What’s the slope of (-5,7) and (5,-3)

Mathematics

1 answer:

Oksana_A [137]3 years ago

7 0

Answer:

slope = - 1

Step-by-step explanation:

m = (y₂ - y₁) / (x₂ - x₁)

m = (-3 - 7) / (5 - (-5)) = -10 / 10 = -1

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The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are

raketka [301]

Answer:

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

Step-by-step explanation:

Let's define

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

$P(A_{i})=0.1$ for i = 1, 2, 3, 4, 5

The complement for $A_{i}$ is given by

$A_{i}^{$c$}$ : the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P( $A_{i}^{$c$}$ )=0.9 for i = 1, 2, 3, 4, 5

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

P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )= $(0.9)^{5}$ =0.59049 because we have independent events.

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

P( $A_{1}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}$ )=5×(0.1)× $(0.9)^{4}$ =0.32805

because we have independent events.

P( $A_{1}$ ∪ $A_{2}$ ∪ $A_{3}$ ∪ $A_{4}$ ∪ $A_{5}$ )=1-P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )=1-0.59049=0.40951