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daser333 [38]
3 years ago
5

Katie operates a machine that makes 150 crayons per minute.

Mathematics
2 answers:
Ainat [17]3 years ago
4 0

Answer:

The number of crayons is dependent on the time.

leva [86]3 years ago
3 0

Answer:

The number of crayons is dependent on the time.

Step-by-step explanation:

Alright here is your answer,

The number of crayons is dependent on the time.

Because to know how many crayons you made you would need to know how long that they were made for!

Hope this helps and plz give my brainiest if it's right!

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bixtya [17]
The answer is C. Opposite sides have the exact same slope, and the shape is a quadrilateral.
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3 years ago
What is the solution to this system of linear equations?
alexandr1967 [171]

Answer:

A. (-1 , 3)

Step-by-step explanation:

          2x + y = 1

plus    3x - y = -6

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

5x = -5

x = -1

y = 1 - (2*-1) = 3

3 0
2 years ago
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Hellllp please
Aneli [31]

Answer:

y = 5/8x - 3

Step-by-step explanation:

(8,2) and (-8,-8)

m=(y2-y1)/(x2-x1)

m = ( -8 -2)/ (-8 - 8)

m = (-10)/(-16)

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y - 2 = 5/8(x - 8)

y - 2 = 5/8x - 5

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3 0
2 years ago
Problem: The mean number of bankruptcies filed per minute in the ` States in a recent year was about two. Find the probability t
tino4ka555 [31]
The number of companies is quite large. That is, n is quite large.
The probability that a company declares bankruptcy is quite small , p is quite small.
np = the mean number of bankruptcies = 2  = a finite number.
Hence we can apply Poisson distribution for the data.
P (x=5 | mean =2) = e-2 25/5! = e-2 * 32/120 = 0.036089
Alternatively 
=poisson(5,2,0) = 0.036089
P(x≥ 5 | mean =2) = 1- P( x ≤ 4) = 1- e-2 (1+2+22/2!+23/3!+24/4!)= 1-e-2 (1+2+2+8/6+16/24)= 1-e-2(7)
  =0.052653
Alternatively
= 1- poisson(4,2,1) =0.052653
P(X > 5 | mean =2) = 1- p(x
 ≤ 5) =1- e-2 (1+2+22/2!+23/3!+24/4!+25/5!)= 1-e-2(7+4/15)
  =0.016564
alternatively=1-poisson(5,2,1)
=0.016564
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3 years ago
Given: BD = BF DE ⊥ BC , FK ⊥ AB Prove: ED ≅ FK
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This is the picture I am not very good at proofs but hopefully someone can solve it.

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