]

Answer:

$x=5,\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

Step-by-step explanation:

1) Move all terms to one side.

$2x^{3} -x^{2} -3x-210=0$

2) Factor $2{x}^{3}-{x}^{2}-3x-210$ using Polynomial Division.

1 - Factor the following.

$2x^{3} -x^{2} -3x-210$

2 - First, find all factors of the constant term 210.

$1,2,3,4,5,6,7,10,14,15,21,30,35,42,70,105,210$

3) Try each factor above using the Remainder Theorem.

Substitute 1 into x. Since the result is not 0, x-1 is not a factor..

$2*1^{3} -1^{2} -3*1-210=-212$

Substitute -1 into x. Since the result is not 0, x+1 is not a factor..

$2(-1)^{3} -(-1)^{2} -3*-1-210=-210$

Substitute 2 into x. Since the result is not 0, x-2 is not a factor..

$2*2^{3} -2^{2} -3*2-210=-204$

Substitute -2 into x. Since the result is not 0, x+2 is not a factor..

$2{(-2)}^{3}-{(-2)}^{2}-3\times -2-210 = -224$

Substitute 3 into x. Since the result is not 0, x-3 is not a factor..

$2\times {3}^{3}-{3}^{2}-3\times 3-210 = -174$

Substitute -3 into x. Since the result is not 0, x+3 is not a factor..

$2{(-3)}^{3}-{(-3)}^{2}-3\times -3-210 = -264$

Substitute 5 into x. Since the result is 0, x-5 is a factor..

$2\times {5}^{3}-{5}^{2}-3\times 5-210 =0$

------------------------------------------------------------------------------------------

⇒ $x-5$

4) Polynomial Division: Divide $2{x}^{3}-{x}^{2}-3x-210$ by $x-5$ .

$2x^{2}$ $9x$ $42$

-------------------------------------------------------------------------

$x-5$ | $2x^{3}$ $-x^{2}$ $-3x$ $-210$

$2x^{3}$ $-10x^{2}$

-----------------------------------------------------------------------

$9x^{2}$ $-3x$ $-210$

--------------------------------------------------------------------------

$42x$ $-210$

-------------------------------------------------------------------------

5) Rewrite the expression using the above.

$2x^2+9x+42$

$(2x^2+9x+42)(x-5)=0$

3) Solve for $x.$

$x=5$

4) Use the Quadratic Formula.

1 - In general, given $a{x}^{2}+bx+c=0$ , there exists two solutions where:

$x=\frac{-b+\sqrt{b^{2} -4ac} }{2a} ,\frac{-b-\sqrt{b^2-4ac} }{2a}$

2 - In this case, $a=2,b=9$ and $c = 42.$

$x=\frac{-9+\sqrt{9^2*-4*2*42} }{2*2} ,\frac{-9-\sqrt{9^2-4*2*42} }{2*2}$

3 - Simplify.

$x=\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

5) Collect all solutions from the previous steps.

$x=5,\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

4 0

3 years ago

Read 2 more answers

Round 1,393 to the nearest thousand

myrzilka [38]

The answer is 1,000 4 or below you round down and 5 or higher you round up

6 0

3 years ago

Read 2 more answers

Which point is located at -7.75−7.75minus, 7, point, 75?

rosijanka [135]

Cccccccccccccccccccc

3 0

3 years ago