There are 200 students in the 7th grade class at Brookside Middle School. Of these students, 12% play volleyball and 15% play ba

Answer:

   

Hence, in this case, the 98% confidence interval would be (163.626;180.374)    

Step-by-step breakdown:

Previous concepts

A confidence interval represents a range that is likely to encompass a population value within a specific confidence level, typically expressed as a percentage whereby a population mean falls between an upper and lower limit.

The margin of errorindicates the span of values surrounding the sample statistic in a confidence interval.

A normal distributionillustrates a probability distribution that is symmetrical around the mean, signifying that values near the mean occur more frequently than those farther away from it.

denote the sample mean

population mean (the variable of interest)

s=16 signifies the sample standard deviation

n=23 represents the sample size  

The solution to the query

The equation for the confidence interval of the mean is given by the following formula:

  (1)

To determine the critical value , we first need to calculate the degrees of freedom, which is expressed as:

Since the confidence level is 0.98 or 98%, we find the value of and using tools like Excel or a calculator, where the Excel command would be: "=-T.INV(0.01,22)". This yields

Having all components ready, we can substitute into formula (1):

   

Thus, for this case, the 98% confidence interval will be (163.626;180.374)    

Answer: 8x - 1200

Step-by-step explanation: I solved the problem

<span>15000 feet The height of Mt. Whitney is listed as 14,505 feet. To round this figure to the closest thousand, we examine the hundreds digit, which is 5. According to the rounding rule, "5 or above, round up, 4 or below, round down". Given that the digit is 5, we round up to yield 15000 feet</span>

Answer:

The hourly rate for lectures is $7.33

Step-by-step explanation:

* Let's break down how to tackle the problem.

- For the level 3 course, examination hours are priced at double that of workshop hours.

- Workshop hours cost twice the rate of lecture hours.

- The total includes examination, workshop, and lecture hours.

- Examination lasts 3 hours, workshops 24 hours, and lectures 12 hours.

* Let’s denote the cost of lecture hours as $x per hour.

∴ The lectures cost $x per hour.

∵ Workshop charge is twice that of lectures

∴ Workshop hours cost 2(x) = 2x per hour.

∵ Examination fees are double that of workshop hours

∵ The workshop cost is 2x

∴ Examination fees are 2(2x) = 4x per hour.

- Combining costs for level 3 gives us the total of lecture, workshop, and examination hours.

∵ 12 hours for lectures

∵ 24 hours for workshops

∵ 3 hours for examinations

∵ Thus the total cost for level 3 = 12(x) + 24(2x) + 3(4x).

∴ Total cost for level 3 = 12x + 48x + 12x.

∵ Therefore, total cost = $528.

∴ 12x + 48x + 12x = 528.

∴ 72x = 528; hence we divide both sides by 72.

∴ x = 7.33.

∵ x represents the cost of lecture hours per hour.

∴ Therefore, the hourly price for lectures is $7.33.

96 soldiers will remain unassigned and the arrangement will consist of 52 rows.