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Ede4ka [16]
3 years ago
13

5/3 (6x + 9) ≤ 2x − 5 step by step please

Mathematics
1 answer:
Sati [7]3 years ago
6 0

Answer:

27.5 (is greater than) -2.5

>

Step-by-step explanation:

Distribute 5/3 to 6x and 9, (5/3•6x)+(5/3•9)=10x+15.

Set 10x+15 equal to 2x-5

10x+15=2x-5

(get rid of the smaller slope)

-2x -2x

8x+15=-5

+15 +15

8x=10

(divide by 8 to get the value of x)

8x divided by 8=x

10 divided by 8=1.25

x=1.25

Plug the known value of x back into the equations.

10(1.25)+15=27.5

2(1.25)-5=-2.5

27.5 (less than or equal to) -2.5 is incorrect. It would be

27.5 (is greater than) -2.5

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A nationwide survey of college seniors by the University of Michigan revealed that almost 70% disapprove of daily pot smoking. I
Whitepunk [10]

Answer:

a) P(X≥6) = 0.9614

b) P(7≤ X ≤9)  = 0.628

c) P(X≤5) = 0.0386

Step-by-step explanation:

This can be solved using the binomial distribution formula:

P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ

Where p = probability of success

           q = probability of failure = 1-p

           n = number of trials

           x = number of successful trials

We have p = 70% = 0.7

               n = 12

a) Find the probability that the number who disapprove of smoking pot daily is no less than 6 i.e. P(X≥6). This can be calculated as:

P(X≥6) = 1 - P(X<6)

        = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)]

        = 1 - [¹²C₀ (0.7)⁰(0.3)¹² + ¹²C₁ (0.7)¹(0.3)¹¹ + ¹²C₂ (0.7)²(0.3)¹⁰ + ¹²C₃ (0.7)³(0.3)⁹ + ¹²C₄ (0.7)⁴(0.3)⁸ + ¹²C₅ (0.7)⁵(0.3)⁷]

         = 1 - (0.000000531 + 0.0000148 + 0.0001909 + 0.00148 + 0.00779 + 0.02911)

         = 1 - 0.0386

P(X≥6) = 0.9614

b) P(7≤ X ≤9) = P(X=7) + P(X=8) + P(X=9)

                     = ¹²C₇ (0.7)⁷(0.3)¹²⁻⁷ + ¹²C₈ (0.7)⁸(0.3)¹²⁻⁸ + ¹²C₉ (0.7)⁹(0.3)¹²⁻⁹

                     = 0.158 + 0.231 + 0.239

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c) P(X≤5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)

               = ¹²C₀ (0.7)⁰(0.3)¹² + ¹²C₁ (0.7)¹(0.3)¹¹ + ¹²C₂ (0.7)²(0.3)¹⁰ + ¹²C₃ (0.7)³(0.3)⁹ + ¹²C₄ (0.7)⁴(0.3)⁸ + ¹²C₅ (0.7)⁵(0.3)⁷

   P(X≤5) = 0.0386

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Simplify to the answer.

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