We have to assume that he does all this with his eyes closed, and so his selections are completely random.
-- There are 26 socks in the drawer all together.
-- 10 of them are black or brown.
-- So the probability that he pulls out
either a black sock or a brown one is
10/26 = 5/13 = about 38.5% .
-- He puts it back, so there are still 26 socks in the drawer.
-- 16 of them are white.
-- So now, the probability of pulling out a white one is
16/26 = 8/13 = about 61.5% .
The probability of the whole process happening
just exactly as you described it is
(10/26) x (16/26)
= (5/13) x (8/13)
= (40) / (13²) = 40/169 = about 23.7% .
Quick check:
We got 38.5% the first time, and 61.5% the second time.
(38.5%) x (61.5%)
= (0.385 x 0.615)
= 0.2367 ==> 23.7% <== yay! that's good enough for me
Answer:
[-1.8], [2.7], 3 2/5, [3.5]
thats the answer
Answer:
B) -350
Step-by-step explanation:
We are given the sequence:
-56, -59, -62, -65...
And we want to determine its 99th term.
First, note that we have an arithmetic sequence. This is because each subsequent term differs from the previous term by a common difference.
In this case, each subsequent term is 3 less than the previous term, so our common difference <em>d</em> is -3.
To find the 99th term, we can write an explicit formula. The explicit formula for an arithmetic sequence is:
![x_n=a+d(n-1)](https://tex.z-dn.net/?f=x_n%3Da%2Bd%28n-1%29)
Where <em>x_n</em> represents the <em>n</em>th term, <em>a</em> is the initial term, and <em>d </em>is the common difference.
Since the first term is -56, <em>a </em>= -56.
By substitution, we acquire:
![x_n=-56-3(n-1)](https://tex.z-dn.net/?f=x_n%3D-56-3%28n-1%29)
The 99th term is when <em>n</em> = 99. Thus:
![x_{99}=-56-3(99-1)](https://tex.z-dn.net/?f=x_%7B99%7D%3D-56-3%2899-1%29)
Evaluate:
![x_{99}=-56-3(98)=-56-294=-350](https://tex.z-dn.net/?f=x_%7B99%7D%3D-56-3%2898%29%3D-56-294%3D-350)
Our answer is B.