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2 years ago

Dimitri drew a right angle. Then he drew an angle that was 20° larger. What is the measure of the second angle Dimitri drew?

Mathematics

2 answers:

marysya [2.9K]2 years ago

4 0

Right angle is 90 degrees. 90 + 20 = 110 degrees

seraphim [82]2 years ago

4 0

The measure would be 110º because a right angle is 90º, and if the angle is 20º larger, you add 90º + 20º and you get 110º

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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

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