Answer:
I know for the fact that it's the first one but I don't know the other answer
Step-by-step explanation:
An equation of a graph is y=mx+b and the B is the y-intercept which is -2 since the line is on top of -2. mx is the rise/run of -2 which is -3/2
Hope this helps, but I don't which other equation represents the graph.
4725-2250= 2475
2475/2250 x100= 110
It increased by 110%
If its less than 90, its a right triangle. Hope this helps!
Using the probability and odds concepts, it is found that the odds against it containing a number greater than or equal to 7 is 
.
- A probability is the <u>number of desired outcomes divided by the number of total outcomes</u>.
- An odd is the <u>number of desired outcomes divided by the number of non-desired outcomes</u>.
In the rack, there are 15 balls, numbered from 1 to 15. Of those, <u>6 are less than 7</u>(against it containing a number greater than or equal to 7 is equivalent to it containing a number less than 7), thus:
- There are 6 desired outcomes.
- There are 9 non-desired outcomes.
The odd is:
The odds against it containing a number greater than or equal to 7 is .
A similar problem is given at brainly.com/question/21094006
For a), this is clearly a given as it is literally to the right of where it says “Given:”
For b), since ON bisects ∠JOH, this means that it splits it into two separate angles - JON and HON, which are similar due to that bisects mean that it splits it equally into two halves
For c), since NO is the same thing as NO, it is equal to itself
For d), since AAS (angle-angle-side) congruence states that if there are two angles that are congruent (proved in a) and b) ) as well as that a side is congruent (proved in c) ), two triangles are congruent
For e), since two triangles are congruent, every side must have one side that it matches up to in the other triangle. As the opposite side of angle H is JO and the opposite side of angle J is OH, and ∠J=∠H, those two are congruent. As JN and HN are the two sides left, they must be congruent.
Feel free to ask further questions!
