Answer:
which inequality? where the question ?
Step-by-step explanation:
Answer:
Given:
I usually walk from home to work. This morning, I walked for 10 minutes until I was halfway to work.
I then realized that I would be late if I kept walking.
I ran the rest of the way. I run twice as fast as I walk.
Find:
The number of minutes in total did it take me to get from home to work
Step-by-step explanation:
Had I kept walking, the second half of my trip would have taken 10 more minutes.
By doubling my speed for the second half of my trip,
I halved the amount of time it took me to finish.
So, the second half of my trip took 5 minutes, for a total trip time of 10+5 = 15 minutes.
The number of minutes in total did it take me to get from home to work is 15 minutes.
12/2=6.
6X0.5=3
so the scale drawing is 3 inches long
10/2=5
5X0.5=2.5
so the width of the scale drawing is 2.5 inches
3+3+2.5+2.5=11
4x - 5y = 24
4(6) - 5(0) = 24
24 - 0 = 24
24 = 24 (correct)....so (6,0) is a solution
4(1) - 5(-4) = 24
4 + 20 = 24
24 = 24 (correct)...and (1,-4) is a solution
Answer:
A= 0,2
B= 0,2
C= 0,4
D=0,2
Step-by-step explanation:
We know that only one team can win, so the sum of each probability of wining is one
P(A)+P(B)+P(C)+P(D)=1
then we Know that the probability of Team A and B are the same, so
P(A)=P(B)
And that the the probability that either team A or team C wins the tournament is 0.6, so P(A)+Pc)= 0,6, then P(C)= 0.6-P(A)
Also, we know that team C is twice as likely to win the tournament as team D, so P(C)= 2 P(D) so P(D) = P(C)/2= (0.6-P(A))/2
Now if we use the first formula:
P(A)+P(B)+P(C)+P(D)=1
P(A)+P(A)+0.6-P(A)+(0.6-P(A))/2=1
0,5 P(A)+0.9=1
0,5 P(A)= 0,1
P(A)= 0,2
P(B)= 0,2
P(C)=0,4
P(D)=0,2
