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2 years ago

Hayley picked 2/3 of a pound of

Mathematics

1 answer:

Ksenya-84 [330]2 years ago

8 0

The answer the this is 0.23 pounds per container

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The quantity demanded x of a certain brand of DVD player is 3000/week when the unit price p is $485. For each decrease in unit p

ANTONII [103]

Answer:

Demand Equation: q = 15125 - 25p.

Supply Equation: 80p - 24000 = 9q.

Equilibrium Price: $525.

Equilibrium Quantity: 2000 units.

Step-by-step explanation:

To solve this question, first, the demand equation has to be calculated. Let price be on the y-axis and quantity by on the x-axis. It is given that p = $485 when q = 3000 units. This can also be written as (q₁, p₁) = (3000, 485). It is also given that when the price goes down by $20, the quantity increases by 500 units. Therefore, (q₂, p₂) = (3500, 465). Due to the statement "For each decrease in unit price of $20 below $485, the quantity demanded increases by 500 units", the demand function will be linear i.e. a straight line. This is because the word "for each" has been used. Therefore, finding the equation of the demand function:

(p - p₁)/(q - q₁) = (q₂ - q₁)/(p₂ - p₁).

(p - 485)/(q - 3000) = (465 - 485)/(3500 - 3000).

(p - 485)/(q - 3000) = (-20)/(500).

(p - 485)/(q - 3000) = -1/25.

25*(p - 485) = -1*(q - 3000).

25p - 12125 = -q + 3000.

25p + q = 15125.

q = 15125 - 25p. (Demand Equation).

Second, find the supply equation. It is given that the suppliers will not sell the product below or at the price of $300. Therefore, (q₁, p₁) = (0, 300). Also, the suppliers will sell 2000 units if the price is $525. Therefore, (q₂, p₂) = (2000, 525). Since it is mentioned that supply equation is linear, therefore:

(p - p₁)/(q - q₁) = (q₂ - q₁)/(p₂ - p₁).

(p - 300)/(q - 0) = (525 - 300)/(2000 - 0).

(p - 300)/(q) = (225)/(2000).

(p - 300)/(q) = 9/80.

80*(p - 300) = 9*(q).

80p - 24000 = 9q. (Supply Equation).

Equilibrium exists when demand = supply. This means that the demand and the supply equations have to be solved simultaneously. Therefore, put demand equation in supply equation:

80p - 24000 = 9(15125 - 25p).

80p - 24000 = 136125 - 225p.

305p = 160125.

p = $525.

Put p = $525 in demand equation:

q = 15125 - 25p.

q = 15125 - 25(525).

q = 15125 - 13125.

q = 2000 units.

To summarize:

Demand Equation: q = 15125 - 25p!!!

Supply Equation: 80p - 24000 = 9q!!!

Equilibrium Price: $525!!!

Equilibrium Quantity: 2000 units!!!

8 0

3 years ago

Let f(x)=5x3−60x+5 input the interval(s) on which f is increasing. (-inf,-2)u(2,inf) input the interval(s) on which f is decreas

o-na [289]

Answers:

(a) f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing at $(-2,2)$ .

(c) f is concave up at $(2, \infty)$

(d) f is concave down at $(-\infty, 2)$

Explanations:

(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that

$f'(x) = 15x^2 - 60
$

So,

$
f'(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \ 0} \text{ (1)}$

The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:

-->> x < -2
-->> -2 < x < 2
--->> x > 2

If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.

If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.

So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since

$f'(x) \ \textgreater \ 0 \Leftrightarrow (x - 2)(x + 2) \ \textgreater \ 0$

Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing only when the derivative of f is negative. Since

$f'(x) = 15x^2 - 60$

Using the similar computation in (a),

$f'(x) \ \textless \ \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \ 0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \ 0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \ 0} \text{ (2)}$

Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.

Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)

(c) f is concave up if and only if the second derivative of f is positive. Note that

$f''(x) = 30x - 60$

Since,

$f''(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 30x - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 30(x - 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow x - 2 \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{x \ \textgreater \ 2}$

Therefore, f is concave up at $(2, \infty)$ .

(d) Note that f is concave down if and only if the second derivative of f is negative. Since,

$f''(x) = 30x - 60$

Using the similar computation in (c),

$f''(x) \ \textless \ 0 
\\ \\ \Leftrightarrow 30x - 60 \ \textless \ 0 
\\ \\ \Leftrightarrow 30(x - 2) \ \textless \ 0 
\\ \\ \Leftrightarrow x - 2 \ \textless \ 0 
\\ \\ \Leftrightarrow \boxed{x \ \textless \ 2}$

Therefore, f is concave down at $(-\infty, 2)$ .

3 0

3 years ago

Merv’s Magic Shop has $56,000.00 in assets and $14,000.00 in liabilities. Merv’s current financial ratio is

Gnom [1K]

4:1.

56,000÷2 = 28,000

14,000÷2 = 7,000

28,000/7,000 = 28/7

28 ÷ 7 = 4
---- -----
7 ÷7 = 1

4:1

Hope this helps!

6 0

2 years ago

Read 2 more answers

ANTHONY SURVEYED 320 STUDENTS AND FOUND OUT THAT 120 OF THEM HAVE A PET BASED ON THIS INFORMATION HOW MANY

lbvjy [14]

I believe it would be 360 students.

6 0