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Nitella [24]
2 years ago
6

Bentano spends $170 to order 30 pounds of nuts from an online company. The company charges $3.50 per pound for peanuts and $7.50

per pound for almonds. It also charges a flat fee of $13 for shipping and handling.
Which equation can be used to find p, the number of pounds of peanuts Bentano buys?



3.5p+7.5(p−30)−13=1703 point 5 p plus 7 point 5 times open paren p minus 30 close paren minus 13 is equal to 170


3.5p+7.5(30−p)−13=1703 point 5 p plus 7 point 5 times open paren 30 minus p close paren minus 13 is equal to 170


3.5p+7.5(p−30)+13=1703 point 5 p plus 7 point 5 times open paren p minus 30 close paren plus 13 is equal to 170


3.5p+7.5(30−p)+13=170
Mathematics
1 answer:
Talja [164]2 years ago
8 0

Given :

Bentano spends $170 to order 30 pounds of nuts from an online company.

The company charges $3.50 per pound for peanuts and $7.50 per pound for almonds. It also charges a flat fee of $13 for shipping and handling.

To Find :

Which equation can be used to find p, the number of pounds of peanuts Bentano buys.

Solution :

Price of peanuts + Price of almonds + Delivery charge = Total amount

3.5p + 7.5( 30 - p ) + 13 = 170

Comparing the answers with given options, the matching answer is option D).

Hence, this is the required solution.

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Use Gaussian elimination to write each system in triangular form
Feliz [49]

Answer:

To see the steps to the diagonal form see the step-by-step explanation. The solution to the system is x =  -\frac{1}{9}, y= -\frac{1}{9}, z= \frac{4}{9} and w = \frac{7}{9}

Step-by-step explanation:

Gauss elimination method consists in reducing the matrix to a upper triangular one by using three different types of row operations (this is why the method is also called row reduction method). The three elementary row operations are:

  1. Swapping two rows
  2. Multiplying a row by a nonzero number
  3. Adding a multiple of one row to another row

To solve the system using the Gauss elimination method we need to write the augmented matrix of the system. For the given system, this matrix is:

\left[\begin{array}{cccc|c}1 & 1 & 1 & 1 & 1 \\1 & 1 & 0 & -1 & -1 \\-1 & 1 & 1 & 2 & 2 \\1 & 2 & -1 & 1 & 0\end{array}\right]

For this matrix we need to perform the following row operations:

  • R_2 - 1 R_1 \rightarrow R_2 (multiply 1 row by 1 and subtract it from 2 row)
  • R_3 + 1 R_1 \rightarrow R_3 (multiply 1 row by 1 and add it to 3 row)
  • R_4 - 1 R_1 \rightarrow R_4 (multiply 1 row by 1 and subtract it from 4 row)
  • R_2 \leftrightarrow R_3 (interchange the 2 and 3 rows)
  • R_2 / 2 \rightarrow R_2 (divide the 2 row by 2)
  • R_1 - 1 R_2 \rightarrow R_1 (multiply 2 row by 1 and subtract it from 1 row)
  • R_4 - 1 R_2 \rightarrow R_4 (multiply 2 row by 1 and subtract it from 4 row)
  • R_3 \cdot ( -1) \rightarrow R_3 (multiply the 3 row by -1)
  • R_2 - 1 R_3 \rightarrow R_2 (multiply 3 row by 1 and subtract it from 2 row)
  • R_4 + 3 R_3 \rightarrow R_4 (multiply 3 row by 3 and add it to 4 row)
  • R_4 / 4.5 \rightarrow R_4 (divide the 4 row by 4.5)

After this step, the system has an upper triangular form

The triangular matrix looks like:

\left[\begin{array}{cccc|c}1 & 0 & 0 & -0.5 & -0.5  \\0 & 1 & 0 & -0.5 & -0.5\\0 & 0 & 1 & 2 &  2 \\0 & 0 & 0 & 1 &  \frac{7}{9}\end{array}\right]

If you later perform the following operations you can find the solution to the system.

  • R_1 + 0.5 R_4 \rightarrow R_1 (multiply 4 row by 0.5 and add it to 1 row)
  • R_2 + 0.5 R_4 \rightarrow R_2 (multiply 4 row by 0.5 and add it to 2 row)
  • R_3 - 2 R_4 \rightarrow R_3(multiply 4 row by 2 and subtract it from 3 row)

After this operations, the matrix should look like:

\left[\begin{array}{cccc|c}1 & 0 & 0 & 0 & -\frac{1}{9}  \\0 & 1 & 0 & 0 &   -\frac{1}{9}\\0 & 0 & 1 & 0 &  \frac{4}{9} \\0 & 0 & 0 & 1 &  \frac{7}{9}\end{array}\right]

Thus, the solution is:

x =  -\frac{1}{9}, y= -\frac{1}{9}, z= \frac{4}{9} and w = \frac{7}{9}

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3 years ago
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DochEvi [55]
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6 0
3 years ago
WILL MARK THE FIRST BRAINLY ANSWER!
ehidna [41]

Answer:

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Step-by-step explanation:

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consider the following scenario: Assume daily protein intake values in a population of athletes are known to have a standard dev
Kipish [7]

Answer:

The 95% confidence interval for the population mean daily protein intake is between 69.97g and 84.03g.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.96*\frac{58.6}{\sqrt{267}} = 7.03

The lower end of the interval is the sample mean subtracted by M. So it is 77 - 7.03 = 69.97g.

The upper end of the interval is the sample mean added to M. So it is 77 + 7.03 = 84.03g.

The 95% confidence interval for the population mean daily protein intake is between 69.97g and 84.03g.

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3 years ago
the ratio of the number of shaded sections to the number of unshaded sections is 4 to 2 what is the value of the ratio of the nu
8090 [49]

Answer:

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Step-by-step explanation:

3 0
2 years ago
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