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3 years ago

Y= -3x+5 please solve

Mathematics

2 answers:

GarryVolchara [31]3 years ago

8 0

The answer is x = - y-5/3

I hope this helps. :)

pychu [463]3 years ago

6 0

X can be from -1.6 to -1.7 or x can be from -3.3 to 3.3

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Hey everyone! A little help will be needed/appreciated

lesya [120]

Answer:

2 and 1/4

Step-by-step explanation:

-5 1/4 - (-7 1/2)=

Rewrite as

-5.25 + 7.5 =2.25

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3 years ago

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David works at a nearby electronic store. He makes commission of 15% on everything he sells. If he sells a laptop for $293.00 ho

Lynna [10]

The answer would be $439.50

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3 years ago

Ms. Green tells you that a right triangle has a hypotenuse of 12 and a leg of 5. She asks you to find the other leg of the trian

Anit [1.1K]

let the other leg be x cm.

(5)^2+(x)^2=(12)

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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

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

6 0

3 years ago

NEED HELP ASAP WILL MARK BRAINLIEST IF RIGHT!!!

Zarrin [17]

Answer: Area = x² + 6x + 8

Step-by-step explanation:

Area = length x width

Area = (x+2)(x+4)

Area = (x)(x) + (x)(4) + (2)(x) + (2)(4)

Area = x² + 4x + 2x + 8

Area = x² + 6x + 8

6 0

3 years ago