Aquality measurement for cabinet manufacturers is whether a drawer slides open and shut easily. historically, 2% of drawers fail the easy slide test. a manufacturer samples 10 drawers from a batch. assuming the chance of failure is independent between drawers, what type of distribution could be used to model the number of failed drawers from the sample of 10?

Binomial distribution could be used to model the number of failed drawers from the sample of 10.

Step-by-step explanation:

We are given the following information:

We treat drawers fail the easy slide test as a success.

P(drawers fail the easy slide test) = 2% = 0.02

The chance of failure is independent between drawers.

A manufacturer samples 10 drawers from a batch.

Since,

Then the number of drawers that fail the easy slide test follows a binomial distribution, where

where n is the total number of trails, x is the number of success, p is the probability of success.

Here, the parameters of binomial distribution are

n = 10, x = number of failed drawers(discrete values, x = 0, 1,...,10), p = 0.02

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.

16 is the absolute value

step-by-step explanation: