For this case we have that a quadratic equation is of the form:
The roots are given by:
We have the following equation:
We look for the roots:
We have to:
So:
We have two imaginary roots:
Answer:
Answer:
There are not 15 failures and 15 successes. A confidence interval can be computed by adding 2 failures and 2 successes .
Step-by-step explanation:
Given that
Survey asked by 40 people
6 said that they were very concerned.
So the following statements correctly describe the above problem
The correct statement is , There are not 15 failures and 15 successes. A confidence interval can be computed by adding 2 failures and 2 successes .
This must be FALSE, because:
7 players x 4(which is the range) = 28 which means that the total number of baskets made is 28.
If the greatest is 8, we must do 28 - 8, which gives us 20. This means that the remaining 6 players (7 - 1 = 6) made a total of 20 baskets.
However, if the fewest is 4 baskets, this means that the remaining 5 players (6 - 1 = 5) must have made a total of 16 baskets (20 - 4 = 16) and that each player must have scored 4 or more baskets each (since 4 was the fewest number of baskets made).
But this is impossible because if the remaining 5 players scored 4 (fewest number possible) each, it would give us 20 (4 x 5). But they should have made 16, or else the total of all the baskets would not be 28 (look back at start).
This means that the statement is impossible and therefore FALSE.
Hope this helped :)
Answer:
No she doesn't have enough paint.
Step-by-step explanation:
You have to find the area of both walls then subtract both areas by 250. It would give you a negative number, meaning there is not enough paint.
Answer:
We have to find the maximum number of solutions of each of the following system:
1)
Two distinct concentric circles:
Since, distinct concentric circles means that the two circles have same center but different radius.
That means they will never intersect each other at any point.
Ans hence we will get zero solutions.
2)
Two distinct parabolas:
Two parabolas can maximum intersect at 2 points this could be seen by the diagrams.
3)
A line and a circle.
A line and a circle can maximum have 2 solutions.
4)
A parabola and a circle.
It can have maximum two solutions it can be seen from the diagram.
