The density of a certain material is such that it weighs 430 kilograms per cubic foot of volume. Express this density in grams p
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There are 32,760 different orders for the first four positions.
This is a permutation where repetition is not allowed; the formula for that is:
For our problem, we have:
This is a permutation where repetition is not allowed; the formula for that is:
For our problem, we have:
Susannah's data are illustration of measure of central tendencies
<u>(a) The mean</u>
To calculate the mean, she needs to divide the total number of turkey sandwiches sold, by the number of days.
The total is:
The number of days is:
So, the mean is:
Hence, the mean number of turkey sandwiches sold is 99.4
<u>(b) The median</u>
To calculate the median, she needs to arrange the data in ascending or descending order, then select the middle item.
In ascending order, we have: 73, 85, 88, 88, 99, 129, 134
The middle item is 88
So:
Hence, the median number of turkey sandwiches sold is 88
<u>(c) The mode</u>
This is the data that occurs most.
From the dataset, 88 appears twice (more than others).
So:
Hence, the mode number of turkey sandwiches sold is 88
<u>(d) The range</u>
This is the difference between the highest and least data
From the dataset,
The highest is 134, and the least is 73
So:
Hence, the range of turkey sandwiches sold is 61
The measure that would be most helpful is the median, because it is not affected by outliers.
Read more about measures of central tendencies at:
A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:
_ _ _ _ a
Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways
b) Since a is before d, we need to account for all of the possible cases.
Case 1: a d _ _ _
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d
Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.
a b c _ _
We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:
_ a b c _
_ _ a b c
So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.
In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).
Let's start by drawing out a representation:
_ _ _ _ a
Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways
b) Since a is before d, we need to account for all of the possible cases.
Case 1: a d _ _ _
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d
Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.
a b c _ _
We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:
_ a b c _
_ _ a b c
So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.
In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).