Math / Algebra

Question57. The atmospheric pressure on an object decreases as altitude increases. If $a$ is the height (in km ) above sea level, then the pressure $P(a)$ (in mmHg ) is approximated by $P(a)=760 e^{-0.13 a}$ . a. Find the atmospheric pressure at sea level. b. Determine the atmospheric pressure at 8.848 km (the altitude of Mt. Everest). Round to the nearest whole unit.

Studdy Solution

STEP 1

What is this asking? We're finding the air pressure at sea level and at the top of Mt.
Everest using a special formula! Watch out! Don't forget that $e$ is a special number, not a variable!
Also, make sure your calculator is in the right mode for exponents.

STEP 2

1. Pressure at Sea Level
2. Pressure at Mt.
Everest

STEP 3

We're given the formula $P(a) = 760e^{-0.13a}$ , where $P(a)$ is the pressure and $a$ is the altitude above sea level.
At sea level, the altitude is zero!
So, we'll substitute $a = 0$ into our formula.

STEP 4

$P(0) = 760e^{-0.13 \cdot 0}$ Any number raised to the power of zero equals one, so $e^{-0.13 \cdot 0} = e^0 = 1$ .

STEP 5

$P(0) = 760 \cdot 1 = 760$ So, the atmospheric pressure at sea level is 760 mmHg!

STEP 6

Mt. Everest is 8.848 km above sea level.
This time, we'll substitute $a = 8.848$ into our pressure formula: $P(a) = 760e^{-0.13a}$ .

STEP 7

$P(8.848) = 760e^{-0.13 \cdot 8.848}$

STEP 8

Let's calculate that exponent first: $-0.13 \cdot 8.848 = -1.14984$ .

STEP 9

Now, we have: $P(8.848) = 760e^{-1.14984}$

STEP 10

Using a calculator to evaluate $e^{-1.14984}$ , we get approximately $0.3166$ .

STEP 11

$P(8.848) = 760 \cdot 0.3166$

STEP 12

Finally, multiplying $760$ by $0.3166$ gives us approximately $240.62$ .
Rounding to the nearest whole number, we get 241 mmHg.

STEP 13

The atmospheric pressure at sea level is 760 mmHg, and the pressure at the top of Mt.
Everest is approximately 241 mmHg.

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Studdy Solution

STEP 1

What is this asking? We're finding the air pressure at sea level and at the top of Mt.
Everest using a special formula! Watch out! Don't forget that $e$ is a special number, not a variable!
Also, make sure your calculator is in the right mode for exponents.

STEP 2

1. Pressure at Sea Level
2. Pressure at Mt.
Everest

STEP 3

We're given the formula $P(a) = 760e^{-0.13a}$ , where $P(a)$ is the pressure and $a$ is the altitude above sea level.
At sea level, the altitude is zero!
So, we'll substitute $a = 0$ into our formula.

STEP 4

$P(0) = 760e^{-0.13 \cdot 0}$ Any number raised to the power of zero equals one, so $e^{-0.13 \cdot 0} = e^0 = 1$ .

STEP 5

$P(0) = 760 \cdot 1 = 760$ So, the atmospheric pressure at sea level is 760 mmHg!

STEP 6

Mt. Everest is 8.848 km above sea level.
This time, we'll substitute $a = 8.848$ into our pressure formula: $P(a) = 760e^{-0.13a}$ .

STEP 7

$P(8.848) = 760e^{-0.13 \cdot 8.848}$

STEP 8

Let's calculate that exponent first: $-0.13 \cdot 8.848 = -1.14984$ .

STEP 9

Now, we have: $P(8.848) = 760e^{-1.14984}$

STEP 10

Using a calculator to evaluate $e^{-1.14984}$ , we get approximately $0.3166$ .

STEP 11

$P(8.848) = 760 \cdot 0.3166$

STEP 12

Finally, multiplying $760$ by $0.3166$ gives us approximately $240.62$ .
Rounding to the nearest whole number, we get 241 mmHg.

STEP 13

The atmospheric pressure at sea level is 760 mmHg, and the pressure at the top of Mt.
Everest is approximately 241 mmHg.

