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kati45 [8]
3 years ago
13

6g+8(v-g) how do I solve this?

Mathematics
1 answer:
Papessa [141]3 years ago
8 0
Let's simplify step-by-step.

<span><span>6g</span>+<span>8<span>(<span>v−g</span>)

</span></span></span>Distribute:

<span>=<span><span><span>6g</span>+<span><span>(8)</span><span>(v)</span></span></span>+<span><span>(8)</span><span>(<span>−g</span>)

</span></span></span></span><span>=<span><span><span><span>6g</span>+<span>8v</span></span>+</span>−<span>8g

</span></span></span>Combine Like Terms:

<span>=<span><span><span>6g</span>+<span>8v</span></span>+<span>−<span>8g

</span></span></span></span><span>=<span><span>(<span><span>6g</span>+<span>−<span>8g</span></span></span>)</span>+<span>(<span>8v</span>)

</span></span></span><span>=<span><span>−<span>2g</span></span>+<span>8v

</span></span></span>Answer:

<span>=<span><span>−<span>2g</span></span>+<span>8v</span></span></span>
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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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Oduvanchick [21]
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x/-9= 1                     multiply by -9 

x= -9

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To get x by itself, you would add 13 to both sides. x=115
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