Mathematics: An article in a magazine discussed the length of time till failure of a particular product. At the end of theâ product s lifetime, the time till failure is modeled using an exponential distribution with a mean of thousand hours. In reliability jargon this is known as theâ wear-out distribution for the product. During its normalâ (useful) life, assume theâ product s time till failure is uniformly distributed over the range thousand to million hours. Complete parts a through c. a. At the end of the

An article in a magazine discussed the length of

Ответ:

AnxiousKid

03.05.2021

Mathematics

Step-by-step explanation:

From the missing information;

Lets assume that;

$\text{the exponential distribution mean = 500 thousand naira}$

$\text{Range 100 thousand}$ → $\text{1 million naira}$

∴

$X \sim exp (\dfrac{1}{500}) \\ \\ P( X \le x) = 1 - e^{\dfrac{-x}{500}} \\ \\ P(X < 700) = 1 - e^{\frac{-7}{5}} \\ \\ \mathbf{P(X< 700) = 0.75340}$

$Y \sim \cup (100,1000) \implies P(Y \le y ) = \dfrac{y - 100}{1000-100} \\ \\ P(Y < 700) = \dfrac{600}{900} = 0.67$

$P(X < 830)= 1 - e^{\frac{-830}{500}} \\ \\ P(X < 830)= 1 - 0.1901 \\ \\ P(X < 830)=0.8099 \\ \\ P(Y < 830) = \dfrac{730}{900} = 0.8111 \\ \\ Thus;\mathbf{ P(X < 0.8099) = P(Y< 0.8111)}$

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Ответ:

si1baalmasri

12.10.2022

Mathematics

XY = 2 units

Step-by-step explanation:

Given that :

ΔEFG ≅ ΔXYZ

Then, segments :

EF = XY, FG = YZ and EG = XZ

GIVEN THAT :

E (0, 0), F (0, 2), and G (3, 0)

We can find the distance between coordinates E (0, 0), F (0, 2) to get the measure of XY since they are both congruent.

EF = sqrt[(x2 - x1)^2 + (y2 - y1)^2]

x1 = 0, y1= 0, x2 = 0, y2 = 2

EF = sqrt[(0 - 0)^2 + (2 - 0)^2]

EF = sqrt[ 0 + 4]

EF = 2

SINCE EF = XY

XY = 2 units

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