Answer:
X = 28+3(6)
X = 28 +18
<u>X = 46</u>
Thus, the <u>value of X </u>will be <u>46</u>.
Hope it helps.
Answer:
The probability that a person with restless leg syndrome has fibromyalgia is 0.183.
Step-by-step explanation:
Denote the events as follows:
<em>F</em> = a person with fibromyalgia
<em>R</em> = a person having restless leg syndrome
The information provided is as follows:
P (R | F) = 0.33
P (R | F') = 0.03
P (F) = 0.02
Consider the tree diagram attached below.
Compute the probability that a person with restless leg syndrome has fibromyalgia as follows:


Thus, the probability that a person with restless leg syndrome has fibromyalgia is 0.183.
FAXT
like i started today but i can't stoppppp
Answer:
Step-by-step explanation:
The population in 2003= 47000
Since the population increase by 1200 every year,
in 2004 the population will be 47000+1200
in 2005 the population will be 47000+(1200+1200) which is the same as
47000+2(1200) where 2 is 2 years after 2003,
Therefore the population x years after 2003 is 47000+x(1200).
P= 47000+1200x
b) The population at 2009 which is 6 years after 2003 will be
47000+(1200)*6=47000+7200= 54200
The population at 2009 is 54200,
Answer:
4×10⁶ is the answer.........