Answer:
The slope is 1.
Step-by-step explanation:
Remember:

Let's plug in!

Answer:
x>1
Step-by-step explanation:
x + 18 > 19
Subtract 18 from each side
x + 18-18 > 19-18
x>1
Answer:
x=13,y=0
Step-by-step explanation:
We are given a system of equations

For equation 1,square all terms to reduce it
√x²+√y²=169
x+y=13
Make x the subject as required by the question to use substitution method
x=13-y
Plug x=13-y into eqn 2
3(13-y)+2y=39
39-3y+2y=39
39-y=39
y=39-39=0
Plug y=0 into equation 1
x+0=13
x=0
5x = 6x-7
subtract 6x from each side
-x=-7
x=7
Answer: 0.0228
Step-by-step explanation:
Given : The mean and the standard deviation of finish times (in minutes) for this event are respectively as :-

If the distribution of finish times is approximately bell-shaped and symmetric, then it must be normally distributed.
Let X be the random variable that represents the finish times for this event.
z score : 

Now, the probability of runners who finish in under 19 minutes by using standard normal distribution table :-

Hence, the approximate proportion of runners who finish in under 19 minutes = 0.0228