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postnew [5]
3 years ago
15

Raj’s bathtub is clogged and is draining at a rate of 1.5 gallons of water per minute. The table shows that the amount of water

remaining in the bathtub, y, is a function of the time in minutes, x, that it has been draining. What is the range of this function? all real numbers such that y ≤ 40 all real numbers such that y ≥ 0 all real numbers such that 0 ≤ y ≤ 40 all real numbers such that 37.75 ≤ y ≤ 40\

Mathematics
2 answers:
olchik [2.2K]3 years ago
7 0

Raj’s bathtub is clogged and is draining at a rate of 1.5 gallons of water per minute. The table shows that the amount of water remaining in the bathtub, y, is a function of the time in minutes, x, that it has been draining.



What is the range of this function?


all real numbers such that y ≤ 40

all real numbers such that y ≥ 0

all real numbers such that 0 ≤ y ≤ 40

all real numbers such that 37.75 ≤ y ≤ 40

grandymaker [24]3 years ago
6 0
Draining rate is at 1.5 gallons per minute
That would be y = 1.5g / x.

From the function or the equation, here is the answer.
<span>The range would be all real numbers such that 0 ≤ y ≤ 40.</span>

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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

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

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

   $=\frac{127.1143}{2}$

   = 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

3 0
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