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2 years ago

6

The seventh grade students are having an end-of-year party in the school cafeteria. There are 754 students attending the party.

Each table seats 6 students. How many tables are needed to seat all of the students? Fill in the blanks to complete the sentences
Determine the number of tables by dividing_______ by_______. If there is a remainder, the answer will need to be rounded_______. There ______a remainder, so________ tables are needed.

1 answer:

valina [46]2 years ago

5 0

Answer:

Determine the number of tables by dividing 754 by 6. If there is a remainder, the answer will need to be rounded tenths or hundredths . There is a remainder, so 125.6 tables are needed.

Step-by-step explanation:

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3 years ago

Jacob Lee is a frequent traveler between Los Angeles and San Diego. For the past month, he wrote down the flight times in minute

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Solution :

We know that

$H_0: \mu_1 = \mu_2=\mu_3$

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We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

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Goust Jet red Cloudtran

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

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$=\frac{51+51+...+49+49}{35}$

= 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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$=\frac{669.8571}{32}$

= 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 :

Source Sum of squares d.f Mean square F

Between 127.1143 2 63.55714 3.036212

Within 669.8571 32 20.93304

Total 796.9714 34

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borishaifa [10]

Answer:

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lawyer [7]

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<u>Explanation:</u>

Given:

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