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zlopas [31]
2 years ago
15

At Freddy's, three steak burgers and two orders of fries cost $18. Two steak burgers and three orders of fries cost $15.75. What

is the cost for one steak burger? What is the cost of one order of fries? Write a system of equations and then solve by the method of your choice. ​
Mathematics
1 answer:
Fudgin [204]2 years ago
6 0

The cost for one steak burger is $4.5. The price/ The cost of one order of fries is $2.25

<h3>Solution</h3>

Let one steak burger cost $x

and one fries order cost $y

Hence, we have

3x+2y = 18-----------1

2x+3y = 15.75-------2

<h3>Let us solve the equation simultaneously</h3>

multiply equation 1 by 2 and equation 2 by 3

6x + 4y = 36 ---------3

6x +  9y = 47.25 ----4

subtraction 4 from 3 we have

0x -5y = -11.25

Divide both sides by -5

y = 11.25/5

y = $2.25

Put y = 2.25 in equation 1 to find x

3x+2(2.25) = 18-----------1

3x + 4.5 = 18

3x = 18 -4.5

3x = 13.5

x = 13.5/3

x = $4.5

Learn more about Algebra here:

brainly.com/question/12602543

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How do you create linear equations from a word problem? Can someone please step me through how to make TWO equations for this??:
storchak [24]
Check out the image attachment for the filled out table. There may be more than one way to fill out the table, but I did it in the way I learned in the past.

Let's work through the table one row at a time
------------------------------------------------------------------------------
So we'll start with row 1

Row 1, column1: The value that goes here is 12 as each adult pays $12

Row 1, column2: You'll write 'a' without quotes here as there are 'a' adults ('a' is just a placeholder for a number)

Row 1, column3: Write 12*a or 12a here. Simply multiply the cost per adult ($12) with the number of adults (a). 
------------------------------------------------------------------------------
Now onto row 2

Row 2, column1: It costs $6 per young adult, so we write 6 here

Row 2, column2: There are y young adults. Write 'y' here without quotes

Row 2, column3: Write 6y here. Multiply the number of young adults with the price per young adult
------------------------------------------------------------------------------
Now onto row 3

Now we add up the values per each column to get the column totals

Row 3, column1: The individual costs 12 and 6 add to 18. We won't use this value but it doesn't hurt to write it in. If it is confusing to add in, then just ignore this cell. The reason why we won't use this is because the number of adults (a) and young adults (y) is not necessarily the same. If we were guaranteed they were the same, then we could use this value. But again there's no guarantee. It's probably best to steer clear of this cell.

Row 3, column2: We have 'a' adults and 'y' young adults. So a+y people total. This total is 8 as we know a family of 8 had been registered. So we write 8 in this box as well. The two expressions a+y and 8 are equal to each other allowing us to form the first equation a+y = 8

Row 3, column3: The cost for all the adults is 12a dollars. Similarly it costs 6y dollars for just the young adults. Adding up the two subtotals we get 12a+6y as the total cost for everyone. We're told that the family paid a total of $66. So like with the previous box, we can equate the two expressions getting us the second equation to be 12a+6y = 66
------------------------------------------------------------------------------

Again everything is summarized in the image attachment. 

The two equations we pull away from that table are
a+y = 8
12a+6y = 66
which is the system of equations to set up

7 0
4 years ago
If 4x+y=h, then x is equal to
lianna [129]
4x + y = h

Rearrange the problem so your equation equals x, not h
y - h = -4x

Divide both sides by -4
-4y - 4h = x

Cheers!
5 0
4 years ago
Help me solve 18.4/2.3*3.4+13.812
dexar [7]
=(8)(3.4)+13.812
=27.2 + 13.812
=41.012
the Answer is 41.012
7 0
3 years ago
A public health official is planning for the supplyof influenza vaccine needed for the upcoming flu season. She wants to estimat
Marizza181 [45]

Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.64})^2}=420.25  

And rounded up we have that n=421

Step-by-step explanation:

We know that the sample proportion have the following distribution:

\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by \alpha=1-0.90=0.1 and \alpha/2 =0.05. And the critical value would be given by:

z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.04 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

We assume that a prior estimation for p would be \hat p =0.5 since we don't have any other info provided. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.64})^2}=420.25  

And rounded up we have that n=421

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%3F%7D%7B5%7D%20%2A%20%5Cfrac%7B10%7D%7B8%7D%20%3D1.5%5C%5C" id="TexFormula1" title=
Serggg [28]

Answer:

\boxed{\sf \huge \boxed{ \sf ?} = 6}

Step-by-step explanation:

\sf Solve  \: for \: \boxed{ \sf ?} :  \\  \sf \implies \frac{\boxed{ \sf ?}}{5}  \times  \frac{10}{8}  = 1.5 \\  \\  \sf  \implies \frac{\boxed{ \sf ?} \times 10}{5 \times 8}  = 1.5 \\  \\  \sf \frac{10}{5}  =  \frac{ \cancel{5} \times 2}{ \cancel{5}}  = 2 :  \\  \sf \implies  \frac{ \boxed{ \sf 2} \times \boxed{ \sf ?} }{8}  = 1.5 \\  \\  \sf \frac{2}{8}  =  \frac{ \cancel{2}}{ \cancel{2}  \times 4}  =  \frac{1}{4}  :  \\  \sf \implies \frac{\boxed{ \sf ?}}{ \boxed{ \sf 4}}  = 1.5 \\  \\  \sf Multiply  \: both  \: sides \:  of \:  \frac{\boxed{ \sf ?}}{4}  = 1.5 \: by \: 4 :  \\  \sf \implies \frac{4 \times \boxed{ \sf ?}}{4}  = 4 \times 1.5 \\  \\  \sf \frac{4 \times \boxed{ \sf ?}}{4} =  \frac{ \cancel{4}}{ \cancel{4}}  \times \boxed{ \sf ?} = \boxed{ \sf ?} :  \\  \sf  \implies \boxed{\boxed{ \sf ?}} = 4 \times 1.5 \\  \\  \sf 4 \times 1.5 = 6 :  \\  \sf \implies \boxed{ \sf ?} = 6

4 0
3 years ago
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