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antoniya [11.8K]
3 years ago
5

Midterm exam pls help

Mathematics
1 answer:
serg [7]3 years ago
4 0
Angles of a pentagon = 108°
angles of a square = 90°
108+90
answer= b
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Help please how many times longer is 3/4 than 2/15
liq [111]
3/4 is longer than 2/15 by 37/60
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3 years ago
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Credit card companies lose money on cardholders who fail to pay their minimum payments. They use a variety of methods to encoura
Ivahew [28]

Answer with  explanation:

A x% confidence interval interprets that a person can be x% confident thatthe true mean lies in it.

Here, Credit card companies is using the collection agency to justify the cost of , the agency must collect an average of at least $200 per customer.

i.e. H_0:\mu \geq200,\ \ \ H_a:\mu

The 90% confidence interval on the mean collected amount was reported as ($190.25, $250.75) .

I recommend that we can be 90% sure that the true mean collected amount  lies in ($190.25, $250.75).

Also, $200 lies in it such that it is more far from $250.75 than $190.25, that means there are large chances of having an average is at least $200 per customer.

8 0
3 years ago
The quality-control department of Starr Communications, the manufacturer of video-game cartridges, has determined from records t
notsponge [240]

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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3 years ago
Are step functions one to one?
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Answer: the function as they change value from one interval to the next

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