K. Brew sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet. Random samples of sales receipts were studied for mail-order sales and internet sales, with the total purchase being recorded for each sale. A random sample of 8 sales receipts for mail-order sales results in a mean sale amount of $76.10 with a standard deviation of $24.75. A random sample of 17 sales receipts for internet sales results in a mean sale amount of $66.00 with a standard deviation of $16.25. Using this data, find the 95% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases. Assume that the population variances are not equal and that the two populations are normally distributed. Step 1 of 3: Find the point estimate that should be used in constructing the confidence interval.

20days louise can read300 pg

step-by-step explanation:

1 day 15 pg 300: 15= 20

x day 300 pg

third choice down

step-by-step explanation:

the first inequality is based on the number of adults and campers, and the second inequality is based on the cost of feeding both the adult and the camper.

we know that there is room for 525 people max.   it's ok to have less than 525, but it's not ok to have more than 525, so the inequality is less than or equal to.   the number combination of adults (x) and campers (y) cannot exceed this 525 max.   therefore:

x + y ≤ 525

now we talk about cost.   it costs $9 to feed one adult, so depending upon the number of adults you have (x) depends upon how much it will cost to feed that number of adults.   same with the campers, who cost $5 to feed.   the adults can be represented by 9x and the campers can be represented by 5y.   since the budget cannot exceed 1,200, the second inequality is:

9x + 5y ≤ 1200