Answer:
Step-by-step explanation:
Given that,
- p ( probability that the child has disease) = 25% = 0.25
- n = number of children = 3
The probability mass function of binomial distribution is,
- (P = X) = (nCx) X (p)^x X (1 - p)^n-x ; x = 0, 1, 2 ,3
- = 3Cx X (0.25)^x X (1 - 0.25)^3-x ; ( n = 3, p = 0.25
a) P ( two will have disease)
p ( X = 2) = 3C2 X (0.250^2 X (1 - 0.25) ^3-2
= 0.1406
b) P ( none will have disease)
p (X = 0) = 3C0 X (0.25)^0 X (1 - 0.25)^3-0
= 0.4219
c) P (neither having the disease nor being a carrier) = 25% = 0.25
The probability that at least one will neither having the disease nor being a carrier ;
P(X> or equals to) = 1 - P(X < 1)
= 1 - P( X = 0)
= 1 - 3C0 X (0.25)^0 X (1 - 0.25)^3-0
= 0.5781
d) p( the first child with the disease will the be 3rd child)
P(X = x) = (1-p)^x-1 X p
p( X= x) = ( 1 - 0.25 )^x -1 X 0.25
for third child = P(X = 3) = (1 - 0.25)^3-1 X (0.25)
= 0.1406
Answer: 5
Step-by-step explanation:
Answer:
You are given:
4Fe+3O_2 -> 2Fe_2O_3
4:Fe:4
6:O_2:6
You actually have the same number of Fe on both sides, The same is true for O_2 so yes this equation is properly balanced.
For added benefit consider the following equation:
CH_4+O_2-> CO_2+2H_2O
ASK: Is this equation balanced? Quick answer: No
ASK: So how do we know and how do we then balance it?
DO: Count the number of each atom type you have on each side of the equation:
1:C:1
4:H:4
2:O:4
As you can see everything is balanced except for O To balance O we can simply add a coefficient of 2 in front of O_2 on the left side which would result in 4 O atoms:
CH_4+color(red)(2)O_2-> CO_2+2H_2O
1:C:1
4:H:4
4:O:4
Everything is now balanced.
Step-by-step explanation:
Y = -3
T = -4
7x - 14
X = 4
Answer:
He worked 9 hours that week.
Step-by-step explanation: