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Nata [24]
3 years ago
15

2y +3<11 What is the largest value of y could be?

Mathematics
2 answers:
Zanzabum3 years ago
6 0

Answer:

Step-by-step explanation:

2y + 3<11                       Subtract 3 from both sides

2y < 11 - 3

2y < 8                           Divide by 2

2y/2 <8/2                  

y < 4

The question is very hard to answer. That's because the answer is very close to 4, but it is not 4.

For example 3.8 is a possible answer. Try it

2*3.8 + 3 < 11

7.6 + 3 < 11

10.6 < 11

But is that the largest value possible? No.

For example

y could be 3.99999999 Well maybe that could be the answer. No it isn't. Because, though very close to 11, it still cannot be 11. It might round to 11, but it does not equal 11.

I don't exactly know what you are expected to answer. I can only say that y cannot be 4, but it can be as close to 4 as you care to make it. If the question means what is the largest integer that y could be, then the answer is 3.

Evgesh-ka [11]3 years ago
5 0
The greatest value of y is 2y+3=11 => y=4
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Which expression is equivalent 5(2x-7)<br> a.2x-25<br> b.7x-12<br> c.7x-12<br> d.10x-7<br> e.10x-35
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Answer:

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5. A company sells small, colored binder clips in packages of 20 and offers a money-back guarantee if two or more of the clips a
SOVA2 [1]

Answer:

a) Binomial.

b) n=20, p=0.01, k≥2

The probability hat a package sold will be refunded is P=0.0169.

Step-by-step explanation:

a) We know that

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  • the sample size is bigger than one subject.

The most appropiate distribution to represent this random variable is the binomial.

b) The parameters are:

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  • Probability of defective clips: p=0.01.
  • number of defective clips that trigger the money-back guarantee: k≥2

The probability of the package being refunded can be calculated as:

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4 years ago
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