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LenaWriter [7]
3 years ago
14

the median of the values in a data set is y. if 48 were subtracted from each of the values in the data set what would be the med

ian of the resulting data
Mathematics
2 answers:
Naddik [55]3 years ago
6 0
Y - 48 -apex , boneless chicken is good as well. Good luck
Nataliya [291]3 years ago
3 0

 Answer: Y-48

<span>If the median of the values in the data set is Y, it means that Y is the value in the middle of the data set. If 48 will be deducted from each of all values in the data set including Y, then the median of the resulting value will be Y-48. </span>

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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)!}}

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2 years ago
Solve this equation for A: A ÷ 2 = 4
Natali5045456 [20]
The answer to the equation is 8
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3 years ago
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Your credit card has a balance of $3300 and an annual interest rate of 14%. You decided to pay off the balance over two years. I
NARA [144]

9514 1404 393

Answer:

  • $137.90 more each month
  • $246.00 less total interest

Step-by-step explanation:

The amortization formula is ...

  A = P(r/12)/(1 -(1 +r/12)^(-12t))

for the monthly payment on principal P at annual rate r for t years. Here, we have P=3300, r = 0.14, and t=1, so the monthly payment is ...

  A = $3300(0.14/12)/(1 -(1 +0.14/12)^-12) ≈ $296.30

The payment of $296.30 is ...

  $295.30 -158.40 = $137.90 . . . more each month

The total amount paid is 12×$296.30 = $3555.60, so 255.60 in interest. This amount is ...

  $501.60 -255.60 = $246.00 . . . less total interest

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During an exit survey after a play, 75 of the first 120 people said they did not like the play.
MA_775_DIABLO [31]
<span>I think its A= 62% and B=37%  and sorry for the answer Brainly asked me</span>
4 0
3 years ago
A student makes money by watching the neighbors' dog. The situation is modeled in the graph below. Money Made 130 120 110 100 90
saul85 [17]

Answer:

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Step-by-step explanation:

4 0
2 years ago
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