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rusak2 [61]
3 years ago
12

Rosario is rollling a six- sided cube. Rosario rolls the number cube twice. What is the probability that Rosario gets the number

1 twice in a row?
A. 1/36
B.1/6
C. 1/3
D.1
Part two: tony flips a coin and gets heads. Ge tells his friend his friend that he can get heads two more times in a row. What is the probability that tony will flip a coin and gets heads three more times in a row?
A.1/8
B.1/2
C.1 & 1/2
D. 2
Whats the answer for each part?
First answer that explains gets to me marked branilest& extra points. I need help ASAP.
Thanks
Mathematics
1 answer:
icang [17]3 years ago
5 0
Part 1:the answer is 1/36
each time we roll the dice, the whole probabilities are 6
if we roll it two times
6*6=36

part 2

it's 1/8

each time we flip a coin, the whole probabilities are 2

if we flip it 3 times

2*2*2=8




good luck
You might be interested in
Please explain the how you got the answers thanks asap
stich3 [128]

Answer:

a) 90 stamps

b) 108 stamps

c) 333 stamps

Step-by-step explanation:

Whenever you have ratios, just treat them like you would a fraction! For example, a ratio of 1:2 can also look like 1/2!

In this context, you have a ratio of 1:1.5 that represents the ratio of Canadian stamps to stamps from the rest of the world. You can set up two fractions and set them equal to each other in order to solve for the unknown number of Canadian stamps. 1/1.5 is representative of Canada/rest of world. So is x/135, because you are solving for the actual number of Canadian stamps and you already know how many stamps you have from the rest of the world. Set 1/1.5 equal to x/135, and solve for x by cross multiplying. You'll end up with 90.

Solve using the same method for the US! This will look like 1.2/1.5 = x/135. Solve for x, and get 108!

Now, simply add all your stamps together: 90 + 108 + 135. This gets you a total of 333 stamps!

5 0
3 years ago
An equilateral ∆ has sides of length 16 cm. Find the length of an altitude.
inn [45]

The length of the altitude is 8\sqrt{3}

Explanation:

Let ABC be an equilateral triangle.

It has sides of length 16 cm

Let AD be the altitude of the triangle.

We need to determine the length of an altitude.

Let AC = 16 cm and CD = 8 cm

Let us consider the right angled triangle ADC

Using the Pythagorean theorem, we have,

AC^2=AD^2+DC^2

Substituting the values, we get,

 16^2=AD^2+8^2

 256=AD^2+64

 192=AD^2

8\sqrt{3}=AD

The length of the altitude is 8\sqrt{3}

5 0
3 years ago
I really need help pls?
Gnesinka [82]

Answer:

See Below

Step-by-step explanation:

Ok, this is just like the systems of equations.

x-7 = 5x-31

Solve

x-7 = 5x-31

-x     -x

-7 = 4x -31

+31       +31

24 = 4x

24/ 4   = 4x/ 4

x=6

Hope this helps!=)

8 0
2 years ago
Math help please i have discalcula
user100 [1]

Answer:

x=4y,3x=4y,y=4x,4x=3y

5 0
3 years ago
Someone please help or i’ll be failing geometry this year
Vlad [161]

Step-by-step explanation:

I don't know what constructions you were taught.

a "similar" triangle is a triangle with exactly the same angles as the other triangle, but the lengths of all sides are stretched or shortened by the same scaling factor f.

by saying 1:2 she means the second triangle should have sides with twice the lengths of the first triangle (f=2).

and the extra challenge - same basic thing. she allows you to pick one of the two triangles as reference. and then you need to draw a third triangle (again with the same angles) with the side lengths extended by the scaling factor f of 4/3.

I would draw the triangles right on top of each other with the same starting corner (let's call it A) for all 3.

we would get the triangles ABC, AMN and AXY.

the points B and C would be then halfway on AM and AN.

and M and N would then a bit before X and Y on AX and AY.

the beauty is, you only need to construct 2 sides of every new triangle down to the new endpoints. the third side is automatically scaled correctly, and you only need to connect these new endpoints.

let's assume you draw a triangle (just very simple) ABC with all side lengths being 3. so, AB=3, AC=3, BC=3.

now you draw AMN by extending AB and AC to a side length of 6 (f=2) creating M and N, and you connect M and N.

and then you can create the third triangle AXY by extending AM and AN by a factor of 4/3 to side lengths of 8 (4/3 × 6 = 8) creating new end points X and Y. and you connect X and Y.

and that is it. all 3 triangles are similar (the same angles), and all sides of a triangle have the same length ratio to the sides of the other triangle(s).

4 0
2 years ago
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