Answer:
The probability that at least 1 car arrives during the call is 0.9306
Step-by-step explanation:
Cars arriving according to Poisson process - 80 Cars per hour
If the attendant makes a 2 minute phone call, then effective λ = 80/60 * 2 = 2.66666667 = 2.67 X ≅ Poisson (λ = 2.67)
Now, we find the probability: P(X≥1)
P(X≥1) = 1 - p(x < 1)
P(X≥1) = 1 - p(x=0)
P(X≥1) = 1 - [ (e^-λ) * λ^0] / 0!
P(X≥1) = 1 - e^-2.67
P(X≥1) = 1 - 0.06945
P(X≥1) = 0.93055
P(X≥1) = 0.9306
Thus, the probability that at least 1 car arrives during the call is 0.9306.
I think it is 1 I am not sure tho
<em>y</em> - 1/<em>z</em> = 1 ==> <em>y</em> = 1 + 1/<em>z</em>
<em>z</em> - 1/<em>x</em> = 1 ==> <em>z</em> = 1 + 1/<em>x</em>
==> <em>y</em> = 1 + 1/(1 + 1/<em>x</em>) = 1 + <em>x</em>/(<em>x</em> + 1) = (2<em>x</em> + 1)/(<em>x</em> + 1)
<em>x</em> - 1/<em>y</em> = <em>x</em> - (<em>x</em> + 1)/(2<em>x</em> + 1) = (2<em>x</em> ² - 1)/(2<em>x</em> + 1) = 1
==> 2<em>x</em> ² - 1 = 2<em>x</em> + 1
==> 2<em>x</em> ² - 2<em>x</em> - 2 = 0
==> <em>x</em> ² - <em>x</em> - 1 = 0
==> <em>x</em> = (1 ± √5)/2
If you start solving for <em>z</em>, then for <em>x</em>, then for <em>y</em>, you would get the same equation as above (with <em>y</em> in place of <em>x</em>), and the same thing happens if you solve for <em>x</em>, then <em>y</em>, then <em>z</em>. So it turns out that <em>x</em> = <em>y</em> = <em>z</em>.
