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algol13
3 years ago
11

What are types of legislature

Mathematics
1 answer:
gulaghasi [49]3 years ago
8 0

Answer:

Two common types of legislature are those in which the executive and the legislative branches are clearly separated, as in the U.S. Congress, and those in which members of the executive branch are chosen from the legislative membership, as in the British Parliament.

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A bag contains 100 marbles which are red, green, and blue. Suppose a student randomly selects a marble without looking, records
babunello [35]

Answer:

Firstly this is all about luck my predictions is impossible to be accurate

ANyways i think ther are 30 red,10 green and 60 blue

Step-by-step explanation:

7 red marbles, 2 green marbles and 11 blue marbles.

Firstly this is all about luck my predictions is impossible to be accurate

ANyways i think ther are 30 red,10 green and 60 blue

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2 years ago
What are the answers since I’m confused an all over the place
dalvyx [7]
3 squares = 4 circles, so (number of squares)/(number of circles) = 3/4.
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2 squares = 5 circles, so (number of squares)/(number of circles) = 2/5.
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7 0
2 years ago
Pregnant women metabolize some drugs at a slower rate than the rest of the population. The half-life of caffeine is about 4 hour
UkoKoshka [18]

Answer:

Husband:

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Woman:

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

Step-by-step explanation:

The amount of caffeine in the body can be modeled by the following equation:

C(t) = C(0)e^{rt}

In which C(t) is the amount of caffeine t hours after 8 am, C(0) is how much coffee they took and r is the rate the the amount of caffeine decreases in their bodies.

110 mg of caffeine at 8 am,

So C(0) = 110

Husband

Half life of 4 hours. So

C(4) = 0.5C(0) = 0.5*110 = 55

C(t) = C(0)e^{rt}

55 = 110e^{4r}

e^{4r} = 0.5

Applying ln to both sides

\ln{e^{4r}} = \ln{0.5}

4r = \ln{0.5}

r = \frac{\ln{0.5}}{4}

r = -0.1733

So for the husband

C(t) = 110e^{-0.1733t}

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

C(t) = 110e^{-0.1733t}

C(11) = 110e^{-0.1733*11} = 16.35

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Pregnant woman

Half life of 10 hours. So

C(10) = 0.5C(0) = 0.5*110 = 55

C(t) = C(0)e^{rt}

55 = 110e^{10}

e^{10r} = 0.5

Applying ln to both sides

\ln{e^{10r}} = \ln{0.5}

10r = \ln{0.5}

r = \frac{\ln{0.5}}{10}

r = -0.0693

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

C(t) = 110e^{-0.0693t}

C(11) = 110e^{-0.0693*11} = 51.33

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

7 0
3 years ago
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