P(boy)=1/2
P(girl)=1/2
P(a girl, a boy , a girl and a boy)=1/2*1/2*1/2*1/2= 1/16
P(2 boys and 2 girls ) = 1/[4!/(2!*2!)] =1/6
hope this helps u
Let c represent the speed of the current.
time = distance/speed
Upriver, the current is subtracted from the canoe speed; downriver, it adds.
6 = 32/(12 -c) +32/(12 +c)
6(12² -c²) = 32(12 +c) +32(12 -c) . . . . . multiply by (12 +c)(12 -c)
0 = 6c² -96 . . . . . . . . . . . . . . . . . . . . . . subtract the left side
96/6 = c² . . . . . . . . . . . . . . . . . . . . . . . . add 96, divide by 6
4 = c . . . . . . . . . . . . . . . . . . . . . . . . . . . . take the square root
The rate of the current is 4 mph.
Answer:
x=y-44 and x+y=410
Step-by-step explanation:
So, you want to use the equations x=y-44 and x+y=410 when x is Ann's score and y is Ruth's score. This is because x (Ann's score) is Ruth's score (y) but 44 less, so you subtract y-44 to get x. Then x+y would also have to equal 410 so that's the other equation. Graphing the 2 equations gets you to the point (183,227) in which Ann's score is 183 points and Ruth's score is 227 points.
Pythagorean thoerem
we imagine the wall as a leg of a right triangle, the ladder is the hypotonuse and the distance from the bottom of the ladder to the wall is the other leg
so if we have a right triangle with leg length a and b and hypotnuse length c then
a²+b²=c²
so
given that one of the legs is 9ft (9 ft is the height of the triangle)
and the hypotnuse is 12 ft (length of ladder)
then say the last leg is x
so
x²+9²=12²
x²+81=144
mius 81 both sides
x²=63
sqrt both sides
x=√63
x=3√7
you should place it 3√7ft or about 7.9 or 8ft away