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3 years ago

PLEASE HELP!!!

Mathematics

2 answers:

frutty [35]3 years ago

8 0

Answer:

1st question: C

2nd question: Increasing (the slope)

3rd question: don't understand what's being asked

Step-by-step explanation:

arsen [322]3 years ago

7 0

Answer:

the answers are

Step-bythe answers are -step explanation:

1.C

2.a

3.B

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Please help!!!!!!!!!!!!!!!!

dedylja [7]

Answer:

b is the answer

Step-by-step explanation:

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3 years ago

<img src="https://tex.z-dn.net/?f=v%20%3D%20%20%5Cfrac%7Ber%7D%7Br%20%2B%20a%7D%20" id="TexFormula1" title="v = \frac{er}{r + a

Mrac [35]

Answer:

r = va/(e - v)

Step-by-step explanation:

v = er/(r + a)

vr + va = er

va = er - vr

va = r(e - v)

r = va/(e - v)

6 0

3 years ago

Read 2 more answers

What is the greatest common factor for 66 and 121

Stels [109]

11 is the greatest common factor

4 0

3 years ago

a man is 13 years older than his wife. The wife is 3 times as old as her daughter,who is 17 years old. How old was the man when

Molodets [167]

Step-by-step explanation:

the man in the question was 64 years old when she was born

5 0

3 years ago

Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a

Bumek [7]

Answer:

0.381 is the probability that the number of drivers will be at most 18.

Step-by-step explanation:

We are given the following information in the question:

The number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter μ = 20.

The Poisson distribution is the discrete probability distribution of the number of events occurring in a given time period, given the average number of times the event occurs over that time period.
The variance of Poisson distribution is equal to the mean of Poisson distribution.

a) P(number of drivers will be at most 18)

Formula:

$P(X =k) = \displaystyle\frac{\mu^k e^{-\mu}}{k!}\\\\ \mu \text{ is the mean of the distribution}$

$P( x \leq 18) =P(x=0) + P(x =1) + P(x = 2) + ... + P(x = 18)\\\\= \displaystyle\frac{20^0 e^{-20}}{0!} + \displaystyle\frac{20^1 e^{-20}}{1!} +...+ \displaystyle\frac{20^{18} e^{-20}}{18!}\\\\ = 0.381$

Thus, 0.381 is the probability that the number of drivers will be at most 18.

3 0

4 years ago