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lara31 [8.8K]
3 years ago
13

Please help me!! thank you

Mathematics
1 answer:
Mama L [17]3 years ago
6 0

9514 1404 393

Answer:

  x = 1/6

Step-by-step explanation:

The relevant rules of exponents are ...

  (a^b)/(a^c) = a^(b-c)

  (a^b)^c = a^(bc)

__

The fraction inside parentheses evaluates to ...

  3^(3/4 -3/8) = 3^(3/8)

Then the whole expression evaluates to ...

  (3^(3/8))^(4/9) = 3^((3/8)(4/9)) = 3^(12/72) = 3^(1/6)

The value of x is 1/6.

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The answer please oh my please
Greeley [361]

96 meters???????????????????

6 0
4 years ago
How do I find the missing value
liberstina [14]
Okay. Since the "y" value and 82* are on the same straight line, their values will always add up to make a sum of 180. So if you subtract 82 from 180 you get 98*. So your "y" value is equal to 98*.

Now, all of the degree values added together should give you a sum of 360, so now we must add the values we know, in order to find the "x" value. So, 112*+82*+98*=292. Now we know that the difference of 360 and 292 will give us the x value. So finally, 360-292=68. So your "x" value is equal to 68.

y=98
x=68
6 0
4 years ago
Use the equation y = 4x to answer the questions.
Nastasia [14]

The slope is 4 and the Y-Intercept is 0.


3 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.

6 0
3 years ago
Y=(x^2-6x+14) Labeling the vertex
zavuch27 [327]
I leaerned this neat trick
you don't need to convert to vertex form

where
y=ax^2+bx+c
the x value of the vertex is -b/2a
the y value is found by inserting the x value of the vertex into the equation


so

1x^2-6x+14
-b/2a=-(-6)/2(1)=6/2=3
plug that in
y=x^2-6x+14
y=3^2-6(3)+14
y=9-18+14
y=5

the vertex is (3,5)
6 0
3 years ago
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