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Mariana [72]
2 years ago
11

Several batches of stew were made yesterday. Each batch required 1 and two-thirds pounds of meat. All together, 10 and StartFrac

tion 5 over 6 EndFraction pounds of meat was used. Janice tried to find the number of batches of stew made. Her work is shown below.
10 and StartFraction 5 over 6 EndFraction divided by 1 and two-thirds = StartFraction 65 over 6 EndFraction divided by five halves = StartFraction 65 over 6 EndFraction times StartFraction 2 over 5 EndFraction = StartFraction 130 over 30 EndFraction = 4 and one-third

When she checked the answer by estimating, it did not make sense. What error did Janice make?
Mathematics
2 answers:
Sav [38]2 years ago
8 0

Answer:

Step 2

1 2/3 was converted as 5/2 instead of 5/3

Step-by-step explanation:

  • Each batch = 1 2/3
  • Total = 10 5/6

<u>Number of batches </u>

  • 10 5/6 : 1 2/3 =
  • 65/6 : 5/3 =
  • 65/6 * 3/5 =
  • 65/10 =
  • 6.5

<u>Solution by Janice:</u>

  1. 10 5/6 : 1 2/3 =
  2. 65/6 : <u>5/2</u> =
  3. 65/6 * 2/5 =
  4. 130/30 =
  5. 4 1/3

<u>Error made at step 2:</u>

  • StartFraction 65 over 6 EndFraction divided by five halves = StartFraction 65 over 6 EndFraction times StartFraction 2 over 5

1 2/3 was converted as 5/2 instead of 5/3

LuckyWell [14K]2 years ago
8 0

Answer:

Step 2

Step-by-step explanation:

Got it right on the Edge test.

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J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

                      P.Q.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~ t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean for mail-order sales = $74.50

\bar X_2 = sample mean for internet sales = $84.40

s_1 = sample standard deviation for mail-order purchases = $17.25

s_2 = sample standard deviation for internet purchases = $21.25

n_1 = sample of sales receipts for mail-order purchases = 16

n_2 = sample of sales receipts for internet purchases = 9

Also,  s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }  =  \sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} } = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by (\mu_1-\mu_2).

Now, 99% confidence interval for (\mu_1-\mu_2) is given by;

             = (\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_)  \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

          = (74.50-84.40) \pm (2.807  \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})

          = [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

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I show an example below of what I mean.

In that diagram, the row and column mentioned intersect at 2.262 (which is approximate). This value then rounds to 2.26

<h3>Answer:  2.26</h3>

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