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Andrew [12]
2 years ago
7

|-9| = |9| true or false.

Mathematics
2 answers:
bogdanovich [222]2 years ago
6 0

Step-by-step explanation:

1) -9 = 9 true/false

the answer is false because negative and positive quite unlike in terms.

2) -(-21) = -21

Using the symbols multiplication rule,

Where,

<h2>+ × + = +</h2><h2>- × - = +</h2><h2>- × + = -</h2><h2>+ × - = -</h2>

Therefore, -(-21) = +21 = 21

⟼ 21 = -21 \sf true / \:  \sout{false}

Sladkaya [172]2 years ago
6 0

     \text{\#1, True       \;\;\;\;\;\;\;\;\;\;\; }            |-9| = |9|

The absolute value (the distance from 0) of -9 and 9:

9 = 9

     This is true.

--------------------

     \text{\#2, False       \;\;\;\;\;\;\;\;\;\;\;}           -(-21) = -21

Given:

-(-21) = -21

Negative times a negative is a positive:

21 ≠ -21

     This is not true.

--------------------

<em>To help you answer the last one about P on the number line, we will need more context.</em>

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Solution :

We know that

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

$H_1 :$ 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

       = 3 - 1 = 2

d.f.D = N - k

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The mean and the variance of each sample :

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

$s_1^2=19.96154$      $s_2^2=14.68681$         $s_3^2=36.57143$

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

$\overline X_{GM}=\frac{\sum \overline X}{N}$

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

        = 52.1714

The variance between the groups

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

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

   $=\frac{127.1143}{2}$

   = 63.55714

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 :

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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