Ask question

Ask question

All categories

3 years ago

14

Write the event as set of outcomes. We flip three coins and obtain more tails than heads.

1 answer:

Vilka [71]3 years ago

7 0

H H H
H H T
H T H
T H H
T T T 1
T T H 2
T H T 3
H T T 4
----------

You might be interested in

Work out<br><br> 1/3 + 2/9<br><br> Give your answer in its simplest form

Soloha48 [4]

Step-by-step explanation:

1/3 + 2/9 = 3/9 + 2/9 = 5/9

Topic: Fractions

If you like to venture further, feel free to check out my insta (learntionary). I'll be constantly posting math tips and notes! Thanks!

4 0

3 years ago

Read 2 more answers

What is 5/3 simplified sorry my brain is not working

grin007 [14]

5/3 is as simplified as it gets :)

8 0

2 years ago

Read 2 more answers

i have 17 coins. N of them are nickels and the rest are dimes. write an expression in two different ways for the amount of money

valkas [14]

Answer:

Step-by-step explanation:

(N)0.05 + (17-N)0.1 = M; M = amount of money I have.

Or 1.7-0.05N = M.

7 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

And the reward for the most beautiful king/queen on this website goes to........................................................

Lemur [1.5K]

Answer:

omg Thank u so much ur so nice

8 0

2 years ago

Read 2 more answers