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PolarNik [594]
3 years ago
14

This is my very first trig problem and I'd like to know how to tackle this. Please explain how I can do problems like these in t

he future!

Mathematics
1 answer:
Paha777 [63]3 years ago
5 0

\tan(50°)  =  \frac{QR}{PQ} \Leftrightarrow QR = PQ \tan(50°) = 8.1 \tan(50°) \approx9.65 \\  \cos(50°)  =  \frac{PQ}{PR} \Leftrightarrow PR =  \frac{PQ}{\cos(50°)}  = \frac{8.1}{\cos(50°)} \approx12.6

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Hansen won the raffle At the zoo and gets to feed the dolphins! The dolphin trainer gives Hansen a bucket of fish to divide even
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Answer:

Number of dolphins in zoo * number of fish per dolphin.

Step-by-step explanation:

Hansen won the raffle At the zoo and gets to feed the dolphins! The dolphin trainer gives Hansen a bucket of fish to divide evenly among five dolphins. Each dolphin gets four fish. Which equation can you use to find the number of fish F in the bucket before hansen feeds them

Given that:

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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

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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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3 years ago
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