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satela [25.4K]
3 years ago
7

A clothing company plans to shear at least 700700700 sheep today. All novices shear the same number of sheep per day, and all ex

perts shear the same number of sheep per day.
Let NNN represent the number of novices and EEE represent the number of experts needed for the company to meet its goal.
7N+12E \geq 7007N+12E≥7007, N, plus, 12, E, is greater than or equal to, 700
According to the inequality, how many sheep does each novice or expert shear per day?
Each novice shears
sheep, and each expert shears
sheep.
Can the company achieve its goal by employing 606060 novices and 151515 experts?
Mathematics
1 answer:
rusak2 [61]3 years ago
6 0
Its A because if u add it all up then divides thats your answer
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Please solve the problem below
mr Goodwill [35]
You would use the cosine rule for this as you have 2 lines and are given the angle between them
The angle can be figured out easily as the clock has 12 equal components, meaning each component is 360/12 which is 30
As there are 5 components between A and B the angle is 30*5 =150
You can plug all of this into the cosine rule
a^2= b^2+c^2-2bcCos(A)
So a = sqrt( 10^2 +7^2 -(2*10*7) cos150)
a=16.4 to 3 sf
5 0
3 years ago
what is 2/a = a/50. These are factions if you already didn't know. I need to know what the answer is. it is a proportion.
Natalija [7]
2/a= a/50

Cross multiply:
a* a= 2*50
⇒ a^2= 100
⇒ a= √100 or a= -√100
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Final answer: a= 10 or a= -10~
3 0
4 years ago
Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 b
jeka94

Answer:

Step-by-step explanation:

From the given information; Let's assume that  R should represent the set of all possible outcomes generated from  a bit string of length 10 .

So; as each place is fitted with either 0 or 1

\mathbf{|R|= 2^{10}}

Similarly; the event E signifies the randomly generated bit string of length 10 does not contain a 0

Now;

if  a 0 bit and a 1 bit are equally likely

The probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit are equally likely is;

\mathbf{P(E) = \dfrac{|E|}{|R|}}

so ; if bits string should not contain a 0 and all other places should be occupied by 1; Then:

\mathbf{{|E|}=1 }   ; \mathbf{|R|= 2^{10}}

\mathbf{P(E) = \dfrac{1}{2^{10}}}

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3 years ago
Record the results of the class demonstrations and discussions in your notes.
lana [24]
You shoulda just paid attention to class my guy, nobody can answer this.
7 0
3 years ago
PLZ HELP WE WILL ALL DIE IF THI IS NOT ANSWERD
stepladder [879]

Answer:

C.

im not gonna dieeee

3 0
3 years ago
Read 2 more answers
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