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butalik [34]
2 years ago
7

Am a little lost so can you help me?​

Mathematics
2 answers:
Scilla [17]2 years ago
8 0
I believe the answer to this question is B . Which is (-♾ ,♾)
Elden [556K]2 years ago
6 0

Answer:

se eu não me engano é 10 graus.

Step-by-step explanation:

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I need to find the limit
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the answer is 1/-4

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PLZ HURRY! WILL GIVE BRAINELST! WORTH 100
Anit [1.1K]

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Could you screen-clip the question? I can't really understand.

Step-by-step explanation:

You can edit your question and I can edit my answer :)

~PumpkinSpice1

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2 years ago
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Solve this: 2-2÷2×2+2​
sergey [27]

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-2 I think

Step-by-step explanation:

6 0
3 years ago
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The problem of matching aircraft to passenger demand on each flight leg is called the flight assignment problem in the airline i
Darina [25.2K]

Answer:

Step-by-step explanation:

Given that the demand for the 6 p.m. flight from Toledo Express Airport to Chicago's O'Hare Airport on Cheapfare Airlines is normally distributed with a mean of 132 passengers and a standard deviation of 42

Let X be the no of passengers who report

X is N(132, 42)

Or Z is \frac{x-132}{42}

a) Suppose a Boeing 757 with a capacity of 183 passengers is assigned to this flight.

the probability that the demand will exceed the capacity of this airplane

=P(X>183) = P(Z>1.21) =

=0.5-0.3869\\=0.1131

b)  the probability that the demand for this flight will be at least 80 passengers but no more than 200 passengers

=P(80\leq x\leq 200)\\= P(-1.23\leq z\leq 1.62)\\

=0.4474+0.3907

=0.8381

c) the probability that the demand for this flight will be less than 100 passengers

=P(x

d) If Cheapfare Airlines wants to limit the probability that this flight is overbooked to 3%, how much capacity should the airplane that is used for this flight have? passengers

=P(Z>c)=0.03\\c=1.88\\X=132+1.88(42)\\=210.96

e) 79th percentile of this distribution

=132+0.81(42)\\= 166.02

7 0
3 years ago
1. Which of the following is a true statement?
san4es73 [151]

Answer:

<u><em>C. To square a number, multiply the number by itself.</em></u>

4 0
1 year ago
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