Answer:
Probability of missing two passes in a row is 0.08.
Step-by-step explanation:
Event E = A football player misses twice in a row.
P(E) = ?
Event X = Football player misses the first pass
P(X) = 0.4
Event Y = Football player misses just after he first miss
P(Y) = 0.2
Both the events are exclusive so the probability of occuring of these two events can be calculated by the formula:
P(E) = P(X).P(Y)
P(E) = 0.4*0.2
P(E) = 0.08
Since both are equivalent to y, the equations must be equivalent.
x^2-x-3= -3x+5
x^2+2x-8=0
(x+4)(x-2)=0
x=-4, x=2
Plug the values of x in to either equation
y=-3(-4)+5
y= 12+5
y=17
y= -3(2)+5
y=-6+5
y=-1
Final answer: (-4,17) and (2,-1)
<h2>•5×7^2</h2>
<em>HOPE</em><em> </em><em>ITS</em><em> </em><em>HELPFUL</em><em> </em>^_^
<h2>
•RHONA</h2>
The radius is 8
Im not sure if this is what your looking for
Answer:
0.21
Step-by-step explanation: edge
