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Anastaziya [24]
3 years ago
14

Alana saved $1,200.00 to buy a pool table, but decided instead to charge it to her credit card. If the credit card had an intere

st rate of 13.5% for 6 months with no other fees, and she made no other purchases, what was her cost of credit of using the credit card instead of paying cash?
The cost of credit is the amount that a person pays over and above the amount borrowed.
P is the principal,
r is the interest rate,
m is the number of monthly payments,
M is the monthly payment

Mathematics
1 answer:
MrRa [10]3 years ago
8 0

Answer:

the answer is $47.70

Step-by-step explanation:

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What is the best solution of the system of equations? y= 3x-4 -3y= -9x +12
Fed [463]

Answer:

C) infinitely many solutions

Step-by-step explanation:

y= 3x-4  

-3y= -9x +12

-3(3x-4 ) = -9x +12

-9x + 12 = -9x + 12

C) infinitely many solutions

7 0
3 years ago
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What is the LCD for ?11/20 31/45<br><br> 315<br><br><br> 105<br><br><br> 66<br><br><br> 945
Mnenie [13.5K]
None of these are choices, because in order to make 20ths equal a 45th, it was to end in a 20,40,60,80, or ending in 00. 
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3 years ago
Brainliest Answer!!!
AleksandrR [38]
Its fairly straightforward. Since the bottom equation only has one unknown,x, because y=1.3, you can plug y in and solve for x. Once you find the value of x, you then have the value for two variables, x and y, and again have one unknown coefficient a. To solve for the coefficient you just plug in your y value (1.3) and your x value (which can be rounded to 0.42). Using a little bit of algebra, you can then solve for a which should be a=2.108. I am not sure if your teacher wants you to solve it this way but you could also use the elimination method or substitution method that you would of learned when discussing system of equations. But no matter which way you do it, the math follows the rules. Hope this helps. I’d suggest you solve it yourself to double check my work.

To verify my credibility,
I am a Mechanical Engineering major w/ minor in mathematics
7 0
3 years ago
Please help!! Will give brainlist is answer right
Drupady [299]

Answer:

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Step-by-step explanation:

8 0
2 years ago
Find the area under the standard normal probability distribution between the following pairs of​ z-scores. a. z=0 and z=3.00 e.
prohojiy [21]

Answer:

a. P(0 < z < 3.00) =  0.4987

b. P(0 < z < 1.00) =  0.3414

c. P(0 < z < 2.00) = 0.4773

d. P(0 < z < 0.79) = 0.2852

e. P(-3.00 < z < 0) = 0.4987

f. P(-1.00 < z < 0) = 0.3414

g. P(-1.58 < z < 0) = 0.4429

h. P(-0.79 < z < 0) = 0.2852

Step-by-step explanation:

Find the area under the standard normal probability distribution between the following pairs of​ z-scores.

a. z=0 and z=3.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 3.00) = 0.9987

Thus;

P(0 < z < 3.00) = 0.9987 - 0.5

P(0 < z < 3.00) =  0.4987

b. b. z=0 and z=1.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 1.00) = 0.8414

Thus;

P(0 < z < 1.00) = 0.8414 - 0.5

P(0 < z < 1.00) =  0.3414

c. z=0 and z=2.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 2.00) = 0.9773

Thus;

P(0 < z < 2.00) = 0.9773 - 0.5

P(0 < z < 2.00) = 0.4773

d.  z=0 and z=0.79

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 0.79) = 0.7852

Thus;

P(0 < z < 0.79) = 0.7852- 0.5

P(0 < z < 0.79) = 0.2852

e. z=−3.00 and z=0

From the standard normal distribution tables,

P(Z< -3.00) = 0.0014  and P(Z< 0) = 0.5

Thus;

P(-3.00 < z < 0 ) = 0.5 - 0.0013

P(-3.00 < z < 0) = 0.4987

f. z=−1.00 and z=0

From the standard normal distribution tables,

P(Z< -1.00) = 0.1587  and P(Z< 0) = 0.5

Thus;

P(-1.00 < z < 0 ) = 0.5 -  0.1586

P(-1.00 < z < 0) = 0.3414

g. z=−1.58 and z=0

From the standard normal distribution tables,

P(Z< -1.58) = 0.0571  and P(Z< 0) = 0.5

Thus;

P(-1.58 < z < 0 ) = 0.5 -  0.0571

P(-1.58 < z < 0) = 0.4429

h. z=−0.79 and z=0

From the standard normal distribution tables,

P(Z< -0.79) = 0.2148  and P(Z< 0) = 0.5

Thus;

P(-0.79 < z < 0 ) = 0.5 -  0.2148

P(-0.79 < z < 0) = 0.2852

8 0
3 years ago
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