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VashaNatasha [74]
3 years ago
13

2) The frequency distribution below shows the scores of 30 Grade 10 learners in their first summative test for Quarter

Mathematics
1 answer:
Anna [14]3 years ago
7 0
A because of I don’t know
You might be interested in
Two consecutive odd integers have a sum of 32. Find the integers.
bezimeni [28]
Let
x = first consecutive odd
x + 2 = second consecutive odd

Based on the problem, we equate
x + (x + 2) = 32
Solving for x,
2x + 2 = 32
2x = 32 - 2
2x = 30
x = 30/2
x = 15
and x + 2 = 15 + 2 = 17

Therefore, the integers are 15 and 17.
6 0
3 years ago
18) -32.-132, -232, -332, ..<br> Find a
Valentin [98]

Answer:

Step-by-step explanation:

next term is -432

5 0
2 years ago
Help a uneducated sis out✊
N76 [4]

Answer & Step-by-step explanation:

The domain of a set of points refers to the input, also known as the x values. To find the domain, record all the x values given (x,y):

domain={-2,0,3,5}

:Done

**The range is the output, aka the y values.

5 0
2 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.

6 0
3 years ago
In the year 2001, a company made $7.2 million in profit. For each consecutive year after that, their profit increased by 11%. Ho
Oksi-84 [34.3K]

Answer:

Who cares

Step-by-step explanation:

5 0
2 years ago
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