Answer: a) P(x=0) = 0.0907, b) P(x≥10) = 0.7986
Step-by-step explanation: the probability mass function of a possion probability distribution is given as
P(x=r) = (e^-λ)×(λ^r) /r!
Where λ = fixed rate at which the event is occurring and each event is independent of each other = 2.4
a) P(x= at least one) = P(x≥1)
P(x≥1) = 1 - P(x<1)
But P(x<1) = P(x=0) { we can not continue to negative values because our values of x can only take positive values of integer}
Hence, P(x≥1) = 1 - P(x=0)
P(x=0) = e^-2.4 * 2.4^0/(0!)
P(x=0) = 0.0907×1/1
P(x=0) = 0.0907
b) if the average number of hits in 1 minutes is 2.4 then for 5 minutes we have 2.4×5 = 12.
Hence λ = 12.
P(x= at least 10) =P(x≥10) = 1 - P(x≤9)
P(x≤9) will be gotten using a cumulative possion probability distribution table whose area is to the left of the distribution.
From the table P(x≤9) = 0.2014.
P(x≥10) = 1 - 0.20140
P(x≥10) = 0.7986
Answer: x = -4, y = 0.5, z = 5 +t
Hi!
The line L whose direction is parallel to vector V a passes through point A
is parametrized
Where t, is a real number, and 
is a any point on line L.
In this case the direction is that of the z-axis , so V = (0, 0, 1)
A is the midpoint between points B = (0, -4, 9) and C=(-8, 5, 1)
The midpoint is A = (B + C)/2 = (-4, 0.5, 5)
Then the line is:
The equations for each coordinate are:
$17
17+11 = $28
28+11 = $39
39+11 = $50
17 + 11(n-1) $$
n-1 because removes the first class of $17
<span>Let r(x,y) = (x, y, 9 - x^2 - y^2)
So, dr/dx x dr/dy = (2x, 2y, 1)
So, integral(S) F * dS
= integral(x in [0,1], y in [0,1]) (xy, y(9 - x^2 - y^2), x(9 - x^2 - y^2)) * (2x, 2y, 1) dy dx
= integral(x in [0,1], y in [0,1]) (2x^2y + 18y^2 - 2x^2y^2 - 2y^4 + 9x - x^3 - xy^2) dy dx
= integral(x in [0,1]) (x^2 + 6 - 2x^2/3 - 2/5 + 9x - x^3 - x/3) dx
= integral(x in [0,1]) (28/5 + x^2/3 + 26x/3 - x^3) dx
= 28/5 + 40/9 - 1/4
= 1763/180 </span>
