An automobile manufacturer obtains the microprocessors used to regulate fuel consumption in its automobiles from three microelectronic firms: a, b, and c. the quality-control department of the company has determined that 6% of the microprocessors produced by firm a are defective, 7% of those produced by firm b are defective, and 5.5% of those produced by firm c are defective. firms a, b, and c supply 45%, 30%, and 25%, respectively, of the microprocessors used by the company. what is the probability that a randomly selected automobile manufactured by the company will have a defective microprocessor?

The probability that a randomly selected automobile manufactured by the company will have a defective microprocessor is 6.2 %.

Step-by-step explanation:

Given:

The probability of getting a microprocessor from firm A is,

The probability of getting a microprocessor from firm B is,

The probability of getting a microprocessor from firm C is,

Now, let event D be having a defective microprocessor at random.

So, as per the question,

Probability of producing a defective microprocessor from firm A is,

Probability of producing a defective microprocessor from firm B is,

Probability of producing a defective microprocessor from firm C is,

Now, probability of having a defective microprocessor when selected at random is given as:

Therefore, the probability that a randomly selected automobile manufactured by the company will have a defective microprocessor is 6.2 %.

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:

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.