Samples of laboratory glass are in small, lightpackaging or heavy, large packaging. Suppose that 2% and 1% of thesample shipped in small and large packages, respectively, breakduring transit. (a) If 60% of the samples are shipped in largepackages and 40% are shipped in small packages, what proportion ofsamples break during shipment? (b) Also, if a sample breaks duringshipment, what is the probability that it was shipped in a smallpackage?

a) 1.4% of the samples break during shipment

b) the probability is 4/7 ( 57.14%)

Step-by-step explanation:

a) defining the event B= the sample of laboratory glass breaks , then the probability is:

P(B)= probability that sample is shipped in small packaging * probability that the sample breaks given that was shipped in small packaging +  probability that sample is shipped in large packaging * probability that the sample breaks given that was shipped in large packaging = 0.40* 0.02 + 0.60*0.01 = 0.014

b) we can use the theorem of Bayes for conditional probability. Then defining the event S= the sample is shipped in small packaging . Thus we have

P(S/B)= P(S∩B)/P(B) = 0.40* 0.02 / 0.014= 4/7 ( 57.14%)

where

P(S∩B)= probability that sample is shipped in small packaging and it breaks

P(S/B)= probability that sample was shipped in small packaging given that is broken

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below:

any options?

step-by-step explanation: