We know that
if the volume of a prism <span>it was increased by a scale factor of 5
then
</span>the original volume is
multiplied by 5*5*5-------> 5³=125
therefore
the new volume compared with the original is 125 times larger
Answer:
3 students are eating lunches other than salads and sandwiches.
Step-by-step explanation:
To solve this you know that there are 18 students in the cafeteria and 1/6 of them are eating salads and 2/3 are eating sandwiches right? So you would have to think about what 1/6 of 18 is so you know how many students are eating salads, and 1/6 of 18 is 3 so there are 3 students eating salads. Now, you have to find out how many students are eating sandwiches, so you need to know what 2/3 of 18 is. 2/3 of 18 is 12 so now you also know that there are 12 students eating sandwiches. Next, you have to add 12 and 3 and you get 15. Since you know that there are 18 students in the cafeteria, you have to subtract 18 by 15, and you should get 3. So 3 students are eating lunches other than salads or sandwiches.
Hope this helps you! :D
Answer:
95891769
Step-by-step explanation:
This is extract from a brain teaser exercise in which sequence is formed to identify the numbers. In this brain teaser we have made combinations of different numbers which lead to the correct answer. The two correct forms of number are given which are used as a base for determining the correct answer.
Answer:
The lower bound for a 90% confidence interval is 0.2033.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of 
, and a confidence level of 
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of 
.
For this problem, we have that:
90% confidence level
So 
, z is the value of Z that has a pvalue of 
, so 
.
The lower limit of this interval is:
The lower bound for a 90% confidence interval is 0.2033.
Answer:
Following are the solution to this question:
Step-by-step explanation:
They provide various boxes or various objects. It also wants objects to be distributed into containers, so no container is empty. All we select k objects of r to keep no boxes empty, which (r C k) could be done. All such k artifacts can be placed in k containers, each of them in k! Forms. There will be remaining 
objects. All can be put in any of k boxes. Therefore, these objects could in the 
manner are organized. Consequently, both possible ways to do this are
Consequently, the number of ways that r objects in k different boxes can be arranged to make no book empty is every possible one
