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Step-by-step explanation:
1. we have to write the system specifications as:
A(x,y) give us the meaning that the consule x can be accessed when y is in a faulty condition
∀y∃A(x,y)
2. B(x,y) shows that users email has sent a message, y. Which is in the archive. C(x) shows the email address of user x is retrievable
∀x∃y[B(x,y)→c(x)]
3. D(x,y) shows that x can detect breach y'' and we have E(z) that tells us there is a compromise of z
∀y∃xD(x,y)↔ ∃zE(z)
4. F(x,y,z)
Y and z are distinct point ends which x connects
We have,
∀y∀z∃x∃a[x ≠a →F(x,y,z)^F(a,y,z)
5. G(x,y)
X knowst the password of y' and H(x) means that we have x to be a system administrator
∀x[H(x)→∀yG(x,y)] ∃x[H(x)^∀yG(x,y)]
Total angle=360=5*72
<span><span>area=5∗16.4π y<span>d2</span>=82π=82∗3.14 y<span>d2</span></span></span>
~ ANSWER=1/2 ~
Simple probability is found by counting all the results which fit requirements and dividing by all possible results.
To find probability of two results in a row, multiply chance of first result by chance of second result.
Since you are replacing the marble before the second draw, we don’t have to figure out the various changes in odds for the different possible first draws. It’s just that simple.
There are 5 white marbles
There are 4 red marbles
There are always 20 marbles in all
5/20*4/20=1/4*1/5=1/20 or 1/2
By coincidence, the same as the chance of drawing the white marble in one draw.
