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cricket20 [7]
3 years ago
12

A city’s population, P, is modeled by the function P(x) = 78,500( 1.02 )x where x represents the number of years after the year

2000.
The population of the city in the year 2000 was . The population increases by % each year. Enter your answers in the boxes.
Mathematics
1 answer:
serg [7]3 years ago
6 0
I presume this is meant to be exponential.

When f(x) = a(b)^x, a represents the starting value, b represents change, and x represents time.

The population began in 2000 (x = 0). This makes the equation equal to 78,500, which is your starting value and population in 2000.

1.02 means that the initial value is multiplied by 1.02 or 102% each year. Therefore, there is a 2% increase.
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Answer:

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Step-by-step explanation:

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En un hotel hay 67 habitaciones entre dobles y sencillas, si el número de camas es 92. ¿Cuántas habitaciones hay de cada tipo?​
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Answer:

25 dobles y 42 sencillas

Step-by-step explanation:

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3 years ago
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4 0
2 years ago
A company employs two shifts of workers. Each shift produces a type of gasket where the thickness is the critical dimension. The
AURORKA [14]

Answer:

(−0.103371 ; 0.063371) ;

No ;

( -0.0463642, 0.0063642)

Step-by-step explanation:

Shift 1:

Sample size, n1 = 30

Mean, m1 = 10.53 mm ; Standard deviation, s1 = 0.14mm

Shift 2:

Sample size, n2 = 25

Mean, m2 = 10.55 ; Standard deviation, s2 = 0.17

Mean difference ; μ1 - μ2

Zcritical at 95% confidence interval = 1.96

Using the relation :

(m1 - m2) ± Zcritical * (s1²/n1 + s2²/n2)

(10.53-10.55) ± 1.96*sqrt(0.14^2/30 + 0.17^2/25)

Lower boundary :

-0.02 - 0.0833710 = −0.103371

Upper boundary :

-0.02 + 0.0833710 = 0.063371

(−0.103371 ; 0.063371)

B.)

We cannot conclude that gasket from shift 2 are on average wider Than gasket from shift 1, since the interval contains 0.

C.)

For sample size :

n1 = 300 ; n2 = 250

(10.53-10.55) ± 1.96*sqrt(0.14^2/300 + 0.17^2/250)

Lower boundary :

-0.02 - 0.0263642 = −0.0463642

Upper boundary :

-0.02 + 0.0263642 = 0.0063642

( -0.0463642, 0.0063642)

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