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MaRussiya [10]
2 years ago
14

If a 12 ounce box of cereal costs $3.84, how many ounces should be in a box marked $2.49?

Mathematics
2 answers:
Ivan2 years ago
6 0

Answer:

  • 7.78 ounces

Step-by-step explanation:

<u>We know that:</u>

  • 12 ounce box of cereal = $3.84

<u>Work:</u>

  • 12/3.84 ounce box of cereal = $1
  • => 12/3.84 x 2.49 box of cereal = $2.49
  • => 7.78 ounces (Rounded to nearest hundredth) = $2.49

Hence, 7.78 ounces cost $2.49.

Nostrana [21]2 years ago
6 0

7.8125 ounces

Step-by-step explanation:

Based on given conditions,

\frac{2.49 \times 12}{3.84} \\  =  >  \frac{ \frac{249}{100} \times 12 }{ \frac{384}{100} }

=  >  \frac{ \frac{249}{25} \times 3 }{ \frac{96}{25} }

=  > \frac{249}{25} \times 3 \times  \frac{25}{96}

=  > 83 \times 3 \times  \frac{1}{32}

=  > \frac{83 \times 3}{32}

=  >  \frac{249}{32}

=  > 7.78125ounces

By rounding off,

=  > 7.9ounces

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Andrej [43]
total weight of both bags = 3 oz + 20.6m oz

Don's bag = 2 oz
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3 years ago
8(-13)-9y=-50<br><br> what does y=
klio [65]

Answer:

y = -6

Step-by-step explanation:

Step 1: Define

8x - 9y = -50

x = -13

Step 2: Substitute and solve for <em>y</em>

8(-13) - 9y = -50

-104 - 9y = -50

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3 years ago
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in 15 seconds, connor counts 7 vehicles passing by his house. At this rate how many vehicles would pass by in 5 hours
lesya [120]
Multiply by 4 to get to one minute —>28.
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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
valkas [14]

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

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Answer:

i did a wild guess and i got  the answer is 2

Step-by-step explanation:

2*7=14

14-4=10  

10-2=8

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