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ira [324]
3 years ago
8

a country's population in 1992 was 103 million. in 1997 it was 108 million. Estimate the population in 2004 using the exponentia

l growth formula. Round your answer to the nearest million.
Mathematics
1 answer:
Stolb23 [73]3 years ago
4 0

115 million ( to nearest million)

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Greg drew a scale drawing of a house. The hall closet is 2 inches wide in the drawing. The actual closet is 6 feet wide. What is
Vlad [161]
The drawing's scale factor is 
Actual Dimensions:Drawing Measurements
= 36:1
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3 years ago
A necklace has and matching bracelet have two types of beads. The necklace has 12 large beads and 8 small beads and weighs 88 gr
miv72 [106K]

Weight of one large bead is 1.5 grams and weight of one small bead is 8.75 grams.

Step-by-step explanation:

Let,

Weight of one large bead = x

Weight of one small bead = y

According to given statement;

12x+8y=88     Eqn 1

5x+2y=25      Eqn 2

Multiplying Eqn 2 by 4

4(5x+2y=25)\\20x+8y=100\ \ \ Eqn\ 3\\

Subtracting Eqn 1 from Eqn 3

(20x+8y)-(12x+8y)=100-88\\20x+8y-12x-8y=12\\8x=12

Dividing both sides by 8

\frac{8x}{8}=\frac{12}{8}\\x=1.5

Putting x=1.5 in Eqn 1

12(1.5)+8y=88\\18+8y=88\\8y=88-18\\8y=70

Dividing both sides by 8

\frac{8y}{8}=\frac{70}{8}\\y=8.75

Weight of one large bead is 1.5 grams and weight of one small bead is 8.75 grams.

Keywords: linear equation, elimination method

Learn more about elimination method at:

  • brainly.com/question/11334714
  • brainly.com/question/11345547

#LearnwithBrainly

5 0
3 years ago
What is the value of 9.05
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Nine point zero tenths and five ones
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3 years ago
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Help! Quick please!
Vesnalui [34]
B..............
c>=3 and n<0 and p<=(9-3 c)/n
8 0
3 years ago
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The antibiotic clarithromycin is eliminated from the body according to the formula A(t) = 500e−0.1386t, where A is the amount re
PolarNik [594]

Answer:

Time(t) = 11.61 hours (Rounded to two decimal place)

Step-by-step explanation:

Given: The antibiotic  clarithromycin is eliminated from the body according to the formula:

A(t) = 500e^{-0.1386t}                 ......[1]

where;

A - Amount remaining in the body(in milligram)

t - time in hours after the drug reaches peak concentration.

Given: Amount of drug in the body is reduced to 100 milligrams.

then,

Substitute the value of A = 100 milligrams in [1] we get;

100= 500e^{-0.1386t}

Divide both sides by 500 we get;

\frac{100}{500}=\frac{ 500e^{-0.1386t}}{500}

Simplify:

\frac{1}{5} = e^{-0.1386t}

Taking logarithm both sides with base e, then we have;

\log_e (\frac{1}{5})= \log_e (e^{-0.1386t})

\log_e (\frac{1}{5})=-0.1386t         [ Using \log_e e^a =a ]

or

\log_e (0.2)=-0.1386t

-1.6094379124341 = -0.1386t

 [using value of \log_e (0.2) = -1.6094379124341 ]

then;

t = \frac{-1.6094379124341}{-0.1386}

Simplify:

t ≈11.61 hours.

Therefore, the time 11.61 hours(Rounded two decimal place) will pass before the amount of drug in the body is reduced to 100 milligrams


6 0
3 years ago
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