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ivanzaharov [21]
2 years ago
15

In a bolt manufacturing factory, machines 4, Band C produce 25%, 30% and 45% of

Mathematics
1 answer:
Ugo [173]2 years ago
7 0

In the first 25% of the produce, 7% of that is defective. Which means that 25%*93% is not defective.

=20.75% not defective.

In the second 30% of the produce, 6% is defective Which means that 30%*94%

=28.2% not defective

Now for the 3rd 45% of the produce, 4% is defective which means that

45%*96%

=43.2% not defective.

Added together, 20.75+28.2+43.2=92.15% not defective.

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tekilochka [14]

n would equal <em>-33</em>.

<em>n=-33</em>

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3 years ago
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faust18 [17]

Answer:

greater than 1

Step-by-step explanation:

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3 years ago
BRAINLIEST WILL I GIVE HURRY
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Answer:

{10,8}

Step-by-step explanation:

-3x + 4y = -62

4x + 5y = 0

let's eliminate the x

-3x + 4y = -62 | x -4 |

4x + 5y = 0 | x 3 |

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-------------------- -

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y = 248/(-31) = 8

since you must do this proble with elimination, we cant use subtitution. so we repeat the way once more to find x (eliminate y)

-3x + 4y = -62 | x 5 |

4x + 5y = 0 | x 4 |

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-------------------- -

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4 0
3 years ago
PLEASE HELP ME ASAP??!!
earnstyle [38]

It says that there was an error with my answer, but I don't know what, so I screenshotted what I typed and attached them as 4 images below:

5 0
3 years ago
Help, I’m stuck. Thank you very much
ZanzabumX [31]
If you notice, the container is really just two circles, with a diameter of 4 each, and a square with sides of 10.5.

now, if we just get the area of each circle, keeping in mind their radius is half of the diameter, namely r = 2, and get the area of the 10.5x10.5 square, sum them up, that'd be the surface area of the container.

\bf \stackrel{\textit{2 circle's area}}{2(\pi r^2)}~~+~~\stackrel{\textit{area of the square}}{10.5\cdot 10.5}&#10;\\\\\\&#10;\stackrel{\textit{2 circle's area}}{2(\pi 2^2)}~~+~~\stackrel{\textit{area of the square}}{10.5\cdot 10.5}\implies 8\pi + 110.25
5 0
3 years ago
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