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Minchanka [31]
2 years ago
14

Arrange in ascending order a) 2⁄5, -1⁄6, -11/30

Mathematics
1 answer:
nikitadnepr [17]2 years ago
8 0

Answer:

-11/30, -1/6, 2/5

Step-by-step explanation:

first, make the question in the same denominator

= 12/30, -5/30, -11/30

= -11/30, -5/30, 12/30

= -11/30, -1/6, 2/5

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Let X = the time between two successive arrivals at the drive-up window of a local bank. If X has an exponential distribution wi
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Answer:

a) E(X)=1

b) σ=1

c) P(X>1)=0.368

d) P(2<X<5)=0.128

Step-by-step explanation:

We have X: "the time between two successive arrivals at the drive-up window of a local bank", exponentially distributed with λ = 1.

a) We have to compute the expected time between two successive arrivals.

The expected value for X as it is exponentially distributed is:

E(X)=\dfrac{1}{\lambda}=\dfrac{1}{1}=1

b) We have to compute the standard deviation of X.

The standard deviation is calculated as:

\sigma=\dfrac{1}{\lambda}=\dfrac{1}{1}=1

c) The probability that X is larger than 1 (P(X>1))

We can express the probability as:

P(X>t)=e^{-\lambda t}\\\\P(X>1)=e^{-\lambda\cdot 1}=e^{-1}=0.368

d) The probability that X is between 2 and 5 (P(2<X<5))

P(X>t)=e^{-\lambda t}\\\\\\P(25)\\\\P(2

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3 years ago
Someone please help me understand how you expand and simplify brackets:(
tatiyna
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3 0
3 years ago
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Grace [21]

Answer:

  • diagram is below
  • 6, 11, 16, 21
  • s[n] = s[n-1] +5
  • 26, 31, 36

Step-by-step explanation:

a) See below for the next diagram in sequence.

__

b) The numbers of straws in each diagram are ...

  6, 11, 16, 21, ...

__

c) Each term is 5 more than the previous one, so the recursive rule for the number of straws is ...

  s[1] = 6

  s[n] = s[n-1] +5

__

d) The next three terms of the sequence are ...

  ..., 26, 31, 36, ...

6 0
3 years ago
Hey everybody just checking in if you say hi I’ll give BRAINLIEST!
nordsb [41]
Hi... how are you ?
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A technician is launching fireworks near the event of a show. Of the remaining 11 fireworks, 4 are blue and 7 are red. If she la
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\displaystyle |\Omega|=\binom{11}{6}=\dfrac{11!}{6!5!}=\dfrac{7\cdot8\cdot9\cdot10\cdot11}{2\cdot3\cdot4\cdot5}=462\\ |A|=\binom{4}{2}=\dfrac{4!}{2!2!}=\dfrac{3\cdot4}{2}=6\\\\ P(A)=\dfrac{6}{462}=\dfrac{1}{77}\approx1\%

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