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Korvikt [17]
3 years ago
14

Choose the solution to this inequality.

Mathematics
1 answer:
natita [175]3 years ago
3 0

Answer:

C

Step-by-step explanation:

If you used the inequality in C, b being able to be as high as -23, it would make 72 is greater than or equal to 72, which is true.

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I need to know how to graph y=-2(x-1)^2-4
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Simplify each expression <br><br> 1. 2x + (x - 4) + (2y - 5)<br><br><br> 2. 3(x - 2) + 4(x + 3)
Shtirlitz [24]

Answer:

1. 3x-9+2y

2. 7x+6

Step-by-step explanation:

1. 2x + (x - 4) + (2y - 5)

  2x+x-4+2y-5

  3x-9+2y

2. 3(x - 2) + 4(x + 3)

  3x+4x-6+12

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2 years ago
Read 2 more answers
What is the measure of angle C A B?
Mamont248 [21]

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8 0
3 years ago
Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2
leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where \lambda = \frac{10.2}{200} = 0.051

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

\frac{1}{4} \times 200 = 50

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004

c)The expected number of grams to be eaten before encountering the first fragments :

E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 grams

7 0
3 years ago
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