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Anit [1.1K]
1 year ago
7

Mrs. Johnson buys 2 chickens. The average weight of the 2 chickens is 4 pounds. One of the chickens is 2 pounds heavier. What is

the weight of the heaviest chicken?
Mathematics
2 answers:
irakobra [83]1 year ago
7 0
Heaviest chicken is 6 pounds
saul85 [17]1 year ago
4 0
The heaviest chicken would actually be 4 pounds. Since the average of both chickens is 4, that means they had to do the weight of the chickens added together than divided by two. The only one that would work is 2. So then you add 2 to that and it would be four.
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A common assumption in modeling drug assimilation is that the blood volume in a person is a single compartment that behaves like
mixas84 [53]

Answer:

a) \mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}

b) \mathbf{x = 2000 - 2000e^{-0.015t}}

c)  the  steady state mass of the drug is 2000 mg

d) t ≅ 153.51  minutes

Step-by-step explanation:

From the given information;

At time t= 0

an intravenous line is inserted into a vein (into the tank) that carries a drug solution with a concentration of 500

The inflow rate is 0.06 L/min.

Assume the drug is quickly mixed thoroughly in the blood and that the volume of blood remains constant.

The objective of the question is to calculate the following :

a) Write an initial value problem that models the mass of the drug in the blood for t ≥ 0.

From above information given :

Rate _{(in)}= 500 \ mg/L  \times 0.06 \  L/min = 30 mg/min

Rate _{(out)}=\dfrac{x}{4} \ mg/L  \times 0.06 \  L/min = 0.015x \  mg/min

Therefore;

\dfrac{dx}{dt} = Rate_{(in)} - Rate_{(out)}

with respect to  x(0) = 0

\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}

b) Solve the initial value problem and graph both the mass of the drug and the concentration of the drug.

\dfrac{dx}{dt} = -0.015(x - 2000)

\dfrac{dx}{(x - 2000)} = -0.015 \times dt

By Using Integration Method:

ln(x - 2000) = -0.015t + C

x -2000 = Ce^{(-0.015t)

x = 2000 + Ce^{(-0.015t)}

However; if x(0) = 0 ;

Then

C = -2000

Therefore

\mathbf{x = 2000 - 2000e^{-0.015t}}

c) What is the steady-state mass of the drug in the blood?

the steady-state mass of the drug in the blood when t = infinity

\mathbf{x = 2000 - 2000e^{-0.015 \times \infty }}

x = 2000 - 0

x = 2000

Thus; the  steady state mass of the drug is 2000 mg

d) After how many minutes does the drug mass reach 90% of its stead-state level?

After 90% of its steady state level; the mas of the drug is 90% × 2000

= 0.9 × 2000

= 1800

Hence;

\mathbf{1800 = 2000 - 2000e^{(-0.015t)}}

0.1 = e^{(-0.015t)

ln(0.1) = -0.015t

t = -\dfrac{In(0.1)}{0.015}

t = 153.5056729

t ≅ 153.51  minutes

4 0
1 year ago
Rework problem 28 from section 3.1 of your text, involving the inspection of refrigerators on an assembly line. You should still
sergeinik [125]
Event: Probability: A. Too much enamel 0.18 B. Too little enamel 0.24 C. Uneven application 0.33 D. No defects noted 0.47 


let P(AC) = x, P(BC) = y, then P(A) + P(B) + P(C) - (x+y) = 1-0.47 = 0.53 x+y = 0.22 
3. The probability of paint defects that results to <span>an improper amount of paint and uneven application? </span>
 P(A U B U C) = 0.53 

4. <span>the probability of a paint defect that results to</span>
<span>the proper amount of paint, but uneven application?</span>
P(C) - P(AC) - P(BC) = 0.47 - 0.22 = 0.25 


A and B are disjoint so P(ABC) = 0, but you can have P(AC) and P(BC). you can't compute these separately here, but you can compute P(AC) + P(BC). By the way, P(AC) eg is just an abbreviated version of P(A∩C).
3 0
2 years ago
A fabric store sells two types of ribbon. One customer buys 3 rolls of the lace ribbon and 2 rolls of the satin ribbon and has a
stepan [7]
<span>Let L be the number of yards on a roll of lace ribbon. Let S be the number of yards on a roll of satin ribbon. We can set up two equations. equation 1: 3L + 2S = 120 yards equation 2: 2L + 4S = 160 yards We can multiply (equation 1) by 2 and subtract (equation 2). equation 1: 6L + 4S = 240 yards equation 2: 2L + 4S = 160 yards 4L = 80 yards L = 20 yards equation 1: 3L + 2S = 120 yards 3(20 yards) + 2S = 120 yards 2S = 60 yards S = 30 yards There are 20 yards on a roll of lace ribbon. There are 30 yards on a roll of satin ribbon.</span>
7 0
2 years ago
A sculpture of a clothespin is 20 feet high. A normal clothespin of this shape is five inches high. Using this scale, how many f
kodGreya [7K]
This is the concept of scale factor. The linear scale factor is given by:
(height of sculpture)/(height of clothespin)
=20/(5/12)
=48
Given that a woman is 5ft 7 in, the sculpture of this woman would be:
height of sculpture=(scale factor)*(height of the woman)
height of the woman=5 ft 7 inches=5 7/12 ft=67/12 ft
hence the height of the sculpture will be:
67/12×48
=268 ft

Answer: 268 ft

7 0
2 years ago
Read 2 more answers
-2(-5x - 10) + 30 = -110
12345 [234]
Answer: X= -6

Hope it helps
6 0
2 years ago
Read 2 more answers
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