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Anna007 [38]
3 years ago
15

Liza wants to have more than $750 in her savings account before she will leave for college. Right now she has $300, and she is g

oing to babysit to earn $12.00 an hour. How many hours will Liza need to babysit before she leaves for college?
Mathematics
1 answer:
Kazeer [188]3 years ago
6 0
She will need a minimum of 38 hours, at 38 hours she will make 456, plus the 300 she already has, so that in total is 756.
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What number is a common multiple of<br>5 and 97?​
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5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100,105,110,115,120,125,130,135,140,145,150,155,160,165,170,175,180,185,190,195,200,205,210,215,220,225,230,235,240,245

Step-by-step explanation:

5 0
4 years ago
A nationwide survey of college seniors by the University of Michigan revealed that almost 70% disapprove of daily pot smoking. I
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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

   P(7≤ X ≤9)  = 0.628

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

4 0
3 years ago
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