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kkurt [141]
3 years ago
8

The number of incarcerated adults N​ (measured in​ thousands) in a certain country can be approximated by the equation Upper N e

quals negative 2.7 x squared plus 72.4 x plus 1911​, where x is the number of years since 2000. In 2013​, the number of incarcerated adults peaked. How many adults were incarcerated in that​ year?
Mathematics
2 answers:
xeze [42]3 years ago
8 0

Answer:

3308.5 thousands or 3,308,500.

Step-by-step explanation:

We have been given a formula that represented the number of incarcerated adults N​ (measured in​ thousands) in a certain country can be approximated by the equation N=2.7x^2+72.4x+1911 where x is the number of years since 2000.

To find the number of incarcerated adults in 2013, we will substitute x=13 (2013-2000=13) in the given equation.

N=2.7(13)^2+72.4(13)+1911

N=2.7*169+941.2+1911

N=456.3+941.2+1911

N=3308.5

Since the number of incarcerated adults N​ is measured in​ thousands, so the value of N would be:

3308.5\text{ thousands}=3308.5\times 1,000

3308.5\text{ thousands}=3,308,500

Therefore, 3,308,500 adults were incarcerated in 2013.

Pavlova-9 [17]3 years ago
4 0

Answer:

2,396,000 adults were incarcerated in 2013.

Step-by-step explanation:

The number of incarcerated adults N​ (measured in​ thousands) in a certain country can be approximated by the equation

N = -2.7x² + 72.4x + 1911

where x is the number of years since 2000.

In 2013, x = 2013 - 2000 = 13, and the number of incarcerated adults was

N = -2.7(13)² + 72.4(13) + 1911

N = -456.3 + 941.2 + 1911

N = 2,396

2,396 thousands (2,396,000) adults were incarcerated in 2013.

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3 years ago
0.12x - 1.5= 0.24x - 0.06
Sophie [7]

To solve this equation, lets move all of the variables to one side of the equation and the constants to the other.

0.12x - 1.5 = 0.24x - 0.06

Subtract 0.12x from both sides of the equation

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Add 0.06 to both sides of the equation

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3 years ago
Log base 9 m - log base 9 (m-4)=-2
choli [55]
Hello,

log base 9(m/(m-4))=-2
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4 0
3 years ago
51% of the tickets sold at a school carnival were early-admission tickets. If the school sold 100 tickets in all, how many early
saul85 [17]

Answer:

51 were the total number of the tickets sold at a school carnival were early-admission tickets.

Step-by-step explanation:

Total number of tickets sold by  = 100

Let x be the early-admission tickets sold by the school.

As 51% of the tickets sold at a school carnival were early-admission tickets.

so

x=\frac{51}{100}\times 100

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Therefore, 51 were the total number of the tickets sold at a school carnival were early-admission tickets.

5 0
3 years ago
In order to conduct an experiment five subjects are randomly selected from a group of 48 subjects how many different groups of f
exis [7]

Answer:

1,712,304 ways

Step-by-step explanation:

This problem bothers on combination

Since we are to select 5 subjects from a pool of 48 subjects, the number of ways this can be done is expressed as;

48C5 = 48!/(48-5)!5!

48C5 = 48!/43!5!

48C5 = 48×47×46×45×44×43!/43!5!

48C5 = 48×47×46×45×44/5!

48C5 = 205,476,480/120

48C5 = 1,712,304

Hence this can be done in 1,712,304ways

3 0
3 years ago
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