Combine like terms, 1+6 is 7 and 2t+4t is 6t. so 6t+7=55. get the t by itself by subtracting 7 from both sides. 6t=48, divide by 6. t=8
P(all 4 are male) = 1/2^4 = 1/16 ( assuming probability of 1 male = 1/2)
P( 3 are male) = 1/2^4*4 = 1/4
The rest will have at least 2 females
So probability of at least 2 females) = 1 - 1/16 - 1/4 = 1 - 5/16 = 11/16 answer
Answer:
Part a) The inequality that represent the situation is
Part b) The possible lengths of the shortest piece of wire are all positive real numbers less than or equal to 
Step-by-step explanation:
Let
x------> the length of the first wire
3x---> the length of the second wire
2(3x)=6x -----> the length of the third wire
Part a) WRITE AN *INEQUALITY* THAT MODELS THE SITUATION
we know that
The inequality that represent the situation is

Part b) WHAT ARE THE POSSIBLE LENGTHS OF THE SHORTEST PIECE OF WIRE?
we know that
The shortest piece of wire is the first wire
so
Solve the inequality


Divide by 10 both sides

The possible lengths of the shortest piece of wire are all positive real numbers less than or equal to 
Answer:
Y=-80
Step-by-step explanation:
X=Ky/z
5=30k/3
30k=15
K=1/2
20=Ky/(-2)
Ky =-40
1/2y=-40
Y=-80