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Nimfa-mama [501]
2 years ago
5

last night, julie's pet hamster zoe kept julie awake for at least an hour running on her exercise wheel and scratching at the co

rner of her cage. zoe ran on the wheel at lease twice as long as she scratched at the corners of her cage, and she spent more than 1/4 hour running on her wheel. Which system of inequalities represents the time zoe kept juile awake last night if x is the number of hours zoe spent running on her wheel and y is the nu=umber of hours spent scratching her cage

Mathematics
1 answer:
borishaifa [10]2 years ago
6 0

Answer: x+y\geq 1, x>\frac{1}{4} ,  x\geq 2y and y\geq 0

Step-by-step explanation:

Here,  x represents the number of hours Zoe spent running on her wheel and y represents the number of hours spent scratching her cage.

Julie was awoke for at least an hour running on her exercise wheel and scratching the of her cage.

⇒ x+y\geq 1

She ran on her wheel at least twice as long as she scratched at the corners of her cage.

⇒ x\geq 2y

Also, She spent more than 1/4 hour running on her wheel.

⇒ x>\frac{1}{4}

And, we know that number of hours can not be negative.

⇒  y\geq 0

Therefore, the complete system of inequality which shows the given situation is,

x+y\geq 1, x>\frac{1}{4} and x\geq 2y, y\geq 0

Note: the feasible region ( covered by the given system) is shown in the below graph.


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A textbook has 500 pages on which typographical errors could occur. Suppose that there are exactly 10 such errors randomly locat
DiKsa [7]

Answer:

The probability of a  selection of 50 pages will contain no errors  is  0.368

The probability that the selection of the random pages will contain at least two errors is 0.2644

Step-by-step explanation:

From the information given:

Let q represent the no of typographical errors.

Suppose that there are exactly 10 such errors randomly located on a textbook of 500 pages. Let \mu be the random variable that follows a Poisson distribution, then mean \mu = \dfrac{10}{500}= 0.02

and the mean that the random selection of 50 pages will contain no error is \lambda = 50 \times 0.02 =1

∴

Pr(q= 0) = \dfrac{e^{-1} (1)^0}{0!}

Pr(q =0) = 0.368

The probability of a  selection of 50 pages will contain no errors  is  0.368

The probability that 50 randomly page contains at least 2 errors is computed as follows:

P(X ≥ 2) = 1 - P( X < 2)

P(X ≥ 2) = 1 - [ P(X = 0) + P (X =1 )]    since it is less than 2

P(X \geq 2) = 1 - [ \dfrac{e^{-1} 1^0}{0!} +\dfrac{e^{-1} 1^1}{1!} ]

P(X \geq 2) = 1 - [0.3678 +0.3678]

P(X \geq 2) = 1 -0.7356

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The probability that the selection of the random pages will contain at least two errors is 0.2644

6 0
2 years ago
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ivann1987 [24]
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。☆✼★ ━━━━━━━━━━━━━━  ☾  

% change = difference / original x 100

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Stay Brainly! ヅ    

- Ally ✧    

。☆✼★ ━━━━━━━━━━━━━━  ☾

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