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4 years ago

14

It is recommended that one fire extinguisher be available for every 4520 square feet in a building. Write and solve an equation

to determine X the number of fire extinguishers needed for a building that has 88, 140 square feet
Plzz help!!I'll give u five stars

1 answer:

stira [4]4 years ago

5 0

X=88,140/4520
x=19.5
Rounded up to the nearest unit, 20
Hope this helps

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Max pays a flat rate of $85 for his cell phone. He is also charged $0.15 for every text that he sends. His cell phone bill can b

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What is the binomial expansion of (2x-3)^5

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(2x - 3)⁵= 32x⁵ - 240x⁴ + 720x³ - 1080x² + 810x - 243

Step-by-step explanation:

We need to write the expansion of Binomial (2x - 3)⁵

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(a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + ⁿC₃aⁿ⁻³b³ + ... + ⁿCₙbⁿ

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+⁵C₄(2x)⁵⁻⁴(-3)⁴ + ⁵C₅(2x)⁵⁻⁵(-3)⁵

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That's the final answer.

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3 years ago

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The question is not complete. The full question and the solution is attached in the file below

Step-by-step explanation:

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> docx </span>

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> docx </span>

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3 years ago

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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

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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

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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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3 years ago