Ask question

Ask question

All categories

3 years ago

12

In strategy games, in addition to the common 6-sided dice, dice having 4, 8, 12, or 20 sides are also used. determine the nu

mber of possibilities if the 1010-sided die is rolled and then the 66-sided die is rolled.

2 answers:

Dovator [93]3 years ago

8 0

There are 1010 possibilities for the 1010 sided die and 66 possibilities for the 66 sided die. Please mark Brainliest!!!

Ahat [919]3 years ago

3 0

Answer:

10

Step-by-step explanation:

Given that in strategy games, in addition to the common 6-sided dice, dice having 4, 8, 12, or 20 sides are also used

When a six sided die is rolled we may get outcomes as

{1,2,3,4,5,6} and hence there are six possibilities

When a ten sided die is thrown, the die can show any one of the 10 faces.

Hence no of possibilities if the 10 sided die is rolled is 10.

You might be interested in

Evaluate the Expression<br> When X=2, 3x-1=

Alex777 [14]

Answer:

5

Step-by-step explanation:

If X equals 2, then 3 times X would be 6. This is because if a number and a variable are put together, you have to multiply them. 6-1 is 5.

4 0

3 years ago

Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

$a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)$

$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

6 0

2 years ago

Help me with this...

marin [14]

i. 171

ii. 162

iii. 297

Solution,

n(U)= 630

n(I)= 333

n(T)= 168

i. Let n(I intersection T ) be X

$333 - x + x + 468 - x = 630 \\ or \: 333 + 468 - x = 630 \\ or \: 801 - x = 630 \\ or \: - x = 630 - 801 \\ or \: - x = - 171 \\ x = 171$

<h3>ii.n(only I)= n(I) - n(I intersection T)</h3><h3> = 333 - 171</h3><h3> = 162</h3>

<h3>iii. n ( only T)= n( T) - n( I intersection T)</h3><h3> = 468 - 171</h3><h3> = 297</h3>

<h3>Venn- diagram is shown in the attached picture.</h3>

Hope this helps...

Good luck on your assignment...

4 0

3 years ago

Which expression is equivalent to?

AlekseyPX

Answer:

D

Step-by-step explanation:

multiply the inside by -0.5

7 0

3 years ago

Read 2 more answers

Michelle is 7 years older than her sister Joan, and Joan is 3 years younger than their brother Ryan. If the sum of their ages is

Zielflug [23.3K]

Answer:

(C) 18

Step-by-step explanation:

We can create a systems of equations. Assuming $m$ is Michelle's age, $j$ is Joan's age, and $r$ is Ryan's age, the equations are:

$m = j + 7$

$j = r-3$

$m+j+r = 64$

We can use substitution, since we know the "values" of m and j.

$(j+7)+(r-3)+r = 64\\(j+7)+(2r-3)=64\\2r + j + 4 = 64\\2r + j = 60\\\\$

$r = 21, j = 18$

So we know that Joan is 18 years old.

Hope this helped!

8 0

3 years ago