Answer:
μ = 235.38
σ = 234.54
Step-by-step explanation:
Assuming the table is as follows:
![\left[\begin{array}{cc}Savings&Frequency\\\$0-\$199&339\\\$200-\$399&86\\\$400-\$599&55\\\$600-\$799&18\\\$800-\$999&11\\\$1000-\$1199&8\\\$1200-\$1399&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7DSavings%26Frequency%5C%5C%5C%240-%5C%24199%26339%5C%5C%5C%24200-%5C%24399%2686%5C%5C%5C%24400-%5C%24599%2655%5C%5C%5C%24600-%5C%24799%2618%5C%5C%5C%24800-%5C%24999%2611%5C%5C%5C%241000-%5C%241199%268%5C%5C%5C%241200-%5C%241399%263%5Cend%7Barray%7D%5Cright%5D)
This is an example of grouped data, where a range of values is given rather than a single data point. First, find the total frequency.
n = 339 + 86 + 55 + 18 + 11 + 8 + 3
n = 520
The mean is the expected value using the midpoints of each range.
μ = (339×100 + 86×300 + 55×500 + 18×700 + 11×900 + 8×1100 + 3×1300) / 520
μ = 122400 / 520
μ = 235.38
The variance is:
σ² = [(339×100² + 86×300² + 55×500² + 18×700² + 11×900² + 8×1100² + 3×1300²) − (520×235.38²)] / (520 − 1)
σ² = 55009.7
The standard deviation is:
σ = 234.54
19 is the answer 4+5(3)= 4+15=19
Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula.
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C.
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space.
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.
Answer:
4
Step-by-step explanation:
The distance from B to C is the same as the distance from A to D. Points B and C will both have the same x-value, as do points A and D. Points B and C will have the same difference in y-values (2 -(-2) = 4) that points A and D have.
The distance from B to C is 4 units.
6. three kids and four adults
