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lutik1710 [3]
3 years ago
11

Ralph buys a desk that regularly sells for $250 and a desk chair that regularly sells for $145. both items are on sale at 15% of

f, and a sales tax of x% is applied to both. Which expression does not represent the amount Ralph pays? Show your work.
A. (395)(0.0085x)+395(0.85)

B. (250+145)(1-0.15)(100+x/100)

C.(0.85+x/100)(250+145)

D.335.75+3.3575x
Mathematics
1 answer:
Lunna [17]3 years ago
6 0
C) (0.85 + x/100)(250+145) does not give the correct answer.

Explanation
A) works; adding the two costs together is 250+145=395.  We multiply this by 0.85 because 100%-15%=85%=0.85.  We also have x% tax, which is represented by x/100, added to 100% of the value, or 1.00.  Altogether this gives us 
395(0.85)(1+x/100) = 395(0.85 + (0.85x/100)) = 395(0.85) + 395(0.85x/100)
= 395(0.85) + 395(0.0085x)
B) works; we have 250+145 for the original price; we have 85% = 0.85 of the value; we also have 100% + x%, which is (100+x)/100.
C) does not work; (0.85+x/100)(395) does not take into consideration that you are finding the tax after taking the 85%.  This will simplify out to
0.85*395 + (x/100)(395) = 335.75 + 395x/100 = 335.75 + 3.95x, which is too much.
D) works; simplifying the expression from A, we have 395(0.85) + 395(0.0085x) = 335.75 + 3.3575x.
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The pirce of an item yesterday was$125 . Today, the price fell to 75. Find the percentage decrease.
Julli [10]

Answer:

Im sure the answer is 40%

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8 0
4 years ago
Which of the following can be modeled with a linear function? Select all that apply.
elena-s [515]

Answer:

Step-by-step explanation:

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6 0
3 years ago
1 Find the value of C to that the function probability density function defined as follow, also calculate the man and variance /
Viefleur [7K]

Answer:

c = \frac{1}{12}

The mean of the distribution is 0.25.

The variance of the distribution is of 0.6875.

Step-by-step explanation:

Probability density function:

For it to be a probability function, the sum of the probabilities must be 1. The probabilities are 3c, 3c and 6c, so:

3c + 3c + 6c = 1

12c = 1

c = \frac{1}{12}

So the probability distribution is:

P(X = -1) = 3c = 3\frac{1}{12} = \frac{1}{4} = 0.25

P(X = 0) = 3c = 3\frac{1}{12} = \frac{1}{4} = 0.25

P(X = 1) = 6c = 6\frac{1}{12} = \frac{1}{2} = 0.5

Mean:

Sum of each outcome multiplied by its probability. So

E(X) = -1(0.25) + 0(0.25) + 1(0.5) = -0.25 + 0.5 = 0.25

The mean of the distribution is 0.25.

Variance:

Sum of the difference squared between each value and the mean, multiplied by its probability. So

V^2(X) = 0.25(-1-0.25)^2 + 0.25(0 - 0.25)^2 + 0.5(1 - 0.25)^2 = 0.6875

The variance of the distribution is of 0.6875.

4 0
3 years ago
What algebraic expression must be added to the sum of
user100 [1]

Answer:

We need to add the expression:  4x^2-16

Step-by-step explanation:

As the first step, add the first two polynomials listed in order to find what their sum is. Do such by combining like terms:

3x^2+4x+8\,+ (2x^2-6x+3) =5x^2-2x+11

now, you need to find another polynomial (let's call it P(x)) such that added to the one we just found above, will render 9x^2-2x-5 as a result.

Putting this phrase in mathematical terms, we write:

5x^2-2x+11\,+ P(x) = 9x^2-2x-5

Now we solve for this unknown P(x) in this equation, by subtracting each of the terms that are accompanying it on the left side of the equation. We do one at a time for clarity:

5x^2-2x+11\,+ P(x) = 9x^2-2x-5\\-2x+11+P(x)=9x^2-5x^2-2x-5\\-2x+11+P(x)=4x^2-2x-5\\11+P(x)=4x^2-2x+2x-5\\11+P(x)=4x^2-5\\P(x)=4x^2-5-11\\P(x)=4x^2-16

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8 0
4 years ago
Enter the number of complex zeros for the polynomial function in the box. f(x)=x3−96x2+400
ioda

Solution:

There is no (zero) complex zeros for the given polynomial.

Explanation:

We apply Descartes' rule of sign to identify the number of complex roots.

The given polynomial is f(x)=x^3-96x^2+400

Let us see the number of sign changes in f(x)

+,-,+

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Now, let us see the number of sign changes in f(-x)

f(-x)=-x^3-96x^2+400

-,-,+

There are only one sign change. Hence, there will be 1 negative roots.

The degree of the polynomial is 3. Hence, there will be exactly 3 zeros.

Therefore, the possible numbers of zeros are

2 positive, 1 negative and 0 complex

0 positive, 1 negative and 2 complex.

Hence, possible number of complex zeros are 0 and 2.

Now, we graph the function in xy plane and see that the graph cuts the x axis at three points. It means all the zeros are real , which is the case of first possibility (2 positive, 1 negative and 0 complex).

Hence, the number of complex zeros for the given polynomial is zero.

5 0
4 years ago
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