Yvaries jointly as x and z, and y = 32 m, x = 6 m, and z = 8 m. what is the value of y when x = 5 m and z = 12 m

y = 40

Step-by-step explanation:

If y varies jointly as x and z, the equation is

y = k* xz

We know y  x  and z  so we can substitute them in

32 = k * 6*8

32 = k*48

Divide by 48

32/48 = k*48/48

32/48 =k

2/3 =

So the equation is y =2/3  x*z

Since we know x=5 and z =12, we can find y

y = 2/3 * 5 * 12

y = 40

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17

w'x is 0.

hope it