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Mathematics

2 answers:

sergiy2304 [10]4 years ago

8 0

Answer:

Gerald Ford: Foreign Affairs | Miller Center

https://millercenter.org/president/ford/foreign-affairs

Step-by-step explanation:

fredd [130]4 years ago

5 0

C. Ford granted Nixon an unconditional pardon for any wrongdoings against the United States.

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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

Use the point-slope form, y - y1 = m(x - x1), to find the linear equation of a line that passes through the points

Andreyy89

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4 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7B%28190%29%5E%7B3%7D%20%7B68%5E%7B2%7D%20%7D%20%20%5Ctimes%20%28%20%5Cfr

mrs_skeptik [129]

Answer:

$\dfrac{\sqrt[3]{95^2}}{17\cdot95^4}=\dfrac{\sqrt[3]{9\,025}}{1\,384\,660\,625}$

Step-by-step explanation:

The applicable rules of exponents are ...

(ab)^c = (a^c)(b^c)

(a^b)/(a^c) = a^(b-c)

$\dfrac{190^3}{68^2}\times\dfrac{34}{95^{\frac{19}{3}}}=\dfrac{(2\cdot 95)^3}{(2\cdot 34)^2}\cdot\dfrac{34}{95^6\cdot 95^{\frac{1}{3}}}=2^{3-2}95^{3-6-\frac{1}{3}}34^{1-2}\\\\=2\cdot 95^{-3\frac{1}{3}}\cdot 34^{-1}=2\cdot 95^{-4+\frac{2}{3}}\cdot 34^{-1}\\\\=\dfrac{2\sqrt[3]{95^2}}{95^4\cdot 34}=\dfrac{\sqrt[3]{95^2}}{17\cdot95^4}\\\\=\dfrac{\sqrt[3]{9\,025}}{1\,384\,660\,625}$