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During the period from 1790 to 1930, the US population P(t) (t in years) grew from 3.9 million to 123.2 million. Throughout this

alex41 [277]

Answer:

Step-by-step explanation:

Given that during the period from 1790 to 1930, the US population P(t) (t in years) grew from 3.9 million to 123.2 million. Throughout this period, P(t) remained close to the solution of the initial value problem.

$\frac{dP}{dt} =0.03135P =0.0001489P^2, P(0) = 3.9$

a) 1930 population is the population at time t = 40 years taking base year as 40

We can solve the differential equation using separation of variables

$\frac{dP}{0.03135P – 0.0001489P^2 } =dt\\\frac{dP}{-P(0.03135 – 0.0001489P } =dt$

Resolve into partial fractions

$\frac{31.8979}{P} -\frac{0.00474}{0.0001489P-0.03135}$

Integrate to get

ln P -0.00474/0.0001489 (ln (0.0001489P-0.03135) = t+C

ln P -31.833 (ln (0.0001489P-0.03135) =t+C

$\frac{P}{( ( (0.0001489P-0.03135)^{31.833} } =Ae^t$

Limiting population would be infinity.

6 0

4 years ago

Answer both questions please

Anna11 [10]

1.) 37/8

2.) .56 X 80 = 44.8

Hope this helps

8 0

3 years ago

Read 2 more answers

A boater travels 532 miles. Assuming the boat averages 6.3 miles per gallon, how many gallons of gasoline(to the nearest then of

Readme [11.4K]

Answer:

84.4 gallons to the nearest tenth.

Step-by-step explanation:

Average usage = miles travelled / gallons used so:

6.3 = 532 / gallons used

Gallons used = 532 / 6.3

= 84.44.

3 0

4 years ago

Determine the remaining sides and angles of the triangle ABC

Lapatulllka [165]

Answer:

The measure of angle B is 41.5°

The length of side a is 30.6 feet ⇒ to the nearest tenth

The length of side b is 32.9 feet ⇒ to the nearest tenth

Step-by-step explanation:

<em>The sum of the measures of the interior angles in a triangle is 180°</em>

In the given Δ ABC

∵ m∠A = 38.1°

∵ m∠C = 100.4°

→ By using the fact above

∵ m∠A + m∠B + m∠C = 180°

∴ 38.1 + m∠B + 100.4 = 180

→ Add the like terms in the left side

∴ 138.5 + m∠B = 180

→ Subtract 138.5 from both sides

∴ m∠B = 41.5°

∴ The measure of angle B is 41.5°

∵ a is the opposite side of ∠A

∵ b is the opposite side of ∠B

∵ c is the opposite side of ∠C

→ By using the sine rule

∴ $\frac{a}{sinA}$ = $\frac{b}{sinB}$ = $\frac{c}{sinC}$

∵ c = 48.4 ft, m∠A = 38.1°, m∠B = 41.5°, and m∠C = 100.4

→ Substitute them in the sine rule above

∴ $\frac{a}{sin38.1}$ = $\frac{b}{sin41.5}$ = $\frac{48.4}{sin100.4}$

→ By using cross multiplication between the 1st and the 3rd fractions

∴ a × sin 100.4 = 48.8 × sin 38.1

→ Divide both sides by sin 100.4

∴ a = 30.61429861

∴ The length of side a is 30.6 feet ⇒ to the nearest tenth

→ By using cross multiplication between the 2nd and the 3rd fractions

∴ b × sin 100.4 = 48.8 × sin 41.5

→ Divide both sides by sin 100.4

∴ b = 32.87596206

∴ The length of side b is 32.9 feet ⇒ to the nearest tenth

4 0

3 years ago

A box at a yard sale contains 3 different china dinner sets, each consisting of 5 plates. A customer will randomly select 2 plat

marshall27 [118]

Answer:

$\dfrac{2}{7}$

Step-by-step explanation:

3 different china dinner sets, each consisting of 5 plates consist of 15 plates.

A customer can select 2 plates in

$C^{15}_2=\dfrac{15!}{2!(15-2)!}=\dfrac{15!}{13!\cdot 2!}=\dfrac{13!\cdot 14\cdot 15}{2\cdot 13!}=7\cdot 15=105$

different ways.

2 plates can be selected from the same dinner set in

$3\cdot C^5_2=3\cdot \dfrac{5!}{2!(5-2)!}=3\cdot \dfrac{3!\cdot 4\cdot 5}{2\cdot 3!}=3\cdot 2\cdot 5=30$

different ways.

Thus, the probability that the 2 plates selected will be from the same dinner set is

$Pr=\dfrac{30}{105}=\dfrac{6}{21}=\dfrac{2}{7}$

7 0

4 years ago