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Dvinal [7]
2 years ago
9

I need help on this question, I also need a for formula ASAP!!!!

Mathematics
2 answers:
Vitek1552 [10]2 years ago
7 0

Answer:

2.25 cm³

Step-by-step explanation:

The volume of a pyramid can be found using the formula:

\boxed{\text{volume of pyramid= ⅓×base ×height}}

Base area

= area of square

= side ×side

= 1.5²

= 2.25 cm²

Volume of paperweight

= ⅓(2.25)(3)

= 2.25 cm³

deff fn [24]2 years ago
7 0
Here,
W= 1.5cm
H=3
L=1.5cm
We have volume of pyramid is lwh/3
=1.5x1.5x3/3=2.25cm
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keeping in mind that parallel lines have the same exact slope, hmmmm what's the slope of the line above anyway?

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\bf (\stackrel{x_1}{18}~,~\stackrel{y_1}{2})~\hspace{10em} \stackrel{slope}{m}\implies \cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{2}=\stackrel{m}{\cfrac{1}{3}}(x-\stackrel{x_1}{18}) \\\\\\ y-2=\cfrac{1}{3}x-6\implies y=\cfrac{1}{3}x-4

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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

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

$H_1 :$ 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

        = 35 - 3 = 32

SO the critical value is 3.295

The mean and the variance of each sample :

Goust                      Jet red                 Cloudtran

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

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

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

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

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

   $=\frac{127.1143}{2}$

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The Variance within the groups

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

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

   $=\frac{669.8571}{32}$

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The F-test  statistics value is :

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

  $=\frac{63.55714}{20.93304}$

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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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Step-by-step explanation:

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