Its depends on how many times you score during the game
in second game the number of points increases as a geometric sequence
with common ratio 2
so for example if you score ten times in first game you get 2000 points
if you score 10 times in Game 2 you score 2 * (2^10) - 1 = 2046 points
so playing 10 or more games is best with Game 2. Any less plays favours gane 1.
Answer
Josh's textbook reached the ground first
Josh's textbook reached the ground first by a difference of 
Step-by-step explanation:
Before we proceed let us re write correctly the height equation which in correct form reads:
Eqn(1).
Where:
: is the height range as a function of time
: is the initial velocity
: is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:
Eqn(2).
Now let us use the given information and set up our equations for Ben and Josh.
<u>Ben:</u>
We know that 
Thus Eqn. (2) becomes:
Eqn.(3)
<u>Josh:</u>
We know that 
Thus Eqn. (2) becomes:
Eqn. (4).
<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em>
<em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>
Thus we have for Ben, Eqn. (3) gives:
Using the quadratic expression to find the roots of the quadratic we have:
Since time can only be positive we reject the 
solution and we keep that Ben's book took 
seconds to reach the ground.
Similarly solving for Josh we obtain
Thus again we reject the negative and keep the positive solution, so Josh's book took 
seconds to reach the ground.
Then we can find the difference between Ben and Josh times as
So to answer the original question:
<em>Whose textbook reaches the ground first and by how many seconds?</em>
- Josh's textbook reached the ground first
- Josh's textbook reached the ground first by a difference of
1) The outcomes for rolling two dice, the sample space, is as follows:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
There are 36 outcomes in the sample space.
2) The ways to roll an odd sum when rolling two dice are:
(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5). There are 18 outcomes in this event.
3) The probability of rolling an odd sum is 18/36 = 1/2 = 0.5
Answer:
It is c = kt.
Step-by-step explanation:
This is direct variation . As the time increases the number of cranes increases.
The equation is c = kt where k is the constant of variation. In this case it will be a positive value because c is increasing with time.
If they make 20 cranes in 10 minutes then we can find the value of k by plugging in these values;
20 = k * 10
g = 20/10
k = 2.
So they make 2 cranes per minute.
