Answer:
MEEEEEEEEEEEEE LOL hahaha
Answer:
For first lamp ; The resultant probability is 0.703
For both lamps; The resultant probability is 0.3614
Step-by-step explanation:
Let X be the lifetime hours of two bulbs
X∼exp(1/1400)
f(x)=1/1400e−1/1400x
P(X<x)=1−e−1/1400x
X∼exp(1/1400)
f(x)=1/1400 e−1/1400x
P(X<x)=1−e−1/1400x
The probability that both of the lamp bulbs fail within 1700 hours is calculated below,
P(X≤1700)=1−e−1/1400×1700
=1−e−1.21=0.703
The resultant probability is 0.703
Let Y be a lifetime of another lamp two bulbs
Then the Z = X + Y will follow gamma distribution that is,
X+Y=Z∼gamma(2,1/1400)
2λZ∼
X+Y=Z∼gamma(2,1/1400)
2λZ∼χ2α2
The probability that both of the lamp bulbs fail within a total of 1700 hours is calculated below,
P(Z≤1700)=P(1/700Z≤1.67)=
P(χ24≤1.67)=0.3614
The resultant probability is 0.3614
First four answers are correct as congruent parts of congruent triangles are congruent but L Is not the opposite to M so they wouldn’t be congruent nor does NL or PQ have the same connection
Answer:
(a) Shown below
(b) There is a positive relation between the number of assemblers and production.
(c) The correlation coefficient is 0.9272.
Step-by-step explanation:
Let <em>X</em> = number of assemblers and <em>Y</em> = number of units produced in an hour.
(a)
Consider the scatter plot below.
(b)
Based on the scatter plot it can be concluded that there is a positive relationship between the variables <em>X</em> and <em>Y</em>, i.e. as the value of <em>X</em> increases <em>Y</em> also increases.
(c)
The formula to compute the correlation coefficient is:
![r=\frac{n\sum XY-\sum X\sum Y}{\sqrt{[n\sum X^{2}-(\sum X)^{2}][n\sum Y^{2}-(\sum Y)^{2}]}} }](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bn%5Csum%20XY-%5Csum%20X%5Csum%20Y%7D%7B%5Csqrt%7B%5Bn%5Csum%20X%5E%7B2%7D-%28%5Csum%20X%29%5E%7B2%7D%5D%5Bn%5Csum%20Y%5E%7B2%7D-%28%5Csum%20Y%29%5E%7B2%7D%5D%7D%7D%20%7D)
Compute the correlation coefficient between <em>X</em> and <em>Y</em> as follows:
![r=\frac{n\sum XY-\sum X\sum Y}{\sqrt{[n\sum X^{2}-(\sum X)^{2}][n\sum Y^{2}-(\sum Y)^{2}]}} }=\frac{(5\times430)-(15\times120)}{\sqrt{[(5\times55)-15^{2}][(5\times3450)-120^{2}]}} =0.9272](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bn%5Csum%20XY-%5Csum%20X%5Csum%20Y%7D%7B%5Csqrt%7B%5Bn%5Csum%20X%5E%7B2%7D-%28%5Csum%20X%29%5E%7B2%7D%5D%5Bn%5Csum%20Y%5E%7B2%7D-%28%5Csum%20Y%29%5E%7B2%7D%5D%7D%7D%20%7D%3D%5Cfrac%7B%285%5Ctimes430%29-%2815%5Ctimes120%29%7D%7B%5Csqrt%7B%5B%285%5Ctimes55%29-15%5E%7B2%7D%5D%5B%285%5Ctimes3450%29-120%5E%7B2%7D%5D%7D%7D%20%3D0.9272)
Thus, the correlation coefficient is 0.9272.