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2 years ago

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The class scores of a history test have a normal distribution with a mean Mu = 79 and a standard deviation Sigma = 7. If Opal’s

test score was 72, which expression would she write to find the z-score of her test score?.

1 answer:

e-lub [12.9K]2 years ago

3 0

Using the normal distribution, we got that expression for the z-score of her test score is

$z= \frac{72-29}{7} \\\\z=-7/7\\\\z=-1$

<h3>What is Normal Probability Distribution?</h3>

it's a type of probability distribution which is symmetric about the mean, it gives data near the mean are more frequent than data far from the mean.

Given that mean = $\mu=79$

and standard deviation = $\sigma = 7$

X=73

we know that formula of z-score is

$z=\frac{X-\mu}{\sigma}$

Hence Z-score can be calculated as

$z= \frac{72-29}{7} \\\\z=-7/7\\\\z=-1$

Using the normal distribution, we got that expression for the z-score of her test score is

$z= \frac{72-29}{7} \\\\z=-7/7\\\\z=-1$

To learn more about the normal distribution visit : brainly.com/question/24663213

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Need help with this ASAP

Mariana [72]

Answer:

Sequence: 13, 20, 27

Rule: Tn = 7n + 6

Step-by-step explanation:

The 3 other numbers that can form an arithmetic progression is 13, 20, 27...

The nth term of an arithmetic progression is expressed as;

Tn = a + (n-1)d

a is the first term = 13

n is the number of terms

d is the common difference = 20 - 13 = 27 - 20

d = 7

Substitute

Tn = 13 + (n-1)(7)

Tn = 13 + 7n - 7

Tn = 7n+13-7

Tn = 7n + 6

This gives the required rule

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3 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

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

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

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2 years ago

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7 bicycles dived by two mopeds = 3.5
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3 years ago

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Step-by-step explanation:

y/x = 75/1 = 75

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3 years ago

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3 years ago

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