9 divided by 883 is 0.010192525481314
How:
i used a calculator
Answer:
Answer : D
Step-by-step explanation:
A school has two kindergarten classes. There are 21 children in Ms. Toodle's kindergarten class. Of these, 17 are "pre-readers" children on the verge of reading. There are 19 children in Mr. Grimace's kindergarten class. Of these, 13 are pre-readers. Using the plus four confidence interval method, a 90% confidence interval for the difference in proportions of children in these classes that are pre-readers is â0.104 to 0.336.
Which of the following statements is correct?
A) This confidence interval is not reliable because the samples are so small.
B)This confidence interval is of no use because it contains 0, the value of no difference between classes.
C)This confidence interval is reasonable because the sample sizes are both at least 5.
D) This confidence interval is not reliable because these samples cannot be viewed as simple random samples taken from a larger population.
The Answer is D - This confidence interval is not reliable because these samples cannot be viewed as simple random samples taken from a larger population.
In this setup, all the students are already involved in the data. This is not a sample from a larger population, but probably, the population itself.
Indegree-The number of inward directed graph edges from a given graph vertex in a directed
graph.
Outdegree-The number of outward directed graph edges from a given graph vertex in a directed
graph.
Step-by-step explanation:
this is a college question ?
oh, my dear ...
we start at -426 ft.
swimming upwards reduces depth by adding hight.
so,
-426 + 212 = -214 ft
basically the same method is used as if we would subtract 212 from 426.
so, the diver is currently at -214 ft (214 ft below the surface).
Answer:
The 95% confidence interval for the population mean span of boys from participating schools in Canada is between 154.3 cm and 160.3 cm.
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 = 100 - 1 = 99
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 99 degrees of freedom(y-axis) and a confidence level of 
. So we have T = 1.9842
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 157.3 - 3 = 154.3 cm
The upper end of the interval is the sample mean added to M. So it is 157.3 + 3 = 160.3 cm.
The 95% confidence interval for the population mean span of boys from participating schools in Canada is between 154.3 cm and 160.3 cm.
