Answer:
The word "ARRANGE" can be arranged in
2!×2!
7!
=
4
5040
=1260 ways.
For the two R's do occur together, let us make a group of R's taking from "ARRANGE" and permute them.
Then the number of ways =
2!
6!
=360.
The number ways to arrange "ARRANGE", where two "R's" will not occur together is =1260−360=900.
Also in the same way, the number of ways where two "A's" are together is 360.
The number of ways where two "A's" and two "R's" are together is 5!=120.
The number of ways where neither two "A's" nor two "R's" are together is =1260−(360+360)+120=660.
Step-by-step explanation:
<h2>
<u>PLEASE</u><u> </u><u>MARK</u><u> ME</u><u> BRAINLIEST</u><u> AND</u><u> FOLLOW</u><u> ME</u><u> LOTS</u><u> OF</u><u> LOVE</u><u> FROM</u><u> MY</u><u> HEART'AND</u><u> SOUL</u><u> DARLING</u><u> TEJASVINI</u><u> SINHA</u><u> HERE</u><u> ❤️</u></h2>
3b - 7u -5
Add them together and put them in alphabetical order.
Answer:
μ ≈ 2.33
σ ≈ 1.25
Step-by-step explanation:
Each person has equal probability of ⅓.
![\left[\begin{array}{cc}X&P(X)\\1&\frac{1}{3}\\2&\frac{1}{3}\\4&\frac{1}{3}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7DX%26P%28X%29%5C%5C1%26%5Cfrac%7B1%7D%7B3%7D%5C%5C2%26%5Cfrac%7B1%7D%7B3%7D%5C%5C4%26%5Cfrac%7B1%7D%7B3%7D%5Cend%7Barray%7D%5Cright%5D)
The mean is the expected value:
μ = E(X) = ∑ X P(X)
μ = (1) (⅓) + (2) (⅓) + (4) (⅓)
μ = ⁷/₃
The standard deviation is:
σ² = ∑ (X−μ)² P(X)
σ² = (1 − ⁷/₃)² (⅓) + (2 − ⁷/₃)² (⅓) + (4 − ⁷/₃)² (⅓)
σ² = ¹⁴/₉
σ ≈ 1.25
Answer: I got 50x+$75 but I’m not too sure for the last step
Step-by-step explanation:
Answer:
7f/z
Step-by-step explanation:
mat h wAyz
