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olga nikolaevna [1]
2 years ago
12

janelle came to the bat 262 times in 91 games. at this rate, how many times should she expect to have at bat in full season of 1

62 games?
Mathematics
1 answer:
adell [148]2 years ago
7 0

Answer:

465 times

Step-by-step explanation:

r=\frac{262}{91} = 2.87

In full season of 162 games

162r=162(2.87) = 465

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andriy [413]
It is 10 to 1 significant figures
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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
Which of these is a unit of density?<br> Please help.
katen-ka-za [31]

Answer:

I think is C.

Step-by-step explanation:

Hope it's help you ❤️❤️❤️❤️❤️❤️❤️

7 0
2 years ago
Read 2 more answers
( Help ASAP ) y varies inversely with x. If x is 6.4 when y is 1.5, what is k, the constant of inverse variation?
DiKsa [7]

Answer:

k = 9.6

Step-by-step explanation:

When dealing with inverse variation, the two variables vary in opposite directions. This means that as the first variable increases, second variable decreases and as the first variable reduces,the second variable increases.

Y varies inversely with x.

So as y increases, x reduces and as y reduces, x increases.

We would proceed by introducing a constant of variation, k

y = k/x

From the information given,

The constant of variation = k

The value if x = 6.4

The value of y = 1.5

To find the value of k

y = k/x

k = yx = 1.5 × 6.4

k = 9.6

7 0
3 years ago
Look at image. reflection across y= -2​
Annette [7]

Answer:

S is -2,-3   r -3,-4   R  0,-5

Step-by-step explanation:

is the reflection points :)

6 0
3 years ago
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