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Cloud [144]
3 years ago
14

Helppppppppppppppppppppppppppp the question is post on my page

Mathematics
1 answer:
Nikitich [7]3 years ago
8 0
The answer is going to be 1650 it’s easy you just gotta read it addition
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F(x)=3x^3-12x^2+9x what the factor<br>​
alexandr1967 [171]

Answer:

f(x) = 3x(x-1)(x-3)

Step-by-step explanation:

f(x) = 3x³-12x²+9x = 3x(x²-4x+3) =

= 3x(x²-x-3x+3) = 3x[x(x-1)-3(x-1)] =

= 3x[(x-1)(x-3)] = 3x(x-1)(x-3)

6 0
3 years ago
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0.00098421, in standard notation to scientific notation
irinina [24]

Answer:Scientific Notation:9.8421*10^-4

Step-by-step explanation:

3 0
3 years ago
The original price for a jacket was 124.95. It is on sale for 20% off. Witch of the following gives the best estimate of the sav
d1i1m1o1n [39]
20% of 124.95 is close to $125 so that is equal of 12.50 of 10% so multiply by 2 which is $25.00 so the estimate discount of the jacket will be $25.00
7 0
3 years ago
Read 2 more answers
$600is what percent of $1000
GrogVix [38]

Answer:

60%

Step-by-step explanation:

$600 as a percentage of $1000

= (600/1000) x 100%

= 60%

Hope this helps!

5 0
3 years ago
Read 2 more answers
a slitter assembly contains 48 blades five blades are selected at random and evaluated each day for sharpness if any dull blade
son4ous [18]

Answer:

P(at least 1 dull blade)=0.7068

Step-by-step explanation:

I hope this helps.

This is what it's called dependent event probability, with the added condition that at least 1 out of 5 blades picked is dull, because from your selection of 5, you only need one defective to decide on replacing all.

So if you look at this from another perspective, you have only one event that makes it so you don't change the blades: that 5 out 5 blades picked are sharp. You also know that the probability of changing the blades plus the probability of not changing them is equal to 100%, because that involves all the events possible.

P(at least 1 dull blade out of 5)+Probability(no dull blades out of 5)=1

P(at least 1 dull blade)=1-P(no dull blades)

But the event of picking one blade is dependent of the previous picking, meaning there is no chance of picking the same blade twice.

So you have 38/48 on getting a sharp one on your first pick, then 37/47 (since you remove 1 sharp from the possibilities, and 1 from the whole lot), and so on.

Also since are consecutive events, you need to multiply the events.

The probability that the assembly is replaced the first day is:

P(at least 1 dull blade)=1-P(no dull blades)

P(at least 1 dull blade)=1-(\frac{38}{48}* \frac{37}{47} *\frac{36}{46}*\frac{35}{45}*\frac{34}{44})

P(at least 1 dull blade)=1-0.2931

P(at least 1 dull blade)=0.7068

5 0
3 years ago
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