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Alinara [238K]
3 years ago
12

I can't find this answer

Mathematics
1 answer:
Julli [10]3 years ago
3 0
The answer I think is C
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3x+2y = -7 solve for x any y using substitution
Naya [18.7K]
The answer is : X=-7/3-2/3y
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3 years ago
Has two or more independent clauses without any punctuation between
n200080 [17]

Answer:

the answer is fused sentence i believe if thats what your asking

Step-by-step explanation:

8 0
4 years ago
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A bag has four green marbles, three red marbles, and three yellow marbles. What is the probability that you pick a yellow marble
valina [46]

Step-by-step explanation:

add all marbles together:

4 (green) + 3 (red) + 3 (yellow) = 10 marbles total

if we pick a yellow marble out of the bag and do not replace it, then we have 2 yellow marbles left.

we need to add marbles again with one yellow marble subtracted.

4 (green) + 3 (red) + 2 (yellow) =

new Total is 9 marbles.

so then we have 2 yellow marbles out of 9 total marbles.

2:9

or 2/9

or .22 (22%)

is the probability

6 0
3 years ago
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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.

6 0
3 years ago
the diagram shows the dimencions. 40yd 20yd 5yd 10yd 30yd. of a new parking lot. what is the area of the parking lot
Furkat [3]
1'200'000 yards because all you have to do is multiply the numbers.
3 0
3 years ago
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