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Stolb23 [73]
1 year ago
13

A worn out bottling machine does not properly apply caps to 5% of the bottles it fills. If you randomly select 20 bottles from t

hose produced by this machine, what is the approximate probability that between 2 and 6 (inclusive) caps have been improperly applied?.
Mathematics
1 answer:
guajiro [1.7K]1 year ago
6 0

Probability would been equal to 0.9841 if we randomly select 20 bottles it fills.

<h3>What is Probability ?</h3>

The ratio of good outcomes to all possible outcomes of an event is known as the probability. The number of the  positive results for the experiment with 'n' outcomes can also be represented by the symbol x. The probability of an event can be calculated using the following formula.

Probability(Event) = Positive Results/Total Results = x/n

Given,

n = 20

p = 0.05

Thia information gives information about binomial distribution with parameter n = 20 , p = 0.05

Let X be the number of bottles have caps that she not properly applied

    X∝ B( n=20 , p= 0.05)

The p.m.f of X is

P( X=0)= (\frac{20}{x}) ( 0.05)ˣ (1 - 0.05)²⁰ ⁻ˣ

                                                       ; x = 0 , 1, 2.......20

                                                         0 < p < 1

                                                         q= 1 - p

= 0                                                    ; otherwise

P {x ≤ 3} = P { X = 0 } + P{ x= 1 } + P { x= 2 } + P { x= 3}

P( X=0)= (\frac{20}{0}) ( 0.05)⁰ (1 - 0.05)²⁰ ⁻⁰  = 0.3585

P( X = 1)= 0.3774

P( X= 2) = 0.1887

P( X= 3) = 0.0596

                = 0.3585+ 0.3774 +  0.1887+ 0.0596

                 = 0.9841

To know more about Probability, Visit:

brainly.com/question/11234923

#SPJ4

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Step-by-step explanation:

1. 4% commission on $1400

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total earnings = 56+24

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= 4/100*pp

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total earnings = 0.04pp+0.03cc

8 0
3 years ago
What is the volume, in cubic ft, of a cylinder with a height of 17ft and a base radius of 5ft, to the nearest tenths place?
sweet [91]

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8 0
4 years ago
If the 5th term of a geometric progression is 162 and the 8th term is 4374, find the (i) 1st three terms of the sequence; (ii) s
Alona [7]

Answer:

see explanation

Step-by-step explanation:

The nth term of a geometric progression is

a_{n} = a₁r^{n-1}

where a₁  is the first term and r the common ratio

Given a₅ = 162 and a₈ = 4374 , then

a₁r^{4} = 162 → (1)

a₁r^{7} = 4374 → (2)

Divide (2) by (1)

\frac{a_{1}r^{7}  }{a_{1}r^{4}  } = \frac{4374}{162}

r³ = 27

r = \sqrt[3]{27} = 3

Substitute r = 3 into (1) and solve for a₁

a₁(3)^{4} = 162

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a₁ = \frac{162}{81} = 2

Then

a₂ = a₁ × 3 = 2 × 3 = 6

a₃ = a₂ × 3 = 6 × 3 = 18

The first 3 terms are 2, 6, 18

(ii)

The sum to n terms of a geometric progression is

S_{n} = \frac{a_{1}(r^{n}-1)  }{r-1} , then

S_{10} = \frac{2(3^{10}-1) }{3-1}

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The volume of this rectangular solid WITHOUT a top is
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Answer:

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Step-by-step explanation:

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This question has a L=10, B= 6 and H=4 .

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Answer:

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divide by 3

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m=-8/3

4 0
3 years ago
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