Answer:
mean=sum of data/no of data
=12/3
=4
Step-by-step explanation:
therefore mean=4
-x^2+2x+3=x^2-2x+3 add x^2 to both sides
2x+3=2x^2-2x+3 subtract 2x from both sides
3=2x^2-4x+3 subtract 3 from both sides
2x^2-4x=0 factor
2x(x-2)=0
So x=0 and 2
The probability that at least two of the missiles hit the arsenal:
P ( x ≥ 2 ) = P ( x = 2 ) + P ( x = 3 )
P ( x = 2 ) = 0.75 · 0.85 · 0.1 + 0.85 · 0.9 · 0.25 + 0.75 · 0.9 · 0.15 =
= 0.06375 + 0.10125 + 0.19125 = 0.35625
P ( x = 3 ) = 0.75 · 0.85 · 0.9 = 0.57375
P ( x ≥ 2 ) = 0.35625 + 0.57375 = 0.93
Answer:
The probability is 0.93 or 93%.
Answer:
m∠M = 79°
m∠N = 66°
Step-by-step explanation:
∠MPN is supplementary to ∠MPQ, so m∠MPN = 35
The sum of the measures of a triangle is 180.
So, m∠M + m∠N + m∠MPN = 180
5y + 4 + 4y + 6 + 35 = 180
9y + 45 = 180
9y = 135
y = 15
m∠M = 5y + 4 = 5(15) + 4 = 75 + 4 = 79
m∠N = 4y + 6 = 4(15) + 6 = 66
Another way to do this problem, which is easier, is to know that an exterior angle of a triangle is equal to the sum of the two remote interior angles.
That means 5y + 4 + 4y + 6 = 145
9y + 10 = 145
9y = 135
y = 15
From knowing the value of y you can now find the measures of angles M and N
The answer is 164. 5x4=20 and so on
