There may be more than one correct answer here.
Note that 48 cubic units is equivalent to (6 units)(8 units)(1 unit). Can you think of any other 3 dimensions whose product is 48 cubic units?
Which data set represents the box plot? A) {56, 66, 93, 90, 57, 66, 69, 68, 76, 64, 48} B) {56, 66, 95, 90, 57, 66, 69, 68, 76,
iVinArrow [24]
Answer:
B) {56, 66, 95, 90, 57, 66, 69, 68, 76, 64, 51}
Step-by-step explanation:
{56, 66, 95, 90, 57, 66, 69, 68, 76, 64, 51} is the only data set that matches the maximum and minimum, upper and lower quartile, and median values.
Answer:
c. 6.2 ± 2.626(0.21)
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 101 - 1 = 100
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 100 degrees of freedom(y-axis) and a confidence level of
. So we have T = 2.626
The confidence interval is:

In which
is the sample mean while M is the margin of error.
The distribution of the number of puppies born per litter was skewed left with a mean of 6.2 puppies born per litter.
This means that 
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
Thus, the confidence interval is:

And the correct answer is given by option c.
We presume your cost function is
c(p) = 124p/((10 +p)(100 -p))
This can be rewritten as
c(p) = (124/11)*(10/(100 -p) -1/(10 +p))
The average value of this function over the interval [50, 55] is given by the integral

This evaluates to
(-124/55)*(ln(65/60)+10ln(45/50)) ≈ 2.19494
The average cost of removal of 50-55% of pollutants is about
$2.19 hundred thousand = $219,000