Mathematics, 15.11.2019 03:31 theflash077

One piece of pvc pipe is to be inserted inside another piece. the length of the first piece is normally distributed with mean value 20 in. and standard deviation .5 in. the length of the second piece is a normal rv with mean and standard deviation 15 in. and .4 in., respectively. the amount of overlap is normally distributed with mean value 1 in. and standard deviation .1 in. assuming that the lengths and amount of overlap are independent of one another, what is the probability that the total length after insertion is between 34.5 in. and 35 in.?

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Answer from: danielwu462

0.159 is the probability that the total length after insertion is between 34.5 inch and 35 inch.

Step-by-step explanation:

We are given the following information in the question:

The length of the first piece is normally distributed with mean value 20 inch and standard deviation 0.5 inch.

The length of the second piece is a normal with mean and standard deviation 15 inch and 0.4 inch respectively.

The amount of overlap is normally distributed with mean value 1 inch and standard deviation 0.1 inch.

The lengths and amount of overlap are independent of one another.

Let the total length of PVC pipe be represented by Y.

The sum of two independent normally distributed random variables is normal, with its mean being the sum of the two means, and its variance being the sum of the two variances

Then, Y follows a normal distribution with

Y = Length of first pipe + Length of second Pipe - Overlap

$\text{Mean} = \text{Mean of first pipe} + \text{Mean of second pipe} - \text{Mean of overlap}\\= 20 + 15 - 1 = 34$

$\text{Variation} = \text{Variation of first pipe} + \text{Variation of second pipe} - \text{Variation of overlap}\\= (0.5)^2 + (0.4)^2 + (0.1)^2 = 0.42$

$\text{Standard deviation} = \sqrt{Variance} = \sqrt{0.42} = 0.648$

P(total length after insertion is between 34.5 inch and 35 inch)

$P(34.5 \leq y \leq 35) = P(\displaystyle\frac{34.5 - 34}{0.648} \leq z \leq \displaystyle\frac{35-34}{0.648}) = P(0.7716 \leq z \leq 1.5432)\\\\= P(z \leq 1.5432) - P(z < 0.7716)\\= 0.939 -0.780 = 0.159 = 15.9\%$

$P(34.5 \leq y \leq 35) = 15.9\%$

0.159 is the probability that the total length after insertion is between 34.5 inch and 35 inch.

Answer from: Quest

step-by-step explanation:

let

we know that

the y-intercept is the y-coordinate of the point when the value of x is equal to zero

the y-coordinate of the point a is the y-intercept

the answer part a) is

the value of y-intercept is

part b) find the slope m

we know that

the value of the slope is equal to

substitute the values

therefore

the answer part b) is

the value of the slope is equal to

the general answer is

the slope is

the value of y-intercept is

Answer from: Quest

solution:

the given function is ,

$f(x) = 4 x^3 - 72 x^2 + 320 x,$

= $4x( x^2- 18 x + 80)\\\\ 4 x(x^2-10 x -8 x + 80) \\\\ 4 x[(x-10)(x-8)]$

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

$f(x) = a x^3+ b x^2 + c x+d,$

sum of the roots = a + b +c= 0+8+10=18= $\frac{-b}{a}$

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 = $\frac{c}{a}$

product of roots = a b c = 0= $\frac{d}{a}$

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function $f(x) = 4 x^3 - 72 x^2 + 320 x,$ is given by entry and exit route of this hall.

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