Answer:
m = - 3
Step-by-step explanation:
a³ + 27 ← is a sum of cubes and factors in general as
a³ + b³ = (a + b)(a² - ab + b²), thus
a³ + 27
= a³ + 3³
= (a + 3)(a² - 3a + 9)
comparing a² - 3a + 9 to a² + ma + 9, then
m = - 3
Answer:
<u>Antonio had a head start of 3 meters.</u>
Step-by-step explanation:
I took the test on Edge and got it correct.
<em>(Sorry, this is late and you probably don't need it anymore.)</em>
Answer:
Step-by-step explanation:
Given that Bill, George, and Ross, in order, roll a die.
The first one to roll an even number wins and the game is ended.
Since Bill starts the game he can win by throwing even number or lose by throwing odd number
P(win) = 0.5, otherwise, the die will go to George. For Bill to win, both George and Ross should throw an odd number so that Bill again gets the chance with game non ending.
Thus we have Prob of Bill winning =P of Bill winning in I throw +P of Bill winning in his II chance of throw +....infinitely
To get back the dice once he loses probability
= p both throws odd = 
Thus Prob for Bill winning
= 
This is an infinite geometric series with I term 0.5 and common ratio 0.125<1
Sum = 
Answer:
a) P(identified as containing explosives)=P(actually contains explosives and identified as containing explosives)+P(actually not contains explosives and identified as containing explosives)
=(10/(4*106))*0.95+(1-10/(4*106))*0.005 =0.005002363
hence probability that it actually contains explosives given identified as containing explosives)
=(10/(4*106))*0.95/0.005002363=0.000475
b)
let probability of correctly identifying a bag without explosives be a
hence a =0.99999763 ~ 99.999763%
c)
No as even if that becomes 1 ; proportion of true explosives will always be less than half of total explosives detected,