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Harlamova29_29 [7]
2 years ago
15

Your friend says that the absolute value equation ∣3x + 8 ∣ − 9 = −5 has no solution because the constant on the right side of t

he equation is negative. Is your friend correct? Explain.
Mathematics
2 answers:
zepelin [54]2 years ago
7 0
No, because u add 9 to both sides to leave what's in the absolute value brackets alone so u get l3x+8l=4.Since the sum is positive then it has a solution.
SashulF [63]2 years ago
5 0
No, ur friend is not correct

| 3x + 8 | - 9 = -5....u have to get the absolute value (and everything in it) on one side of the equal sign, and everything else on the other side. So we have to get that -9 across the equal sign by adding 9 to both sides
| 3x + 8 } = -5 + 9
| 3x + 8 | = 4...so that number is not negative anymore
now u can set up ur 2 equations and solve for x...ur 2 equations would be: 
3x + 8 = 4 and 3x + 8 = -4...so there will be 2 solutions

** if that -9 would not have been there, then ur friend would have been correct
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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.
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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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