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alexira [117]
3 years ago
9

Fatores 2m^4 - 18n^6

Mathematics
1 answer:
exis [7]3 years ago
6 0
2m⁴ - 18n⁶
2(m⁴) - 2(9n⁶)
2(m⁴ - 9n⁶)
2(m⁴ - 3m²n³ + 3m²n³ - 9n⁶)
2[m²(m²) - m²(3n³) + 3n³(m²) - 3n³(3n³)]
2[m²(m² - 3n³) + 3n³(m² - 3n³)]
2(m² + 3n³)(m² - 3n³)
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The appearance of text is called hyperlink.<br> True<br> False
Lerok [7]
The answer is True hope this helps
8 0
3 years ago
[Q71 Suppose that the height, in inches, of a 25-year-old man is a normal random variable with parameters g = 71 inch and 02 = 6
viktelen [127]

Answer: (a) Percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

              (b) Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

Step-by-step explanation:

Given that,

                  Height (in inches) of a 25 year old man is a normal random variable with mean g=71 and variance o^{2} =6.25.

To find:  (a) What percentage of 25 year old men are 6 feet, 2 inches tall

               (b) What percentage of 25 year old men in the 6 footer club are over 6 feet. 5 inches.

Now,

(a) To calculate the percentage of men, we have to calculate the probability

P[Height of a 25 year old man is over 6 feet 2 inches]= P[X>74in]

                           P[X>74] = P[\frac{X-g}{o} > \frac{74-71}{2.5}]

                                         = P[Z > 1.2]

                                         = 1 - P[Z ≤ 1.2]

                                         = 1 - Ф (1.2)

                                         = 1 - 0.8849

                                         = 0.1151

Thus, percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

(b) P[Height of 25 year old man is above 6 feet 5 inches gives that he is above 6 feet] = P[X, 6ft 5in - X, 6ft]

     P[X > 6ft 5in I X > 6ft] = P[X > 77 I X > 72]

                                          = \frac{P[X > 77]}{P[ X > 72]}

                                          = \frac{P[\frac{X - g}{o}>\frac{77-71}{2.5}]  }{P[\frac{X-g}{o} >\frac{72-71}{2.5}] }

                                          = \frac{P[Z >2.4]}{P[Z>0.4]}

                                          =  \frac{1-P[Z\leq2.4] }{1-P[Z\leq0.4] }

                                          = \frac{1-0.9918}{1-0.6554}

                                          = \frac{0.0082}{0.3446}

                                          = 0.024

Thus, Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

4 0
3 years ago
Asking again, will give brainiest !
Dima020 [189]

Answer:

add them

Step-by-step explanation:

got it righttt

5 0
3 years ago
Read 2 more answers
A new car is purchased for $50,000 and over time it’s value depreciates by one half every 5 years. What is the value of the car
Setler79 [48]

The value of the car after 5 years would be $23914.85

<u>Explanation:</u>

Given:

Worth of the new car = $50,000

Every 5 years the value depreciates by 1/2

i.e., every 5 years the value depreciates by 50%

So, we can say that every year the value depreciates by 10%

The worth of the car after 7 years would be

W = C (1 - \frac{D}{100} )^t

where,

W is the depreciated value

C is the current value

D is the depreciation rate

t is the time

Substituting the value in the equation we get:

W = 50000( 1 - \frac{10}{100} )^7\\\\W = 50000 (\frac{9}{10} )^7\\\\W = 23914.85

Therefore, the value of the car after 5 years would be $23914.85

7 0
3 years ago
What is the answered for 9 divided by982?
ella [17]
0.00916496945 Its The Answer
6 0
3 years ago
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