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xxTIMURxx [149]
2 years ago
5

An experiment was done to look at whether there is an effect of the number of hours spent practising a musical instrument and ge

nder on the level of musical ability. A sample of 30 (15 men and 15 women) participants who had never learnt to play a musical instrument before were recruited. Participants were randomly allocated to one of three groups that varied in the number of hours they would spend practising every day for 1 year (0 hours, 1 hours, 2 hours). Men and women were divided equally across groups. All participants had a one-hour lesson each week over the course of the year, after which their level of musical skill was measured on a 10-point scale ranging from 0 (you can’t play for toffee) to 10 (‘Are you Mozart reincarnated?’). An ANOVA was conducted on these data, and the results revealed significant main effects of gender and the number of hours spent practising. Which of the following contrasts could we use to break down the effect of gender?
A- Helmert contratst
B- Simple contrast
C- Repeated contrast
D- None
Mathematics
1 answer:
AVprozaik [17]2 years ago
8 0

Answer: Repeated contrast

Step-by-step explanation:

In the two way ANOVA that was performed above, there were 3 groups of which 30 people (15 men and 15 women) participants who had never learnt to play a musical instrument before were recruited and divided equally.

The ANOVA is of repeated measures. There is within group effect, between group effect and interaction effect. The result also shows significant main effect for the gender and the hours used practicing. Hence, to breakdown the effect of gender, the repeated contrast method will be utilized. The repeated contrast method compares the mean of every level with the succeeding level except the last level.

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Charles has a collection of dimes and quarters worth $1.25. He has 8 coins. Write a systems of equations to represent this situa
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Answer:

* The systems of equations are:

# d + q = 8 ⇒ (1)

# 10d + 25q = 125 ⇒ (2)

Charles has 5 dimes and 3 quarters

Step-by-step explanation:

* Lets explain how to solve the problem

- Charles has a collection of dimes and quarters worth $1.25

- He has 8 coins

* To solve the problem remember that:

# 1 dim = 10 cents

# 1 quarter = 25 cents

# 1 dollar = 100 cents

- Assume that the number of dimes is d and the number of

 quarter is q

∵ Charles has 8 coins

- The number of dimes and the number of quarters equal the

  number of the coins

∴ d + q = 8 ⇒ (1)

∵ 1 dime = 10 cents

∴ The value of dimes = 10 × d = 10d

∵ 1 quarter = 25 cents

∴ The value of quarters = 25 × q = 25q

∵ The collection worth $1.25

∵ 1 dollar = 100 cents

∴ The collection worth = 1.25 × 100 = 125 cents

∴ 10d + 25q = 125 ⇒ (2)

* The systems of equations are:

# d + q = 8 ⇒ (1)

# 10d + 25q = 125 ⇒ (2)

* Lets solve the equations

- Multiply equation (1) by (-10) to eliminate d

∴ -10d + -10q = -80 ⇒ (3)

- Add equations (2) and (3)

∴ 15q = 45

- Divide both sides by 15

∴ q = 3

- Substitute the value of q in equation (1) to find the value of d

∴ d + 3 = 8

- Subtract 3 from both sides

∴ d = 5

∵ d represents the number of dimes and q represents the number

  of quarters

∴ Charles has 5 dimes and 3 quarters

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Step-by-step explanation:

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James conducted an experiment with 4 possible outcomes. He determined that the experimental probability of event A happening is
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Power series of y''+x^2y'-xy=0
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Assuming we're looking for a power series solution centered around x=0, take

y=\displaystyle\sum_{n\ge0}a_nx^n
y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}
y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}

Substituting into the ODE yields

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}+\sum_{n\ge1}na_nx^{n+1}-\sum_{n\ge0}a_nx^{n+1}=0

The first series starts with a constant term; the second series starts at x^2; the last starts at x^1. So, extract the first two terms from the first series, and the first term from the last series so that each new series starts with a x^2 term. We have

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=2a_2+6a_3x+\sum_{n\ge4}n(n-1)a_nx^{n-2}

\displaystyle\sum_{n\ge0}a_nx^{n+1}=a_0x+\sum_{n\ge1}a_nx^{n+1}

Re-index the first sum to have it start at n=1 (to match the the other two sums):

\displaystyle\sum_{n\ge4}n(n-1)a_nx^{n-2}=\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}

So now the ODE is

\displaystyle\left(2a_2+6a_3x+\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}\right)+\sum_{n\ge1}na_nx^{n+1}-\left(a_0x+\sum_{n\ge1}a_nx^{n+1}\right)=0

Consolidate into one series starting n=1:

\displaystyle2a_2+(6a_3-a_0)x+\sum_{n\ge1}\bigg[(n+3)(n+2)a_{n+3}+(n-1)a_n\bigg]x^{n+1}=0

Suppose we're given initial conditions y(0)=a_0 and y'(0)=a_1 (which follow from setting x=0 in the power series representations for y and y', respectively). From the above equation it follows that

\begin{cases}2a_2=0\\6a_3-a_0=0\\(n+3)(n+2)a_{n+3}+(n-1)a_n=0&\text{for }n\ge2\end{cases}

Let's first consider what happens when n=3k-2, i.e. n\in\{1,4,7,10,\ldots\}. The recurrence relation tells us that

a_4=-\dfrac{1-1}{(1+3)(1+2)}a_1=0\implies a_7=0\implies a_{10}=0

and so on, so that a_{3k-2}=0 except for when k=1.

Now let's consider n=3k-1, or n\in\{2,5,8,11,\ldots\}. We know that a_2=0, and from the recurrence it follows that a_{3k-1}=0 for all k.

Finally, take n=3k, or n\in\{0,3,6,9,\ldots\}. We have a solution for a_3 in terms of a_0, so the next few terms (k=2,3,4) according to the recurrence would be

a_6=-\dfrac2{6\cdot5}a_3=-\dfrac2{6\cdot5\cdot3\cdot2}a_0=-\dfrac{a_0}{6\cdot3\cdot5}
a_9=-\dfrac5{9\cdot8}a_6=\dfrac{a_0}{9\cdot6\cdot3\cdot8}
a_{12}=-\dfrac8{12\cdot11}a_9=-\dfrac{a_0}{12\cdot9\cdot6\cdot3\cdot11}

and so on. The reordering of the product in the denominator is intentionally done to make the pattern clearer. We can surmise the general pattern for n=3k as

a_{3k}=\dfrac{(-1)^{k+1}a_0}{(3k\cdot(3k-3)\cdot(3k-2)\cdot\cdots\cdot6\cdot3\cdot(3k-1)}
a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^k(k\cdot(k-1)\cdot\cdots\cdot2\cdot1)\cdot(3k-1)}
a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^kk!(3k-1)}

So the series solution to the ODE is given by

y=\displaystyle\sum_{n\ge0}a_nx^n
y=a_1x+\displaystyle\sum_{k\ge0}\frac{(-1)^{k+1}a_0}{3^kk!(3k-1)}

Attached is a plot of a numerical solution (blue) to the ODE with initial conditions sampled at a_0=y(0)=1 and a_1=y'(0)=2 overlaid with the series solution (orange) with n=3 and n=6. (Note the rapid convergence.)

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the cost of one soccer ball is 150/6 or 25 dollars.

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