CyberCafe charges a computer station rental fee of $5 plus $0.20 for each quarter hour spent surfing. Which expression represent

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

babunello [11817]

The 99% confidence interval for the actual mean difference between average mail-order and internet purchase amounts falls within [$(-31.82), $12.02].

Step-by-step clarification:

We know a random sample of 16 mail-order sales receipts shows a mean sale amount of $74.50 and a standard deviation of $17.25.

For internet sales, a random sample of 9 receipts gives a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal value utilized for constructing a 99% confidence interval for the true mean difference is given by;

P.Q. =

~

where,

= sample mean of mail-order sales = $74.50 $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ $t__n_1_+_n_2_-_2$

= sample mean of internet sales = $84.40

$\bar X_1$ = standard deviation for mail-order sales = $17.25

$\bar X_2$ = standard deviation for internet sales = $21.25

$s_1$ = number of mail-order sales receipts = 16

= number of internet sales receipts = 9 $s_2$

Furthermore,

= $n_1$ = 18.74

$n_2$

The actual mean difference between average mail-order and internet purchases is denoted by ( $s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ $\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }$

).

Thus, the 99% confidence interval for ( $\mu_1-\mu_2$ ) is expressed as;

= $\mu_1-\mu_2$ Here, the t critical value at the 0.5% significance level with 23 degrees of freedom is 2.807. =

= [$-31.82, $12.02] $(\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

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

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

5 0

2 months ago

1. What is the probability of rolling a number greater than a 4 on a standard dice? Remember to reduce your fraction.

babunello [11817]

1. 1/3. 2. 7.

6 0

1 month ago

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A company that is selling computer tablets has determined that it regularly makes a net profit of $180 for each computer tablet

PIT_PIT [12445]

Response:

Based on the anticipated number of individuals completing the form to claim the refund, the company ought to foresee its net profit decreasing to:

$174.6 per computer tablet.

Detailed explanation:

Initially, the profit per tablet is $180, but after initiating the rebate program, it is expected that 18% of purchasers will fill out the forms and obtain the VISA card, which will lower the profit as demonstrated below:

Calculation of the reduced benefit for 18% of computer tablets = original benefit - refund amount.
Reduced benefit for 18% of tablets = $180 - $30 = $150

From here, we find that 18% of tablets will yield a benefit of $150, while the other 82% will maintain a benefit of $180. To ascertain an overall figure for the benefit, we must compute it as follows:

Total benefit = Percentage of reimbursed tablets * Benefit of reimbursed tablets + Percentage of non-reimbursed tablets * Benefit of non-reimbursed tablets.
Overall profit = 18% * 150 + 82% * 180
Overall profit = $174.6

Thus, after the $30 rebate on 18% of the tablets, the total benefit amounts to $174.6.

6 0

1 month ago

Finally, you meet a group of six natives, U, V, W, X, Y, and Z, who speak to you as follows. U says: None of us is a knight. V s

tester [12383]

Answer:

w and y are the knights

Step-by-step explanation:

6 0

2 months ago

A factory produced 150 tons of steel in one year. The next year, the output increased to 180 tons. What is the percentage increa

Svet_ta [12734]

The increase in production is found by calculating 180 - 150 = 30 tons. The percentage rise is computed by taking 30 divided by 150, which equals 0.2 or a 20% rise.

3 0

2 months ago

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