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2 years ago

Find the value of a from the following data whose mean is 23 29,33, 27, a, 19, 17, 10

Mathematics

1 answer:

ollegr [7]2 years ago

5 0

Answer:

a = 26

Step-by-step explanation:

29 + 33 + 27 + a + 19 + 17 + 10 / 7 = 23

135 + a / 7 = 23

135 + a = 161

a = 26

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Mary is twice as old as Helen’s age, If 8 is subtracted from Helen’s Age and 4 is added to Mary Age? Mary will be 4 times as old

erik [133]

Answer:

set mary is x years old, and helen is y years old.

first sentence:x=2y

second one:4(y-8)=x+4

bring x=2y in 4(y-8)=x+4

=>4y-32=2y+4

=>2y=36

=>y=18

=>x=36

mary:36

helen:18

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3 years ago

A carpenter uses 10 1/2 ft of cedar for every 12 1/3 ft of redwood for a construction project. If the carpenter uses 5 1/4 ft of

motikmotik

Answer: 6 1/6

Step-by-step explanation:

1) First divide 10 1/2 = 5 1/4.

2) Then, since we divide 10 1/2 then we need to divide 12 1/3.

3) 12 1/3 divided by 2 = 6 1/6.

And there is your answer; 6 1/6

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3 years ago

Solve this inequality.<br> 8x + 6 x 54

sergeinik [125]

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3 years ago

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3 years ago

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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

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

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

4 0

2 years ago

Find the value of a from the following data whose mean is 23 29,33, 27, a, 19, 17, 10 ​

Find the value of a from the following data whose mean is 23 29,33, 27, a, 19, 17, 10