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ad-work [718]
3 years ago
11

A company that makes cartons finds that the probability of producing a carton with a puncture is 0.03​, the probability that a c

arton has a smashed corner is 0.08​, and the probability that a carton has a puncture and has a smashed corner is 0.002. Answer parts​ (a) and​ (b) below. ​(a) Are the events​ "selecting a carton with a​ puncture" and​ "selecting a carton with a smashed​ corner" mutually​ exclusive? Explain. A. ​No, a carton cannot have a puncture and a smashed corner. B. ​No, a carton can have a puncture and a smashed corner. Your answer is correct.C. ​Yes, a carton cannot have a puncture and a smashed corner. D. ​Yes, a carton can have a puncture and a smashed corner. ​(b) If a quality inspector randomly selects a​ carton, find the probability that the carton has a puncture or has a smashed corner. The probability that a carton has a puncture or a smashed corner is nothing.
Mathematics
1 answer:
matrenka [14]3 years ago
4 0

Answer:

(a) Having a smashed corner need to now not depend on whether or not the carton had a puncture or not,

so these occasions have to be independent.

One manner to check it the usage of the given probabilities:

If P(A)P(B) = P(A and B), then the events are independent.

Here we have : (0.05)(0.08) which is 0.004, that is P(A and B), so the events are  independent.

(b) P(A or B) = P(A)+P(B)-P(A and B) = 0.05 + 0.08 – 0.004 = 0.126  

Step-by-step explanation:

Please mark me brainliest!

I can help with any other questions you may have!

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Complete question is;

The levels of mercury in two different bodies of water are rising. In one body of water the initial measure of mercury is 0.05 parts per billion (ppb) and is rising at a rate of 0.1 ppb each year. In the second body of water the initial measure is 0.12 ppb and the rate of increase is 0.06 ppb each year.

Which equation can be used to find y, the year in which both bodies of water have the same amount of mercury?

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B) 0.05y + 0.1 = 0.12y + 0.06

C) 0.05 + 0.1y = 0.12 + 0.06y

D) 0.05y – 0.1 = 0.12y – 0.06

Answer:

Option C: 0.05 + 0.1y = 0.12 + 0.06y

Step-by-step explanation:

In the first body, the initial measure of mercury is 0.05 parts per billion (ppb) while it's rising at a rate of 0.1 ppb each year. We are told to use y for the number of years.

Thus, amount of mercury for y years in this first body is;

A1 = 0.05 + 0.1y

Now, for the second body, we are told that;the initial measure is 0.12 ppb and the rate of increase is 0.06 ppb each year.

Thus, amount of mercury for y years in this second body is;

A2 = 0.12 + 0.06y

Since we want to find the year in which both bodies of water have the same amount of mercury. Thus, it means;

A1 = A2

Thus;

0.05 + 0.1y = 0.12 + 0.06y

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