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1 year ago

A single die is rolled twiceFind the probability of rolling a 6 the first time and a 1 the second time.

Mathematics

1 answer:

Westkost [7]1 year ago

3 0

Answer:

1/36

Explanation:

In a single die, the total number of outcomes = 6

• The probability of rolling a 6 the first time = 1/6

• The probability of rolling a 1 the second time = 1/6

Thus, the probability of rolling a 6 the first time and a 1 the second time is:

$\begin{gathered} =\frac{1}{6}\times\frac{1}{6} \\ =\frac{1}{36} \end{gathered}$

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a set v is given, together with definitions of addition and scalar multiplication. determine which properties of a vector space

agasfer [191]

The properties of a vector space are satisfied Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes aren't legitimate are if $v = x ^ 2 1× v=1^ ×x ^ 2 = 1 #V$

Property three does now no longer follow: Suppose that Property three is legitimate, shall we name $v = a * x ^ 2 +bx +c$ the neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently $0 = O + v = (O x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a$

= 0 If O is the neuter, then it ought to restore x², but $0+ x² = (0x²+0x+zero) + (x²+0x+zero) = 1.$ This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant

have additive inverse

$Let r= v ×2x ^ 2 + v × 1x +v0 , w= w ×2x ^ 2 + w × 1x +w0 . We have that\\v+w= (vO + wO) ^ x^ 2 +(vl^ × wl)^ x+ ( v 2^ × w2)• w+v= (wO + vO) ^x^ 2 +(wl^ × vl)x+ ( w 2^ ×v2)$

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could use z = 1 then $v = x ^ 2 w = x ^ 2 + 1\\(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1$

v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3

Since 3x ^ 2 +1 ne x^ 2 +3. then the associativity rule doesnt hold.

$(1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 3\\1^ × (x^ 2 +x)+2^ × (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )\\(1^ ×2)^ ×(x^ 2 +x)=2^ × (x ^ 2 + x) = 2x + 2\\1^ × (2^ × (x ^ 2 + x) )=1^ × (2x+2)=2x^ 2 +2x( ne2x+2)$

Property f doesnt observe because of the switch of variables. for instance, if v = x ^ 2 1 × v=1^ × x ^ 2 = 1 #V

Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes arent legitimate.

Step-with the aid of using-step explanation:

Note that each sum and scalar multiplication entails in replacing the order from that most important coefficient with the impartial time period earlier than doing the same old sum/scalar multiplication.

Property three does now no longer follow: Suppose that Property three is legitimate, shall we name v = a × x ^ 2 +bx +c the neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently $0 = O + v = (O × x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a$

= zero If O is the neuter, then it ought to restore x², but zero $+ x² = (0x²+0x+zero) + (x²+0x+zero) = 1.$ This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant have additive inverse

Let $r= v × 2x ^ 2 + v × 1x +v0 , w= w2x ^ 2 + w × 1x +w0 . \\We have thatv+w= (vO + wO) ^ x^ 2 +(vl^ wl)^x+ ( v 2^ w2)w+v= (wO + vO) ^ x^ 2 +(wl^ vl)x+ ( w 2^v2)$

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could use $z = 1 then v = x ^ 2 w = x ^ 2 + 1(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3\\Since 3x ^ 2 +1 ne x^ 2 +3.$ then the associativity rule doesnt hold.

Note that each expressions are same because of the distributive rule of actual numbers. Also, you could be aware that his assets holds due to the fact in each instances we 'switch variables twice.

$· (1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 31^ * (x^ 2 +x)+2^ * (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )(1^ * 2)^ * (x^ 2 +x)=2^ * (x ^ 2 + x) = 2x + 21^ * (2^ * (x ^ 2 + x) )=1^ ×* (2x+2)=2x^ 2 +2x( ne2x+2)$

Read more about polynomials :

brainly.com/question/2833285

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8 0

10 months ago

What do you know about parallel slopes; perpendicular slopes?

sukhopar [10]

Answer:

Parallel slopes: two lines that has the same slopes but have different y-intercept

Perpendicular slope: two lines that are negative reciprocals(common) of each other

7 0

3 years ago

Khianna is trying to help her neighbor Mrs. Johnson design and estimate the cost of a new square patio to be made from 16 inch s

Gnoma [55]

Part I: Scale Drawing

Decide on a scale factor to represent the distance covered by the patio. Then, use the space below to design Mrs. Johnson’s patio to be a square that is at least 8 feet on each side.

a) Scale Factor: 1 in/ 2 ft

Use a straightedge and a ruler to draw to scale a design for Mrs. Johnson’s patio

b) see the picture attached

Mrs. Johnson’s patio to be a square that is 10 ft x 10 ft

c) What are the dimensions of Mrs. Johnson’s patio? 
the dimensions of Mrs. Johnson’s patio are 10 ft x 10 ft

d) Calculate the area of Mrs. Johnson’s patio. Show all work.
 
area of the square=b²
where b is the length side of the square
b=10 ft
so
Area=10²-----> area =100 ft²

e) How many pavers will be needed? Show all work. 

we know that
1 paver is 16 in x 16 in dimensions
convert to ft
1 ft----------->12 in
x ft-----------> 16 in
x=16/12-----> x=4/3 ft
so
1 paver is (4/3) ft x (4/3) ft dimensions

area of one paver=(4/3)²----> 16/9 ft²

if one paver has an area of----------------> 16/9 ft²
x pavers-----------------------> 100 ft²
x=100/(16/9)------> x=100*9/16-----> x=56.25 pavers

if one box --------------> 12 pavers
x box---------> 56.25 pavers
x=56.25/12-----> x=4.68 box-------> x=5 boxes
5 boxes of pavers will be needed

f) What will it cost to build the patio? Show all work.
 
the cost of one box is--------> $99.99
5 boxes-----------> x
x=5*$99.99------>x=$499.95

the cost to build the patio is $499.95

Part II: Bigger Design

There is a saying that bigger is better, so why not double the dimensions of Mrs. Johnson’s patio to make the side measurement twice as big? Mrs. Johnson and I think that it would better meet her needs. After seeing the original estimation, she thinks that she could afford to double the size. I explained that making the patio twice as big would mean twice the cost. Mrs. Johnson says, “Let’s do it!”

a) What would be the new dimensions of Mrs. Johnson’s patio?
the new dimensions of Mrs. Johnson’s patio are 20 ft x 20 ft

b) Calculate the new area of Mrs. Johnson’s patio. Show all work.
area of the square=b²
where b is the length side of the square
b=20 ft
so
Area=20²-----> area =400 ft²

c) How many pavers will be needed for the new design? Show all work.
 
1 paver is (4/3) ft x (4/3) ft dimensions

area of one paver=(4/3)²----> 16/9 ft²

if one paver has an area of----------------> 16/9 ft²
x pavers-----------------------> 400 ft²
x=400/(16/9)------> x=400*9/16-----> x=225 pavers

if one box --------------> 12 pavers
x box---------> 225 pavers
x=225/12-----> x=18.75 box-------> x=19 boxes
19 boxes of pavers will be needed

d) What will it cost to build the bigger patio? Show all work.
the cost of one box is--------> $99.99
19 boxes-----------> x
x=19*$99.99------>x=$1899.81
the cost to build the bigger patio is $1899.81

e) Is Khianna right? Will doubling the size of the patio, double the cost?

Khianna is wrong to double the dimensions the cost quadruples

5 0

3 years ago

What are the points

Goryan [66]

Answer:

There are 9 dots. So 9.

Step-by-step explanation:

Just take my word for it.

7 0

3 years ago

Figure A

Amiraneli [1.4K]

<h2>Answer:</h2>

Figure B

<h2>Step-by-step explanation:</h2>

The Pythagorean Theorem is $a^2 + b^2 = c^2$ , where c is the longest side of the triangle (the hypotenuse).

To find the side length of each square, you have to square root the area of each square. This means that Figure A has side lengths of 3, 6 and 8 units. Figure B has side lengths of 5, 12 and 13 units.

In Figure A, if the triangle is right-angled, the equation $3^2 + 6^2 = 8^2$ must be correct. 9 + 36 = 45. 45 is not equal to 64, so the triangle is not right-angled.

In Figure B, if the triangle is right-angled, the equation $5^2 + 12 ^2 = 13^2$ must be correct. 25 + 144 = 169. 169 is 13 squared, so the triangle is right-angled.

Alternatively, as you are already given the square values for each side length, there is no need to square root and square again. You can just test if the two smaller areas equal the larger area, but the explanation above uses a more detailed example of the Pythagorean Theorem.

8 0

2 years ago

Read 2 more answers