The probability that at least one of them has an injection-site reaction is: 0.5356.
<h3>
Probability</h3>
Using this formula
Probability of atleast one= 1- (1- Proportion)^n
Where:
Proportion=.12
Sample size=n=6
Let plug in the formula
Probability of atleast one=1-(1-.12)^6
Probability of atleast one=1-(.88)^6
Probability of atleast one=1-.4644041
Probability of atleast one=0.53559
Probability of atleast one=0.5356 (Approximately)
Therefore the probability that at least one of them has an injection-site reaction is: 0.5356.
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Answer:
y=3 - x/9
Step-by-step explanation:
Answer: {(5,0), (3,4)}
Step-by-step explanation:
You can apply the Substitution method:
- Solve for y from the second equation.
- Substitute into the first equation and solve for x.
Then:
(Remember that: )
Substitute the values obtained into one of the original equation and solve for y:
The solution set is: {(5,0), (3,4)}
Answer:
Eq: (x+a/2)²+(y+1)²=(a²-8)/4
Center: O(-a/2, -1)
Radius: r=0.5×sqrt(a²-8)
Mandatory: a>2×sqrt(2)
Step-by-step explanation:
The circle with center in O(xo,yo) and radius r has the equation:
(x-xo)²+(y-yo)²=r²
We have:
x²+y²+ax+2y+3=0
But: x²+ax=x²+2(a/2)x+a²/4-a²/4= (x+a/2)²-a²/4
And
y²+2y+3=y²+2y+1+2=(y+1)²+2
Replacing, we get:
(x+a/2)²-a²/4+(y+1)²+2=0
(x+a/2)²+(y+1)²=a²/4-2=(a²-8)/4
By visual inspection we note that:
- center of circle: O(-a/2, -1)
- radius: r=sqrt((a²-8)/4)=0.5×sqrt(a²-8). This means a²>8 or a>2×sqrt(2)