A chemical engineer is interested in determining whether a certain trace impurity is present in a product. An experiment has a probability of 0.75 of detecting the impurity if it is present. The probability of not detecting the impurity if it is absent is 0.85. The prior probabilities of the impurity being present and being absent are 0.30 and 0.70, respectively. Three separate experiments result in only two detections. What is the posterior probability that the impurity is present

5x+40

5(x + 8)

7x - 49

7(x - 7)

18x + 9

9(2x + 1)

15x - 10

5(3x - 2)

step-by-step explanation:

for number one all you have to do is factor 5 out of 5x + 40

for number 2 factor 7 out of 7x - 49

for number three factor 9 out of 18x + 9

for number four factor 5 out of 15x - 10

heudhduxhdnshenebdutbbt

step-by-step explanation:

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54 km / hours

step-by-step explanation:

we need to convert seconds to hours

60 seconds = 1 minutes

60 minutes = 1hour

15 m         60 second   60 minutes

  * *         = 54000 meters/ hour

second     1 minute             1 hour

now we need to convert meters to km

1000 m = 1 km

54000 m       1 km

*   = 54 km / hours

hour               1000m

20%

step-by-step explanation: