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3 years ago

The mean of 5,7,9 show that the mean of these numbers is 7

Mathematics

1 answer:

olasank [31]3 years ago

3 0

Answer: 7

Step-by-step explanation:

5+7+9=21

3 total numbers to start with so;

21/3=7

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4 x ? is close to 105

Serga [27]

Do you have a picture to your problem ?

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3 years ago

Read 2 more answers

Given P = x^0.3 y^0.7 is the chicken lay eggs production function, where P is the number of eggs lay, x is the number of workers

lora16 [44]

Answer:

Part A)

$\displaystyle \frac{dy}{dx}=-\frac{3}{7}P^\frac{10}{7}x^{-\frac{10}{7}}$

Part B)

The daily operating cost decreases by about $143 per extra worker.

Step-by-step explanation:

We are given the equation:

$\displaystyle P=x^{\frac{3}{10}}y^{\frac{7}{10}}$

Where P is the number of eggs laid, x is the number of workers, and y is the daily operating budget (assuming in US dollars $).

We want to find dy/dx.

So, let’s find our equation in terms of x. We can raise both sides to 10/7. Hence:

$\displaystyle P^\frac{10}{7}=\Big(x^\frac{3}{10}y^\frac{7}{10}\Big)^\frac{10}{7}$

Simplify:

$\displaystyle P^\frac{10}{7}=x^\frac{3}{7}y$

Divide both sides by the x term to acquire:

$\displaystyle y=P^\frac{10}{7}x^{-\frac{3}{7}}$

Take the derivative of both sides with respect to x:

$\displaystyle \frac{dy}{dx}=\frac{d}{dx}\Big[P^\frac{10}{7}x^{-\frac{3}{7}}\Big]$

Apply power rule. Note that P is simply a constant. Hence:

$\displaystyle \frac{dy}{dx}=P^\frac{10}{7}(-\frac{3}{7})(x^{-\frac{10}{7}})$

Simplify. Hence, our derivative is:

$\displaystyle \frac{dy}{dx}=-\frac{3}{7}P^\frac{10}{7}x^{-\frac{10}{7}}$

Part B)

We want to evaluate the derivative when x is 30 and when y is $10,000.

First, we will need to find P. Our original equations tells us that:

$P=x^{0.3}y^{0.7}$

Hence, at x = 30 and at y = 10,000, P is:

$P=(30)^{0.3}(10000)^{0.7}$

Therefore, for our derivative, we will have:

$\displaystyle \frac{dy}{dx}=-\frac{3}{7}\Big(30^{0.3}(10000^{0.7})\Big)^\frac{10}{7}\Big(30^{-\frac{10}{7}}\Big)$

Use a calculator. So:

$\displaystyle \frac{dy}{dx}=-\frac{1000}{7}=-142.857142...\approx-143$

Our derivative is given by dy/dx. So, it represents the change in the daily operating cost over the change in the number of workers.

So, when there are 30 workers with a daily operating cost of $10,000 producing a total of about 1750 eggs, the daily operating cost decreases by about $143 per extra worker.

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3 years ago

Please help i don’t understand this

avanturin [10]

Answer:

Step-by-step explanation:

The solution is in the attached file

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2 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

3 years ago

Other term to write 2x+6

eimsori [14]

You could factor it as 2(x + 3)

7 0

2 years ago