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3 years ago

The sums that appear when two fair? four-sided dice? (tetrahedrons) with sides 1?, 2?, 3?, and 4 are tossed

Mathematics

1 answer:

Karolina [17]3 years ago

5 0

Answer:

The possible sums are 2, 3, 4, 5, 6, 7 & 8 (but they're not all equally likely)

Step-by-step explanation:

For example, the only way to get a sum of 2 is to roll a 1 and then a 1

And the only way to get a sum of 8 is to roll a 4 and then a 4 (imagine you don't roll the two dice simultaneously; this will help you to see the "other possibilities" that aren't obvious at first)

The probability of rolling a sum of 2 = probability of rolling sum of 8 (1/16 probability for each)

If you roll a 1 then a 2, or a 2 then a 1, there are 2 ways to get a sum of 3

If you roll a 3 then a 1, a 2 then a 2, or a 1 then a 3, there are 3 ways to get a sum of 4

If you roll a 4 then a 1, a 3 then a 2, a 2 then a 3, or a 1 then a 4, there are 4 ways to get a sum of 5 (5 is the most likely sum)

If you roll a 4 then a 2, a 3 then a 3, or a 2 then a 4, there are 3 ways to get a sum of 6

If you roll a 4 then a 3, or a 3 then a 4, there are 2 ways to get a sum of 7

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What is the value of x, the acid concentration of the first solution?

bezimeni [28]

Answer:

A) 0.05

Step-by-step explanation:

Let's summarize into an equation the information we can get from that table.

We have 4 liters of an acid of an unknown concentration, let's call it (4x).

We have 10 liters of an acid with known concentration of 0.40.

And we have a total of 14 liters overall with a concentration of 0.3o.

That's like a weighted average formula: 4x + 10 y = 14z

Let's replace the concentration values we know and solve this:

4x + 10 (0.4) = 14 * 0.3

4x + 4 = 4.2

4x = 0.2

x = 0.05

So, the concentration of the 4 liters of acid on concentration X are in fact of concentration of 0.05.

6 0

3 years ago

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Ben made a model (shown below) of the square pyramid he plans to build when he grows up.

elena-s [515]

Answer:

$336 m^2$

Step-by-step explanation:

The surface area of a square pyramid is the sum of the area of the squared base + 4 times the area of each triangular face, therefore:

$A=A_b + 4A_t$

where:

$A_b=L^2$ is the area of the base, where

L is the length of the base

$A_t=\frac{1}{2}Lh$ is the area of each triangular face, where

h is the height of the face

Substituting,

$A=L^2+2Lh$

For the model in this problem,

L = 12

h = 8

Therefore, the surface area here is:

$A=12^2 +2(12)(8)=336$

4 0

3 years ago

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G(x) = 3x2 + 14x – 5?

Alla [95]

Answer:

x= $-\frac{1}{14}$

Step-by-step explanation:

im going to assume the 3x2 mean 3 times 2

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3 years ago

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A total of 50,000 contestants are chosen for a survey. A

ankoles [38]

I am pretty sure it is linear decrease,

4 0

3 years ago

Can some show the steps and the answer please I really need help tell me how you get it and the answer

Lana71 [14]

Answer:

Step-by-step explanation:

so this is about triangles.. sooo the following is a bit of helpful reminders that I keep on my computer to help me remember how to fit the trig functions to triangles.. I strongly suggest you copy it and keep it where you can look at it often.

Use SOH CAH TOA to recall how the trig functions fit on a triangle

SOH: Sin(Ф)= Opp / Hyp

CAH: Cos(Ф)= Adj / Hyp

TOA: Tan(Ф) = Opp / Adj

I use this anytime I run into triangles or need some help with sin or cos

now the problem , 3 ladder 10, 12, & 15 feet. Alex wants to get to 8 feet.

the problems is also telling you that you can use t..... and then , the words are cut off.. but I know they were going to say Tan ... next.. :P

b/c Tan is how you figure out problems with the adjacent side and the opposite side. like this problem. Look at TOA above. use that to recall how the parts fit in the formula

Tan(∅) = Opp / Adj

they give us the Opp side of 8 feet in the problem

then they also tell us the Hyp of the triangle which is each of the ladders length. Then they ask us what is the Adj sides length?

So we also need to solve the triangle with the know hyp (ladder length).. uggg, this problem is long. Then we can solve the dist. from the wall or Adj side length.

it's two steps, if you want to think of it that way. You're supposed to be pretty confident with trig functions. I'm guessing this is a trig class.. right?

let's solve for the 3 different angles that the ladders make , each going to 8 feel. Obviously, nobody would really do this with a ladder they would just lean it against the wall . and if it's taller than where they want to climb, they would just go up part way. so anyway, find the 3 different angles.

look above to see which formula to use.

I like SOH b/c it seems to have all the pieces of the triangle we want to work with.

ladder 1 ( 10')

Sin(∅) = Opp / Hyp

Sin(∅) = 8 / 10

∅ = arcSin (4/5)

[ first, yes, I just reduced the fraction, then I did the arcSin on both sides, I think you might know how to do that already ? ]

∅ = 53.13010 °

( yes, I used my calculator to find that, calculators are okay to use when figuring out non standard angles )

ladder 2 (12')

Sin(∅) = 8/12

∅ = arcSin (2/3)

∅ = 41.81031°

ladder 3 (15')

Sin(∅) = 8/15

∅ = arcSin (8/15)

∅ = 32.230952°

now use our Tan function to find the Adjacent side which is the distance from the wall

Tan(∅)= Opp / Adj

Adj = Opp / Tan(∅)

( I did some quick algebra to move the side we want to solve for, now plug and chug all 3 angles )

ladder 1

Adj = 8 / Tan(53.13010)

Adj = 6.0000005

ladder 2

Adj = 8 / Tan(41.81031)

Adj = 8.94427

ladder 3

Adj = 8 / Tan(32.230952)

Adj = 12.688577

so the 10' ladder is 6 feet from the wall

the 12' ladder is 8.9 feet from the wall

the 15 foot ladder is 12.7 feet from the wall.

I really don't think that 15' ladder is going to stay on the wall.. if Alex climbs it... it's way way too far out... it will just fall straight down the wall :/ Maybe another math problem for the forces involved :P

5 0

2 years ago