Based on the calculation below, the likelihood that if someone died they were a heavy smoker is 42.11%.
<h3>How do we calculate the likelihood of an occurrence?</h3>
Let:
Likelihood that a non-smoker will die = x
Therefore, we have:
Likelihood that a light smoker will die = 2 * x = 2x
Likelihood that a heavy smoker will die= 2 * 2x = 4x
From the above, we have:
Expected number of non smokers that will die = 50% * x = 0.5x
Expected number of light smokers that will die = 30% * 2x = 60%x = 0.6x
Expected number of heavy smokers that will die = 20% * 4x = 80%x = 0.8x
Total expected = 0.5x + 0.6x + 0.8x = 1.9x
Therefore, we have:
Likelihood that a heavy smoker died = Expected number of heavy smokers that will die / Total expected = 0.8x / 1.9x = 0.4211, or 42.11%
Learn more about likelihood here: brainly.com/question/14832179.
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Answer:
3/10
Step-by-step explanation:
8 blue marbles and 12 yellow marbles. = 20 marbles
P( yellow) = yellow/total =12/20 =3/5
P(plastic) =1/2
P(yellow, plastic) = 3/5 * 1/2 = 3/10
Answer:
0.06 and to change that to a percentage all you do is move the decimal point 2 places to the right so the answer would be <u>6</u><u> </u>percent
Answer: 45 ft²
<u>Step-by-step explanation:</u>
A = 
= 
= 9 * 5
= 45
Answer:
a) The set of solutions is 
y b) the set of solutions is 
.
Step-by-step explanation:
a) Let's first find the echelon form of the matrix ![\left[\begin{array}{ccc}5&2&6\\-2&1&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%262%266%5C%5C-2%261%263%5Cend%7Barray%7D%5Cright%5D)
.
- We add
from row 1 to row 2 and we obtain the matrix ![\left[\begin{array}{ccc}5&2&6\\0&\frac{9}{5} &\frac{27}{5}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%262%266%5C%5C0%26%5Cfrac%7B9%7D%7B5%7D%20%26%5Cfrac%7B27%7D%7B5%7D%5Cend%7Barray%7D%5Cright%5D)
- From the previous matrix, we multiply row 1 by
and the row 2 by 
and we obtain the matrix ![\left[\begin{array}{ccc}1&\frac{2}{5} &\frac{6}{5} \\0&1&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26%5Cfrac%7B2%7D%7B5%7D%20%26%5Cfrac%7B6%7D%7B5%7D%20%5C%5C0%261%263%5Cend%7Barray%7D%5Cright%5D)
. This matrix is the echelon form of the initial matrix.
The system has a free variable (x3).
- 0=x1+x2+
x3=
x1+(-3x3)+x3=
x1-x3+x3
then x1=0.
The system has infinite solutions of the form (x1,x2,x3)=(0,-3x3,x3), where x3 is a real number.
b) Let's first find the echelon form of the aumented matrix ![\left[\begin{array}{ccccc}1&-2&1&-4&1\\1&3&7&2&2\\0&-12&-11&-16&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%26-2%261%26-4%261%5C%5C1%263%267%262%262%5C%5C0%26-12%26-11%26-16%265%5Cend%7Barray%7D%5Cright%5D)
.
- To row 2 we subtract row 1 and we obtain the matrix
![\left[\begin{array}{ccccc}1&-2&1&-4&1\\0&5&6&6&1\\0&-12&-11&-16&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%26-2%261%26-4%261%5C%5C0%265%266%266%261%5C%5C0%26-12%26-11%26-16%265%5Cend%7Barray%7D%5Cright%5D)
- From the previous matrix, we add to row 3,
of row 2 and we obtain the matrix ![\left[\begin{array}{ccccc}1&-2&1&-4&1\\0&5&6&6&1\\0&0&\frac{17}{5}&\frac{-8}{5}&\frac{37}{5} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%26-2%261%26-4%261%5C%5C0%265%266%266%261%5C%5C0%260%26%5Cfrac%7B17%7D%7B5%7D%26%5Cfrac%7B-8%7D%7B5%7D%26%5Cfrac%7B37%7D%7B5%7D%20%20%20%5Cend%7Barray%7D%5Cright%5D)
.
- From the previous matrix, we multiply row 2 by and the row 3 by
and we obtain the matrix ![\left[\begin{array}{ccccc}1&-2&1&-4&1\\0&1&\frac{6}{5} &\frac{6}{5}&\frac{1}{5}\\0&0&1&\frac{-8}{17}&\frac{37}{17} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%26-2%261%26-4%261%5C%5C0%261%26%5Cfrac%7B6%7D%7B5%7D%20%26%5Cfrac%7B6%7D%7B5%7D%26%5Cfrac%7B1%7D%7B5%7D%5C%5C0%260%261%26%5Cfrac%7B-8%7D%7B17%7D%26%5Cfrac%7B37%7D%7B17%7D%20%5Cend%7Barray%7D%5Cright%5D)
. This matrix is the echelon form of the initial matrix.
The system has a free variable (x4).
- x3-
x4=
, then x3=+ ![\frac{8}{17}x4.x2+[tex]\frac{6}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B8%7D%7B17%7Dx4.%3C%2Fli%3E%3C%2Ful%3E%3Cul%3E%3Cli%3Ex2%2B%5Btex%5D%5Cfrac%7B6%7D%7B5%7D)
x3+x4=, x2+(+![\frac{8}{17}x4)+[tex]\frac{6}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B8%7D%7B17%7Dx4%29%2B%5Btex%5D%5Cfrac%7B6%7D%7B5%7D)
x4=, then
x2=
x4.
- x1-2x2+x3-4x4=1, x1+
+
x4++x4-4x4=1, then x1=
The system has infinite solutions of the form (x1,x2,x3,x4)=(-6,x4,+ x4,x4), where x4 is a real number.
