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Iteru [2.4K]
3 years ago
14

Susan started her homework at 1{:}59\text { p.M.}1:59 p.M.1, colon, 59, start text, space, p, point, m, point, end text and fini

shed her homework 969696 minutes later. Susan had volleyball practice at 4{:}00\text{ p.M.}4:00 p.M.
Mathematics
1 answer:
inn [45]3 years ago
7 0

Answer:

Step-by-step explanation:

Susan started her homework at = 1:59 pm

She took time = 96 mins

Time at which she had volleyball practice = 4:00 pm

So, time between her starting of homework and starting of volleyball is given by

1:59 pm to 4:00 pm = 2 hrs and 1 mins

2 hr 1 min = 120 mins + 1 min = 121 mins

But she took 96 mins to do her homework, so

Out of 121 mins, 96 mins she took for homework

Remaining time = 121 mins - 96 mins = 25 mins

Therefore, time Susan had between finishing her homework and the beginning of volleyball practice is 25 mins

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The quality-control department of Starr Communications, the manufacturer of video-game cartridges, has determined from records t
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Answer: (A) The probability that a cartridge purchased will have a video or audio defect is 1.9%

(B) The probability that a cartridge purchased will not have a video or audio defect is zero.

Step-by-step explanation: The data given shows that 1.2% (or 120) cartridges have video defects, 0.9% have audio defects (or 90) and 0.2% (or 20) have both audio and video defects.

The possible outcomes for all events (audio defects and video defects) is derived as 120 plus 90 which is equals 210 possibilities (or possible outcomes).

Therefore the probability of having an audio defect is calculated as follows;

P(Audio) = Number of required outcomes/Number of all possible outcomes

P(Audio) = 90/210

P(Audio) = 3/7

Also the probability of having a video defect is derived as follows;

P(Video) = Number of required outcomes/Number of all possible outcomes

P(Video) = 120/210

P(Video) = 4/7

However we should take note of the fact that 0.2% or 20 of the cartridges in the sample size has both audio and video defects. Hence the probability that a cartridge has both audio and video defects is calculated as;

P(Audio and Video) = Number of required outcomes/Number of all possible outcomes

P(Audio and Video) = 20/210

P(Audio and Video) = 2/21

To calculate the probability that a cartridge bought would have either an audio or a video defect would mean to add both probabilities together, but we MUST SUBTRACT the probability of having both an audio defect and video defect (that is P{Audio and Video}). The reason is that this is already included in both probabilities and we need to avoid double counting. Hence we have;

(A); P(Video OR Audio defect) = P(Audio) + P(Video) - P(Audio and Video)

P(Video OR Audio defect) = (3/7 + 4/7) - 2/21

P(Video OR Audio defect) = 1 - 2/21

P((Video OR Audio defect) = 19/21

Therefore the probability that a cartridge purchased will have a video or audio defect is 190, or better still 1.9%.

(B): From all possibilities shown, which is 210 possibilities of either events, we have determined that 120 will be the probability of having an audio defect and 90 will be the probability of having a video defect. Therefore the probability that a cartridge purchased will not fall into any of either possibilities is zero.

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Ans:
18min and 30 sec
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2 years ago
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