A manufacturing process that produces electron tubes is known to have a 10% defective rate. Suppose a random sample of 15 tubes is selected from the manufacturing process. a) Find the probability that no more than two defectives are found?

Probability of obtaining no more than two defective tubes = 0.816

Step-by-step explanation:

The Probability of obtaining no more than two defective tubes in a randomly selected sample of 15 tubes is obtained using the binomial distribution formula: nCr × p^r × q^(n -r).

Where n is number of samples;

r is maximum number of defective tubes, r ≤ 2;

p is probability of defective tubes = 10% or 0.1

q is probability of non-defective tubes, q = 1 - p

Further explanations and calculations are given in the attachment below:

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1

at least two

step-by-step explanation: