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luda_lava [24]
3 years ago
11

Factorize​ab(ab-bc)+bc(bc-ab)​

Mathematics
2 answers:
EleoNora [17]3 years ago
6 0

Answer:

(ab - bc)^2.

Step-by-step explanation:

ab(ab - bc) + bc(bc - ab)​

= ab(ab - bc) - bc(ab - bc)​

= (ab - bc)(ab - bc)

= (ab - bc)^2

Bond [772]3 years ago
5 0

Answer:

(ab-bc)²

Step-by-step explanation:

  • ab(ab-bc)+bc(bc-ab)​ =                 ⇒ distribution
  • (ab)² - ab*bc + (bc)² - ab*bc =     ⇒ regrouping
  • (ab)² - 2ab*bc + (bc)² =                ⇒ applying (x-y)² = x² -2xy +y² formula
  • (ab-bc)²                                        ⇒ final result
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The mean incubation time for a type of fertilized egg kept at 100.2100.2â°f is 2323 days. suppose that the incubation times are
zlopas [31]

To solve this problem, what we have to do is to calculate for the z scores of each condition then find the probability using the standard normal probability tables for z.

The formula for z score is:

z = (x – u) / s

where,

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u = sample mean = 23 days

s = standard deviation = 1 day

 

A. P when x < 21 days

z = (21 – 23) / 1

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Using the table,

P = 0.0228

Therefore there is a 2.28% probability that the hatching period is less than 21 days.

 

B. P when 23 ≥ x ≥ 22

<span>z (x=22) = (22 – 23)  / 1 = -1</span>

P (z=-1) = 0.1587

 

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P (z=0) = 0.5

 

P = 0.5 - 0.1587 = 0.3413

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P true = 0.0228

<span>Therefore there is a 2.28% probability that the hatching period is more than 25 days.</span>

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