Ten kids line up for recess. there are five boys and five girls. how many ways are there for the 10 kids to line up so that all the boys are ahead of all the girls?

14400

Step-by-step explanation:

We have 2 groups of 5 elements each

there is a condition all boys will be ahead always, therefore  in the group of boys there are

5!  =  5*4*3*2*1  =  120 differents form

And the girls at the same time can change theirs places according to

5! =  120

Then total ways for the line up in the conditions of the statement of the problem are

T = 120*120  

14400

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)

69.64%

step-by-step explanation: