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Nikitich [7]
3 years ago
14

On a particular production line, the likelihood that a light bulb is defective is 10%. seven light bulbs are randomly selected.

What is the probability that at most 4 of the light bulbs will be defective
Mathematics
1 answer:
amid [387]3 years ago
6 0

Answer:

0.9995

Step-by-step explanation:

10% = 0.10

1 - 0.10 = 0.9

n = number of light bulbs = 7

we calculate this using binomial distribution.

p(x) = nCx × p^x(1-p)^n-x

our question says at most 4 is defective

= (7C0 × 0.1⁰ × 0.9⁷) + (7C1 × 0.1¹ × 0.9⁶) + (7C2 × 0.1² × 0.9⁵) + (7C3 × 0.1³ × 0.9⁴) + (7C4 × 0.1⁴ × 0.9³)

= 0.478 + 0.372 + 0.1239 + 0.023 + 0.0026

= 0.9995

we have 0.9995 probability that at most 4 light bulbs are defective.

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PLEASE HELP WITHIN 5 MINUTES! Be as accurate as possible!
Nookie1986 [14]

Answer:

2,75in

4,25in

33in²

Step-by-step explanation:

shorter base is 2,75in

larger base is 4,25in

S=11+10,5+9+2,5=33in²

8 0
2 years ago
I need help with this question
Murrr4er [49]

The number of cows is given by

14\cdot 2^{\frac{y}{5}}

So, after k years, the number of cows will be

14\cdot 2^{\frac{y+k}{5}}

We want this number to be twice as much as the original:

14\cdot 2^{\frac{y+k}{5}} = 2(14\cdot 2^{\frac{y}{5}})

First of all, we can cancel 14 from both sides:

2^{\frac{y+k}{5}} = 2\cdot 2^{\frac{y}{5}}

Finally, on the right hand side, we can use the exponent rule

a^b\cdot a^c=a^{b+c}

to get

2^{\frac{y+k}{5}} = 2^{\frac{y}{5}+1}

To solve this equation, we must impose that the two exponents are the same:

\dfrac{y+k}{5} = \dfrac{y}{5}+1 \iff \dfrac{y+k}{5} = \dfrac{y+5}{5}

And clearly this is true if and only if k=5. So, it will take 5 years for the cow heard to double in number.

You can do the exact same steps to find the doubling time for the sheeps.

8 0
3 years ago
Need the answer Thanks
zlopas [31]
What is the question its to blurry
8 0
3 years ago
What Times what equals 66
tankabanditka [31]
This is a question that has several answers to it. There is no fixed way to find the answer to this question. Only thing is either your intuition or trial and error method. Let us now get down to the problem.

1 x 66 = 66
2 x 33 = 66
3 x 22 = 66
6 x 11 = 66

I hope the answer is clear to you and this is the answer that has actually come to your desired help. <span />
4 0
3 years ago
Read 2 more answers
Hospital sells raffle tickets to raise fund for new medical equipment. Last year, 2000 tickets were sold for $24 each. The fund-
Zanzabum

Answer:

1) Price decrease = $4; 2) new price = $20; 3) maximum revenue = $50 000

Step-by-step explanation:

The hospital sold 2000 tickets for $24 each

Revenue = price per ticket × number of tickets sold

Let x = change in price

New price = 24 - x

New number of tickets sold = 2000 + 125x

1) Calculate change in price to maximize revenue

y = (24 - x)(2000 + 125x)

y = 48 000 + 1000x - 125x²

y = -125x² + 1000x + 48 000

a = -125; b = 1000; c = 48 000

The vertex is at

x = -\dfrac{b}{2a} = -\dfrac{1000}{2(-125)}= \dfrac{1000}{250} = \mathbf{4}

A price decrease of $4 will maximize revenue.

2) New ticket price

Original price = $24

Price change =  <u>  - 4 </u>

New price =       $20

A ticket price of $20 will maximize revenue.

3) Maximum revenue

         Ticket price = $20

No of tickets sold = 2000 + 125(4) = 2000 + 500  = 2500

             Revenue = 2500 × $20 = $ 50 000

The maximum revenue is $50 000.

The graph below slows the relation between the price drop and total revenue.  A price drop of $4 results in a maximum revenue of $50 000.

8 0
3 years ago
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