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3 years ago

Which of these data sets is symmetrical?

Mathematics

2 answers:

Licemer1 [7]3 years ago

5 0

My answers C. I dunno if it's correct tho

fenix001 [56]3 years ago

5 0

Answer:

13 is the data set for the symmetrical

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The table below shows the results of a screening program organized by Level 300 students of the department physiotherapy of the

Vladimir [108]

The correct answer is C because they are asking for the specific screening kit marks

7 0

2 years ago

Write the point-slope form of the line that passes through (5, 5) and is PARALLEL to a line with a slope of 1/4. Include all of

lesya692 [45]

Parallel lines will have the same slope

y - y1 = m(x - 1)
slope(m) = 1/4
(5,5)...x1 = 5 and y1 = 5
now we sub
y - 5 = 1/4(x - 5) <==== ur parallel line
==================
perpendiculr lines will have a negative reciprocal slope. To find the negative reciprocal, u flip the number and change the sign. So the negative reciprocal of 1/4 is -4....see how I flipped 1/4 and made it 4/1...then changed the sign making it -4/1 or just -4. That will be ur slope of the perpendicular line.

y - y1 = m(x - x1)
slope(m) = -4
(5,5)...x1 = 5 and y1 = 5
now we sub
y - 5 = -4(x - 5) <=== ur perpendicular line

4 0

3 years ago

An education researcher claims that 58% of college students work year-round. In a random sample of 400 college students, 232

Hatshy [7]

Answer:

The proportion of college students who work year-round is 58%.

Step-by-step explanation:

The claim made by the education researcher is that 58% of college students work year-round.

A random sample of 400 college students, 232 say they work year-round.

To test the researcher's claim use a one-proportion z-test.

The hypothesis can be defined as follows:

H₀: The proportion of college students who work year-round is 58%, i.e. p = 0.58.

Hₐ: The proportion of college students who work year-round is 58%, i.e. p ≠ 0.58. C

Compute the sample proportion as follows:

$\hat p=\frac{232}{400}=0.58$

Compute the test statistic value as follows:

$z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.58-0.58}{\sqrt{\frac{0.58(1-0.58)}{400}}}=0$

The test statistic value is 0.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected.

Compute the p-value for the two-tailed test as follows:

$p-value=2\times P(z$

*Use a z-table for the probability.

The p-value of the test is 1.

The p-value of the test is very large when compared to the significance level.

The null hypothesis will not be rejected.

Thus, it can be concluded that the proportion of college students who work year-round is 58%.

6 0

3 years ago

Kelly paid $45.36 for a sweater that was on sale for 25% off the original price. She paid 8% sales tax on the sale price.

Shalnov [3]

Take 45.36 *.08 to get the price of the sweater 45.36 *.08 =3.6288 so take 45.36-3.63= 41.73 and take 41.73 *.25 = 10.4325 take 41.73+10.43=52.16 is the original price of the sweater

5 0

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