Suppose a bank charges a base rate of $5 plus $0.10 per check for your checking account. You can switch to a different account t
1 answer:
Answer:
10 number of checks.
Cost = $6
Step-by-step explanation:
Given that:
Fixed Charges of first bank = $5
Charges of first bank per check = $0.10
Fixed Charges of second bank = $4
Charges of second bank per check = $0.20
To find:
Number of checks such that the charges for the banks become the same?
Solution:
Let the number of checks =
Cost for the first bank for checks = $5 + $0.10
Cost for the second bank for checks = $4 + $0.20
As per question statement, both the costs are the same.
Comparing the values:
So, for 10 number of checks the cost will be same.
The cost = 4 + 0.20 10 = <em>$6</em>
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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).
X - the number of questions he answered correctly
y - the number of questions he answered incorrectly
There were 40 questions on the test.
The correct answers were given 5 points each, the incorrect answers were given -2 points each. He made a score of 95.
Set 40-x and (-95/2)+(5/2)x equal to each other:
Johnny answered correctly 25 questions.
y - the number of questions he answered incorrectly
There were 40 questions on the test.
The correct answers were given 5 points each, the incorrect answers were given -2 points each. He made a score of 95.
Set 40-x and (-95/2)+(5/2)x equal to each other:
Johnny answered correctly 25 questions.