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Grace [21]
3 years ago
14

Someone pls help plsssss?? ASAP pls

Mathematics
1 answer:
larisa [96]3 years ago
5 0
1)First, understand that whatever your equation is it’s always gonna be less than 100, because she doesn’t wanna get more than 100 ft.
*>100
2) remember that Lainey already dropped 20 feet. The amount of minutes is your unknown variable (x)
3) so, note that the rate is 20 ft/minute. If you multiply 20 by the amount of minutes she has to descend (your unknown variable: x) you cancel out the “minutes” in the 20 ft/minute fraction.
(Ex. 20ft/min *multiply* x [unknown]minutes= 20ftx)
This is what you want since you need all of the terms to be the same unit to solve with algebra.
4) so, to put it all together you have, 20 times x (representing the unknown amount of minutes) plus 20 (representing the amount she has already fallen) is greater than 100

* 20x+20>100
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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.

6 0
3 years ago
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