This is something you would do through trial and error. At least, that's the approach I took. I'm not sure if there is any algorithm to solve. The solution I got is shown in the attached image below. There are probably other solutions possible. The trick is to keep each number separate but not too far away so that the other numbers to be filled in later don't get too crowded to their neighbor.
Side note: any mirror copy of what I posted would work as well since you can flip the page around and it's effectively the same solution.
No of outcomes =16
favourable outcome=4
<span>probability of getting exactly 3 heads=0.25</span>
the assumption being that the first machine is the one on the left-hand-side and the second is the one on the right-hand-side.
the input goes to the 1st machine and the output of that goes to the 2nd machine.
a)
if she uses and input of 6 on the 2nd one, the result will be 6² - 6 = 30, if we feed that to the 1st one the result will be √( 30 - 5) = √25 = 5, so, simply having the machines swap places will work to get a final output of 5.
b)
clearly we can never get an output of -5 from a square root, however we can from the quadratic one, the 2nd machine/equation.
let's check something, we need a -5 on the 2nd, so
so if we use a "1" as the output on the first machine, we should be able to find out what input we need, let's do that.
so if we use an input of 6 on the first machine, we should be able to get a -5 as final output from the 2nd machine.
Answer:
The answer is "Options A, B, and E represent mutually exclusive events".
Step-by-step explanation:
Two occurrences that can happen immediately called mutually incompatible. Let's now glance at our options and figure out where the statements are mutually incompatible events.
In Option A: You could see that landing on an unwanted portion and arriving on 2 are events that are locally incompatible, even as undesirable portion contains 3 and 4, and 2 were shaded.
In Option B: Arriving on a shaded part and falling on 3 are also mutually incompatible because there are 3 on a windows azure.
In Option C: A darkened portion and an increasing amount can land while 2 would be an even number as well as on the shaded portion. That number is very much the same.
In Option D: At the same time as 4 is greater than 3 and it is situated upon an undistressed section, landing and attracting a number larger than 3 can happen.
In Option E: Landing on a shaded part and landing on even a shaded part is an excluding event, since shaders may either be shaded or unlit.
Answer:
9 gram/cm³
Step-by-step explanation:
Volume of a square (L³) = 2³ = 8cm³
Volume of a square (L³) = 2³ = 8cm³
But Density = mass/volume
But Density = Mass/Volume
Density = 72grams/8cm³
Density = 9grams/cm³
