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3 years ago

15

There are 9 players on a baseball team. After the game each player high-fives once with everyone else. How many high-fives were

given?

1 answer:

Mekhanik [1.2K]3 years ago

3 0

Answer:

32

Step-by-step explanation:

Give each player a letter.

a, b, c, d, e, f, g, h, i

Each player high fives each other once.

a: b, c, d, e, f, g, h, i

b: c, d, e, f, g, h, i

c: d, e, f, g, h, i

d: e, f, g, h, i

f: g, h, i

g: h, i

h: i

Remember that you do not recount high-fives (as each player only gives it to another player once). Add the amount of high fives by counting.

32 is your answer.

~

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Answer:

Step-by-step explanation:

We want a line parallel to 3x − 4y = 7. Any such line will have an equation of the same form but a different constant.

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We know that

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$H_1 :$ At least one mean is different form the others (claim)

We need to find the critical values.

We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

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$\overline X_1 =50.5$ $\overline X_2 =50.07143$ $\overline X_3 =55.71429$

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The grand mean or the overall mean is(GM) :

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= 52.1714

The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

$=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$

$=\frac{127.1143}{2}$

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The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

$=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

$=\frac{669.8571}{32}$

= 20.93304

The F-test statistics value is :

$F=\frac{s_B^2}{s_W^2}$

$=\frac{63.55714}{20.93304}$

= 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

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Between 127.1143 2 63.55714 3.036212

Within 669.8571 32 20.93304

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