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Studdy Solution

STEP 1

What is this asking? We're finding the air pressure at sea level and at the top of Mt.Everest using a special formula! Watch out! Don't forget that eee is a special number, not a variable!Also, make sure your calculator is in the right mode for exponents.

STEP 2

1. Pressure at Sea Level2. Pressure at Mt.Everest

STEP 3

We're given the formula P(a)=760e−0.13aP(a) = 760e^{-0.13a}P(a)=760e−0.13a, where P(a)P(a)P(a) is the pressure and aaa is the altitude above sea level.At sea level, the altitude is **zero**!So, we'll **substitute** a=0a = 0a=0 into our formula.

STEP 4

P(0)=760e−0.13⋅0P(0) = 760e^{-0.13 \cdot 0}P(0)=760e−0.13⋅0 Any number raised to the power of zero equals **one**, so e−0.13⋅0=e0=1e^{-0.13 \cdot 0} = e^0 = 1e−0.13⋅0=e0=1.

STEP 5

P(0)=760⋅1=760P(0) = 760 \cdot 1 = 760P(0)=760⋅1=760 So, the atmospheric pressure at sea level is **760 mmHg**!

STEP 6

Mt. Everest is **8.848 km** above sea level.This time, we'll **substitute** a=8.848a = 8.848a=8.848 into our pressure formula: P(a)=760e−0.13aP(a) = 760e^{-0.13a}P(a)=760e−0.13a.

STEP 7

P(8.848)=760e−0.13⋅8.848P(8.848) = 760e^{-0.13 \cdot 8.848}P(8.848)=760e−0.13⋅8.848

STEP 8

Let's **calculate** that exponent first: −0.13⋅8.848=−1.14984-0.13 \cdot 8.848 = -1.14984−0.13⋅8.848=−1.14984.

STEP 9

Now, we have: P(8.848)=760e−1.14984P(8.848) = 760e^{-1.14984}P(8.848)=760e−1.14984

STEP 10

Using a calculator to **evaluate** e−1.14984e^{-1.14984}e−1.14984, we get approximately 0.31660.31660.3166.

STEP 11

P(8.848)=760⋅0.3166P(8.848) = 760 \cdot 0.3166P(8.848)=760⋅0.3166

STEP 12

Finally, **multiplying** 760760760 by 0.31660.31660.3166 gives us approximately 240.62240.62240.62.Rounding to the nearest whole number, we get **241 mmHg**.

STEP 13

The atmospheric pressure at sea level is **760 mmHg**, and the pressure at the top of Mt.Everest is approximately **241 mmHg**.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We're finding the air pressure at sea level and at the top of Mt.
Everest using a special formula! Watch out! Don't forget that $e$ is a special number, not a variable!
Also, make sure your calculator is in the right mode for exponents.

1. Pressure at Sea Level
2. Pressure at Mt.
Everest

We're given the formula $P(a) = 760e^{-0.13a}$ , where $P(a)$ is the pressure and $a$ is the altitude above sea level.
At sea level, the altitude is zero!
So, we'll substitute $a = 0$ into our formula.

$P(0) = 760e^{-0.13 \cdot 0}$ Any number raised to the power of zero equals one, so $e^{-0.13 \cdot 0} = e^0 = 1$ .

$P(0) = 760 \cdot 1 = 760$ So, the atmospheric pressure at sea level is 760 mmHg!

Mt. Everest is 8.848 km above sea level.
This time, we'll substitute $a = 8.848$ into our pressure formula: $P(a) = 760e^{-0.13a}$ .

$P(8.848) = 760e^{-0.13 \cdot 8.848}$

Let's calculate that exponent first: $-0.13 \cdot 8.848 = -1.14984$ .

Now, we have: $P(8.848) = 760e^{-1.14984}$

Using a calculator to evaluate $e^{-1.14984}$ , we get approximately $0.3166$ .

$P(8.848) = 760 \cdot 0.3166$

Finally, multiplying $760$ by $0.3166$ gives us approximately $240.62$ .
Rounding to the nearest whole number, we get 241 mmHg.

The atmospheric pressure at sea level is 760 mmHg, and the pressure at the top of Mt.
Everest is approximately 241 mmHg.