The probability is 1% that an electrical connector that is kept dry fails during the warranty period of a portable computer. if the connector is ever wet, the probability of a failure during the warranty period is 5%. if 90% of the connectors are kept dry and 10% are wet, what proportion of connectors fail during the warranty period? let: d = electrical connector is dry d' = electrical connector is not kept dry (i. e., is wet) f = connector fails during warranty period f' = connector does not fail during warranty period match the correct probability description to each given probability value

0.014 or 1.4% of connectors fail during the warranty period.

Step-by-step explanation:

Consider the provided information.

Let: D = electrical connector is dry

D' = electrical connector is not kept dry (i.e., is wet)

F = connector fails during warranty period

F' = connector does not fail during warranty period

90% of the connectors are kept dry

That means P(D)=0.9

10% are wet,

P(D')=0.1

The probability is 1% that an electrical connector that is kept dry fails during the warranty period, i.e.

P(F|D)=0.01

If the connector is ever wet, the probability of a failure during the warranty period is 5%, i.e.

P(F|D')=0.05

Hence, 0.014 or 1.4% of connectors fail during the warranty period.

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

a reflection across the x-axis , followed by the by a reflection across the y-axis , and then a scale dilation of 0.5 .