1 decigram is 100 milligrams so the flea weighs 800 milligrams and 800-3=797 so the flea weighs 727 milligrams more than the ant
Answer:
x ≅ 20.10 torr
Step-by-step explanation:
Given that the equation which models blood pH in the question is;
pH(x)=6.1+log(800/x)
where;
pH = 7.7
x = partial pressure of carbon dioxide in arterial blood, measured in torr.
we are asked to find (x)
In order to do that, we use the given equation:
pH(x)=6.1+log(800/x)
since pH = 7.7
7.7 = 6.1 + log (800/x)
7.7 - 6.1 = log (800/x)
1.6 = log (800/x)


x = 20.09509145
x ≅ 20.10 torr
Since there is no number in the thousandths place which would make the number in the hundredths place round up or down, we keep the number the same:
314.16
Hope it helps! Good luck. :)
Answer:
a) No
b) 42%
c) 8%
d) X 0 1 2
P(X) 42% 50% 8%
e) 0.62
Step-by-step explanation:
a) No, the two games are not independent because the the probability you win the second game is dependent on the probability that you win or lose the second game.
b) P(lose first game) = 1 - P(win first game) = 1 - 0.4 = 0.6
P(lose second game) = 1 - P(win second game) = 1 - 0.3 = 0.7
P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%
c) P(win first game) = 0.4
P(win second game) = 0.2
P(win both games) = P(win first game) × P(win second game) = 0.4 × 0.2 = 0.08 = 8%
d) X 0 1 2
P(X) 42% 50% 8%
P(X = 0) = P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%
P(X = 1) = [ P(lose first game) × P(win second game)] + [ P(win first game) × P(lose second game)] = ( 0.6 × 0.3) + (0.4 × 0.8) = 0.18 + 0.32 = 0.5 = 50%
e) The expected value 
f) Variance 
Standard deviation 
1/2(12+8)=10 Answer is 12