I don’t know what some of that means but I can do the start for you? You rearrange
2x-y=1.
To do this, you +y to both sides, giving you 2x=1+y and then you minus 1, giving you 2x-1=y
Which can be rewritten the other way round to make it slightly easier
y=2x-1
You also have y=5x-5
These are both straight line equations and are now in the form y=mx+c
To sketch these graphs I would do two tables.
X -3 -2 -1 0 1 2 3
Y
For this, you now substitute each of the values for X into one of the equations you have. This is 2x-y=1 (2x-1=y)
X -3 -2 -1 0 1 2 3
Y -7 -5 -3 -1 1 3 5
You may have noticed a pattern there, the y values increased by two each time. This makes it linear. You would plot that line, onto an axis, using the coordinates you now have.
So, (-3, -7), (-2,-5), (-1,-3), (0,-1), (1,1), (2,3), (3,5)
Then I would do the same for the second equation, and plot that too.
X -3 -2 -1 0 1 2 3
Y -20 -15 -10 -5 0 5 10
You may have spotted this time the values increased by 5.
Then again plot this line using the coordinates shown.
I honestly have no idea what it means by “the line system on a corporate” but if that means on an axis then there’s your answer. If not then I do not know.
Hope this helps?
The correct box plot will have the "whiskers" extending to 24 and 49. The first quartile will be around 28.5 and the third quartile will be around 41.8. The median will be at 34.
**your question didn't include the box plots to choose from, so use this information to find which one it is :)
Answer:
0.070
Step-by-step explanation:
Y = number on trial
Y has a negative binomial distribution
r = 3
P = 30% = 0.3 probability of positive indication.
P(Y = 11) probability of 11 employees that must be tested to get 3 positives
Y-1Cr-1*p^r*q^(y-r)
Y-1 = 11-1 = 10
r-1 = 3 -1 = 2
10C2 x 0.3³x0.7⁸
45x0.027x0.05764801
= 0.070
This is the probability that 11 employees must be tested to get 3 positives.
You have to distribute the radical negative 18 in first. Simplifying from there, you’ll prob have to get into imaginary numbers since the square root of a negative is imaginary
