3(1)+4<span>≥ 13
7</span><span>≥ 13
No
3(2.5)+4</span><span>≥ 13
11.5</span><span>≥ 13
No
3(3)+4</span><span>≥ 13
13</span><span>≥ 13
Yes
{3, 4.5, 5}</span>
Answer:
$159.09
Step-by-step explanation:
Math
Answer:
1/11
Step-by-step explanation
There are 12 marbles in the bag. When we first pick we have 4 blue marbles. So 4 blue marbles/12 random marbles. When we pick blue and noted, there are 3 marbles in the bag because of we didn't put it back. So when we choose again there are 11 marbles and 3 blue marbles in the bag. Choosing a blue one case is 3/11.
The last part of this case is happening as a chain. So we need to multiply our two answers.
=4/12*3/11
=1/3*3/11
=1/11
Answer:
The equation of the line that passes through the points (0, 3) and (5, -3) is 
.
Step-by-step explanation:
From Analytical Geometry we must remember that a line can be formed after knowing two distinct points on Cartesian plane. The equation of the line is described below:
(Eq. 1)
Where:
- Independent variable, dimensionless.
- Dependent variable, dimensionless.
- Slope, dimensionless.
- y-Intercept, dimensionless.
If we know that 
and 
, the following system of linear equations is constructed:
(Eq. 2)
(Eq. 3)
The solution of the system is: , 
. Hence, we get that equation of the line that passes through the points (0, 3) and (5, -3) is .
Tossing a die will have 6 possible outcomes. Those are having sides that are number 1 to 6. The sample space of tossing 3 dice is equal to 6³ which is equal to 216. Now for the calculation of probabilities,
P(two 5s) = (1 x 1 x 5)/216
As we have to have the 5 in the die for two times, then for the 1 time, we can have all other numbers except 5. The answer is 5/216.
P(three 5s) = (1 x 1 x 1)/216 = 1/216
P(one 5 or two 5s) = (1 x 5 x 5)/216 + (1 x 1 x 5)/216 = 5/36
