A warehouse contains ten printing machines, four of which are defective. A company selects five of the machines at random, thinking all are in working condition. The company repairs the defective ones at a cost of $20 each. In what interval would you expect the repair costs on these five machines in dollars to lie 75% of the time

Step-by-step explanation:

There from ypergemetric distribution

mean number of defective in sample =n×m/N =5×4/10=2

and std deviation =sqrt(mn/N*(1-m/N)*(N-n)/(N-1))

=sqrt(5*4/10*(1-4/10)*(10-5)/(10-1))=0.816

from Tchebysheff's theorem ; 75% values fall within 2 standard deviation from mean

therefore interval in terms of defective

=2-2*0.816 ; 2+2*0.816 =0.367 ; 3.633

interval in terms of  repair costs =20*0.367 ; 20*3.633 =$7.34 ; $ 72.66

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116

the answer is a if my teacher taught me correct