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2 years ago

You are going to roll two dice. Let the variable x = the sum of the numbers rolled. What is the probability that x = 9 ?

1 answer:

7 0

The first cube can land in any one of 6 ways.
The second cube can land in any one of 6 ways.
Total number of ways that 2 dice can land = (6 x 6) = 36 ways.

For any of these ways, the sum of the numbers rolled is 9 :

3, 6
4, 5
5, 4
6, 3

There are 4 ways to roll a 9, out of a total of 36 ways that
the dice can land. So the probability of rolling a 9 is

4 / 36 = 1 / 9 = <em>11-1/9 percent</em>

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Perform the indicated operation. 15b/4 * 8/9a^2b^2

Sergio [31]

Answer:

The simplified expression is $\frac{10}{3 a^2 b}$

Step-by-step explanation:

The given expression is:

$\frac{15b}{4} * \frac{8}{9a^2 b^2}$

Multiply the items in the numerator together ( 15b * 8 = 120 b). Also multiply the items in the denominator together ( 4 * 9a²b² = 36a²b²). The expression thus becomes:

$= \frac{120b}{36 a^2 b^2} \\\\$

Divide both the numerator and the denominator by 12b:

$= \frac{120b /12b}{36 a^2 b^2/12b}$

The expression finally becomes:

$= \frac{10}{3 a^2 b}$

7 0

3 years ago

Plsss help me with this question!!!

Molodets [167]

There are 27 men in the train car.
69-15=54
54 divided by 2 = 27

5 0

2 years ago

Please could you find the answers to the questions in the attachment.

Fudgin [204]

( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ )we need 3 equations
1. midpoint equation which is ( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ ) when you have 2 points

2. distance formula which is D= $\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

3. area of trapezoid formula whhic is (b1+b2) times 1/2 times height

so

x is midpoint of B and C
B=11,10
c=19,6
x1=11
y1=10
x2=19
y2=6
midpoint=( $\frac{11+19}{2} , \frac{10+6}{2}$ )
midpoint=( $\frac{30}{2} , \frac{16}{2}$ )
midpoint= (15,8)

point x=(15,8)

y is midpoint of A and D
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
midpoint=( $\frac{5+21}{2} , \frac{8+0}{2}$ )
midpoint=( $\frac{26}{2} , \frac{8}{2}$ )
midpoint=(13,4)

Y=(13,4)

legnths of BC and XY
B=(11,10)
C=(19,6)
x1=11
y1=10
x2=19
y2=6
D= $\sqrt{(19-11)^{2}+(6-10)^{2}}$
D= $\sqrt{(8)^{2}+(-4)^{2}}$
D= $\sqrt{64+16}$
D= $\sqrt{80}$
D= $4 \sqrt{5}$
BC= $4 \sqrt{5}$

X=15,8
Y=(13,4)
x1=15
y1=8
x2=13
y2=4
D= $\sqrt{(13-15)^{2}+(4-8)^{2}}$
D= $\sqrt{(-2)^{2}+(-4)^{2}}$
D= $\sqrt{4+16}$
D= $\sqrt{20}$
D= $2 \sqrt{5}$
XY= $2 \sqrt{5}$

the thingummy is a trapezoid
we need to find AD and BC and XY
we already know that BC= $4 \sqrt{5}$ and XY= $2 \sqrt{5}$

AD distance
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
D= $\sqrt{(21-5)^{2}+(0-8)^{2}}$
D= $\sqrt{(16)^{2}+(-8)^{2}}$
D= $\sqrt{256+64}$
D= $\sqrt{320}$
D= $4 \sqrt{2}$
AD= $4 \sqrt{2}$

so we have
AD= $4 \sqrt{2}$
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$

AD and BC are base1 and base 2
XY=height
so
(b1+b2) times 1/2 times height
( $4 \sqrt{2}+4 \sqrt{5}$ ) times 1/2 times $2 \sqrt{5}$ =
( $4 \sqrt{2}+4 \sqrt{5}$ ) times \sqrt{5} [/tex] =
$4 \sqrt{10}+4*5$ = $4 \sqrt{10}+20$ =80 $\sqrt{10}$ =252.982

X=(15,8)
Y=(13,4)
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$
Area=80 $\sqrt{10}$ square unit or 252.982 square units

7 0

3 years ago

the length of a rectagle is 5 in longer than its width. if the perimeter of the rectangle is 56 in, find its length and width

lisabon 2012 [21]

Answer:

Step-by-step explanation:

Let the width of the rectangle = x

As length is 5 inches longer than width, we have to add 5 to width

Length = x + 5

Perimeter of ractangle = 56 in

2* (length + width) = 56

2*( x + 5 + x) = 56

2* (2x + 5) = 56

Use distributive property: a*(b +c) =(a*b) + (a * c)

2*2x + 2*5 = 56

4x + 10 = 56

Subtract 10 from both sides

4x = 56- 10

4x = 46

Divide both sides by 4

x = 46/4

x = 11.5

Width = 11.5 in

length = 11.5 + 5

= 16.5 in

3 0

2 years ago

How to solve 6x+3=5x+10

laila [671]

$6x+3=5x+10 \\\\ 6x-5x=10-3 \\\\ \boxed{x=7}$

7 0

3 years ago

Read 2 more answers