The probability that a lab specimen contains high levels of contamination is 0.10. five samples are checked, and the samples are independent. (a) what is the probability that none contain high levels of contamination? (b) what is the probability that exactly one contains high levels of contamination? (c) what is the probability that at least one contains high levels of contamination?

(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

: the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

for i = 1, 2, 3, 4, 5

The complement for is given by

: the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P()=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P(∩∩∩∩)==0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P(∩∩∩∩)+P(∩∩∩∩)+P(∩∩∩∩)+P(∩∩∩∩)+P(∩∩∩∩)=5×(0.1)×=0.32805

because we have independent events.

(c) The probability that at least one contains high levels of contamination is

P(∪∪∪∪)=1-P(∩∩∩∩)=1-0.59049=0.40951

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

i dont know

step-by-step explanation: