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Alla [95]
3 years ago
5

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

hat charges $4 plus $0.20 per check. For what number of checks is the monthly cost of the two accounts the same? What is the cost?
Mathematics
1 answer:
wolverine [178]3 years ago
8 0

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 = x

Cost for the first bank for x checks = $5 + $0.10x

Cost for the second bank for x checks = $4 + $0.20x

As per question statement, both the costs are the same.

Comparing the values:

5 + 0.10x = 4 + 0.20x\\\Rightarrow 0.10x = 1\\\Rightarrow x = 10

So, for 10 number of checks the cost will be same.

The cost = 4 + 0.20 \times 10 = <em>$6</em>

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Answer:

6

Step-by-step explanation:

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3 years ago
Graph y = tanx for -pi/4 ≤ x ≤ pi/4. What is the range?
Debora [2.8K]

Answer:

( -1, 1 )

Step-by-step explanation:

For f ( x ) = tan x ;   Range = R (real no.)

Range in interval = ( -1, 1 )

3 0
3 years ago
Read 2 more answers
How many different linear arrangements are there of the letters a, b,c, d, e for which: (a a is last in line? (b a is before d?
inna [77]
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.
\text{Case 1: } 4 \cdot 4!

Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
\text{Case 2: } 3 \cdot 4!

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.
\text{Case 3: } 2 \cdot 4!

\text{Case 4: } 1 \cdot 4!

\text{Total arrangements}: 4 \cdot 4! + 3 \cdot 4! + 2 \cdot 4! + 1 \cdot 4! = 240

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).

\text{Total arrangements: } 3 \cdot 3! \cdot 2! = 36
7 0
3 years ago
Francine has a piece of wood that is five 5 5/12 feet long she uses three 3 1/4 feet of the wood from a science project how much
zhannawk [14.2K]

Answer:

4/3 ft

Corrected question;

Francine has a piece of wood that is five 5/12 feet long she uses three 1/4 feet of the wood from a science project how much for just Francine have left.

Step-by-step explanation:

Given;

Francine has a piece of wood that is five 5/12 feet long.

Total amount of wood T = 5 × 5/12 ft = 25/12 ft

she uses three 1/4 feet of the wood from a science project.

Amount of wood Used U = 3 × 1/4 = 3/4 ft

The amount of wood left equals the total amount of wood minus the amount of wood used;

The amount of wood left L = T - U = 25/12 - 3/4

L = 25/12 - 9/12

L = 16/12

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3 years ago
On a math test, Johnny made a raw score of 95 on a 40 question test. The correct answers were given 5 points each while incorrec
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X - the number of questions he answered correctly
y - the number of questions he answered incorrectly

There were 40 questions on the test.
x+y=40 \\ y=40-x

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Set 40-x and (-95/2)+(5/2)x equal to each other:
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Johnny answered correctly 25 questions.
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3 years ago
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