To solve that question we can think by the following way:
- The first letter can be any of the letters (A, C, T or G). Therefore there are 4 possibilites.
- To choose the second letter of the sequence, any letter also can be chose. So, one more time, there are 4 possibilites.
Using the fundamental principle of counting, we will find out that can be formed differents sequences (for example, AA, AG, TT, TG, CT...).
Know we know that, we can generalize. If I have elements and I must choose one of them times to do a sequence, the total number of sequence that I can form will be .
In our question, we want to know how many DNA sequence are exactly 29 letters long. As indicated above, the answer for that is (29 times) .
Therefore, there are DNA sequences with exactly 29 letters long.
I hope I've helped you. =D
Enjoy your studies. \o/
Answer:
Francine is 13
Sonya is 23
Joey is 23
Step-by-step explanation:
J = 10 + F
S=2F-3
J=S
10+f=2F-3
10+F-F= 2F-F-3
10+3=F-3+3
F=13
S=2(13)-3
S=26-3
S=23
Answer:
6y - 48z + 42
Step-by-step explanation:
6 ( y - 8z + 7 )
= 6y - 48z + 42
Answer: A) I think hope this helps
Answer:
a= 1/2 and b= 1/2
Step-by-step explanation:
The given probability can be expressed in the form of binomial expansion where a= p and b= q
Therefore a= 1/2 and b= 1/2
For n= 8 trials the binomial expansion can be written as
[ p+q] ^n = [ 1/2+1/2]^8
=(1/2)^0(1/2)^8 +8C1(1/2)^7(1/2)+8C2(1/2)^6(1/2)^2+8C3(1/2)^5(1/2)^3+8C4(1/2)^4(1/2)^4+8C5(1/2)^3(1/2)^5+8C6(1/2)^2(1/2)^6+8C7(1/2)(1/2)^7+8C8(1/2)^0(1/2)^8
=
1(1/2)^0(1/2)^8 +8(1/2)^7(1/2)+28(1/2)^6(1/2)^2+56(1/2)^5(1/2)^3+70(1/2)^4(1/2)^4+56(1/2)^3(1/2)^5+28(1/2)^2(1/2)^6+8(1/2)(1/2)^7+1(1/2)^0(1/2)^8
=
1/(2)⁸ { 1+ 8+28+56+70+56+28+8+1}
= 1/(2)⁸ {256}
= 1
The total probability is always 1.
Also P+q= 1
1/2+1/2= 1