Answer:
For this case, the first thing we must do is define variables:
x: number of hammers
y: number of wrenches
We write the system of inequations:
10x + 6y <= 120
x + y> = 14
Step-by-step explanation:
Okay,
Both games are: 35
First game: ?
Second game: 6 more than the first.
So,
first we subtract 6 from 35.
35 - 6 = 29
Divide by 2.
49 divided by 2 = 14.5
Add the 6 point= 14.5 + 6 = 20.5
To make sure add.
First game: 14.5
Second game:20.5
14.5 + 20.5 = 35
0.96 or 19/20
0.96 as fraction = 96/100 = 19.2 / 20 5 into 96 is 19.2 and 5 into 100 is 20
So 0.96 is the same as 19.2/20
Since 19.2 / 20 and 19 / 20 have the same denominator, therefore the 19.2 / 20 is bigger than 19 / 20.
So the 0.96 is bigger than the 19 / 20.
Answer:
1/3
Step-by-step explanation:
Lets start with the red ball because it is easier. You have 1/10 chances to pick the same number that is drawn randomly. Thus your base probability is .1
Now going to the white balls, you need atleast 2 numbers picked as the numbers drawn. Let's say we pick numbers 1-3. In order to win, the numbers 1 and 2, 2 and 3, 1 and 3, or 1 2 and 3 must be drawn. You can calculate the probability for each of these cases. In the case only 2 are drawn, the probability for each is 0.075. Since there are three cases that this could happen, you multiply this probability by 3, becoming 1/4. However, there is also the chance that you get all three. The probability of this is 3/10*2/9*1/8.
.225 + .1 + 0.008333 = 33.333...% or 1/3
