Answer:
your answer is 10/3 (fraction)
Step-by-step explanation:
120-40=80
4 x 6=24
80/24=
- first divide both sides by four and get 20/6
- then divide by 2 to get 10/3 (fraction form)
Answer:
A. 5.3
Step-by-step explanation:
While both of these numbers are close to 7, 5.3 is the closest and we can find this out with subtraction. So, 9.9 - 7 = 2.9, but 7 - 5.3 = 1.7. Which one can we determine is closer? Well, 2.9 > 1.7, so our answer must be A. 5.3!
Have a nice day and I hope this helped!
Answer:
a) 3.47% probability that there will be exactly 15 arrivals.
b) 58.31% probability that there are no more than 10 arrivals.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given time interval.
If the mean number of arrivals is 10
This means that 
(a) that there will be exactly 15 arrivals?
This is P(X = 15). So


3.47% probability that there will be exactly 15 arrivals.
(b) no more than 10 arrivals?
This is 














58.31% probability that there are no more than 10 arrivals.
The table tells us that the x coordinate. It also tells us that y is always x + 1.
For #1 you plot the coordinate (0, 1).
0 (the x coordinate) is given to us already.
1 (the y coordinate) is needed to be found by the equation.
You would then need to fill in the equation given with the x coorident.
y = 0 + 1
Then, solve for y.
0 + 1 = 1
The y coordinate is 1
Go to the horizontal line (x) and find 0.
Then go to the veridical line (y) and find 1.
Then match up the the x and y to plot the coordinate.
You would continue with this equation with the rest of the xs.
This is a hard concept to explain in just words, so feel free to comment with any more questions. :D
Step-by-step explanation:
Divide two on both sides to get rid of it and the make the equation in the form y = mx + c

y = 3/2x + 5
since both lines are parallel they must have the same gradient which is 3/2
y = 3/2x + c
All you have to do now is to replace x and y with (2, -5) to find c
x = 2
y = -5
-5 = 3/2 × 2 + c
-5 = 3 + c
c = -5-3
c = -8
