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seraphim [82]
3 years ago
14

Plz guys help me plz

Mathematics
1 answer:
Verizon [17]3 years ago
4 0

Answer:

B

Step-by-step explanation:

✔️Rate of change using two points on the line, say (2, 60) and (3, 75);

Rate of change (m) = ∆y\/∆x = (75 - 60) / (3 - 2) = 15/1

Rate of change = 15

This means each class costs $15

✔️Initial value is the y-intercept (b). It represents the initial start-up fee.

To find initial value/y-intercept (b), substitute m = 15 and (x, y) = (2, 60) into y = mx + b.

Thus:

60 = 15(2) + b

60 = 30 + b

60 - 30 = b

30 = b

b = 30

Therefore, initial start-up fee is $30.

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Answer:

12x - 4.2

4.2 - (12+x)

4.2 - 12x

(12+x) - 4.2

6 0
3 years ago
Craig has $1850 dollars in a bank account that he uses to make automatic payments of $400.73 on his car loan. If Craig stops mak
Ivenika [448]

Given:

Amount in the bank account = $1850

Monthly payment of can loan = $400.73

To find:

When would automatic payments make the value of the account zero?

Solution:

Craig stops making deposits to that account. So, amount $1850 in the bank account is used to make monthly payment of can loan.

On dividing the amount by monthly payment, we get

\dfrac{1850}{400.73}=4.61657

It means, the amount is sufficient for 4 payment but for the 5th payment the amount is not sufficient.

Therefore, the 5th automatic payments make the value of the account zero.

6 0
3 years ago
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Triss [41]
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Answer: 9 years ago
7 0
3 years ago
Simplify the expression by combining like terms.<br><br> 3.6x+5.9−2.2−1.7x
makvit [3.9K]

Answer:

1.9x+3.7

Step-by-step explanation:

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7 0
3 years ago
Target Costing
Lynna [10]

The target cost for Model J20 is $2000.

The required cost reduction is $30.

The total saving will be $30.3.

<h3>How to compute the cost?</h3>

The target cost will be:

= Sales price - (100% × Target cost)

= 400 - (100% × Y)

Y + Y = 400

2Y = 400

Y = 400/2 = 200

Target cost = $200

Therefore, the target cost is $200.

The required cost reduction will be:

= $230 - $200

= $30

The required cost reduction is $30.

The three engineering improvements together will be:

Direct labor reduction = 7.5

Additional inspection = 17

Injection molding = 5.8

Total savings = 30.3

The total savings is $30.3.

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6 0
2 years ago
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