Multiply both sides by -8.
Then add the one to both side to cancel it out and get 7x by itself
Then divide the whole thing by 7. So it’ll look like this: 32+1=7x and then add the one then divide by 7
Answer:
Let X be the number of times the target is hit. The probability P(X≥1) then equals 1 minus the probability of missing the target three times:
P(X≥1) = 1− (1−P(A)) (1−P(B)) (1−P(C))
= 1−0.4*0.3*0.2
= 0.976
To find the probability P(X≥2) of hitting the target at least twice, you can consider two cases: either two people hit the target and one does not, or all people hit the target. We find:
P(X≥2)=(0.4*0.7*0.8)+(0.6*0.3*0.8)+(0.6*0.7*0.2)+(0.6*0.7*0.8) = 0.788
Step-by-step explanation:
Answer:
No, h = (-23)/5
Step-by-step explanation:
Solve for h:
3 (4 - 6 h) - 7 h = 127
3 (4 - 6 h) = 12 - 18 h:
12 - 18 h - 7 h = 127
-18 h - 7 h = -25 h:
-25 h + 12 = 127
Subtract 12 from both sides:
(12 - 12) - 25 h = 127 - 12
12 - 12 = 0:
-25 h = 127 - 12
127 - 12 = 115:
-25 h = 115
Divide both sides of -25 h = 115 by -25:
(-25 h)/(-25) = 115/(-25)
(-25)/(-25) = 1:
h = 115/(-25)
The gcd of 115 and -25 is 5, so 115/(-25) = (5×23)/(5 (-5)) = 5/5×23/(-5) = 23/(-5):
h = 23/(-5)
Multiply numerator and denominator of 23/(-5) by -1:
Answer: h = (-23)/5
