Solve the inequality 2(3x - 4) > 9x - 10

Mathematics

2 answers:

jenyasd209 [6]3 years ago

7 0

Answer:

x<-6

Step-by-step explanation:

USPshnik [31]3 years ago

3 0

it is b please give me brainlyest

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There are currently 16 teams in the National Football Conference (NFC). Of those teams, 11 have won at least one super bowl and

svp [43]

You are given 16 teams, 11 have won at least one super bowl and 5 have not.

A. The probability that both selected teams have won at least 1 super bowl is

$Pr(A)=\dfrac{C_{11}^2}{C_{16}^2}=\dfrac{\dfrac{11!}{2!(11-2)!}}{\dfrac{16!}{2!(16-2)!}}=\dfrac{\dfrac{10\cdot 11}{2}}{\dfrac{15\cdot 16}{2}}=\dfrac{55}{120}=\dfrac{11}{24}.$

B. The probability that neither selected team has won at least 1 super bowl is

$Pr(B)=\dfrac{C_{5}^2}{C_{16}^2}=\dfrac{\dfrac{5!}{2!(5-2)!}}{\dfrac{16!}{2!(16-2)!}}=\dfrac{\dfrac{4\cdot 5}{2}}{\dfrac{15\cdot 16}{2}}=\dfrac{10}{120}=\dfrac{1}{12}.$

C. The probability that at least one selected team has won at least 1 super bowl is

$Pr(C)=1-Pr(B)=1-\dfrac{1}{12}=\dfrac{11}{12}.$

D. to find the probability that the second team selected has won at least 1 super bowl given that the first team selected has not won a super bowl, consider such events:

P - the second team selected has won at least 1 super bowl;

Q - the first team selected has not won a super bowl.

Then

$Pr(P|Q)=\dfrac{Pr(P\cap Q)}{Pr(Q)}=\dfrac{\dfrac{5\cdot 11}{C_{16}^2}}{\dfrac{C_5^1\cdot C_{15}^1}{C_{16}^2}}=\dfrac{55}{75}=\dfrac{11}{15}.$

E. To find the probability that the second team selected has won at least 1 super bowl given that the first team selected has won at least 1 super bowl, consider events:

M - the second team selected has won at least 1 super bowl;

N - the first team selected has won at least 1 super bowl.

Then

$Pr(M|N)=\dfrac{Pr(M\cap N)}{Pr(N)}=\dfrac{\dfrac{11\cdot 10}{C_{16}^2}}{\dfrac{C_{11}^1\cdot C_{15}^1}{C_{16}^2}}=\dfrac{110}{165}=\dfrac{2}{3}.$