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poizon [28]
3 years ago
10

For a field trip, the school bought 47 sandwiches for $4.60 each and 39 bags of chips for $1.25 each. How much did the school sp

end in all?
Mathematics
2 answers:
NNADVOKAT [17]3 years ago
8 0

Answer:  \$264.95

Step-by-step explanation:

You know that 1 sandwich costs $4.60 and the school bought 47 sandwiches.

Therefore, in order to calculate the total cost of 47 sandwiches, you need  to multiply this number of sandwiches bought by the schoole, by the cost of 1 sandwich. Then:

sandwiches=(47)(\$4.60)\\\\sandwiches=\$216.2

You know that 1 bag of chip costs $1.25 and the school bought  39 bags of chips.

So, in order to calculate the total cost of  39 bags of chips, you must  multiply the total number of them bought by the school, by the cost of 1 bag of chips.

Then:

bag\ of\ chips=(39)(\$1.25)\\\\bag\ of\ chips=\$48.75

Therefore, the total amount school spent was:

Total\ spent=\$216.2+\$48.75\\\\Total\ spent=\$264.95

dmitriy555 [2]3 years ago
7 0

Multiply the price per each by the quantity of each:

Sandwiches: 4.60 x 47 = $216.20

Chips: 1.25 x 39 = $48.75

Total spent = 216.20 + 48.75 = $264.95

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guapka [62]

Given:

The fees for a day student are $600 a term.

The fees for a boarding student are $1200 a term.

The school needs at least $720000 a term.

To show:

That the given information can be written as x + 2y\geq 1200.​

Solution:

Let x be the number of day students and y be the number of boarding students.

The fees for a day student are \$600 a term.

So, the fees for x day students are \$600x a term.

The fees for a boarding student are \$1200 a term.

The fees for y boarding student are \$1200y a term.

Total fees for x day students and y boarding student is:

\text{Total fees}=600x+1200y

The school needs at least $720000 a term. It means, total fees must be greater than or equal to $720000.

600x+1200y\geq 720000

600(x+2y)\geq 720000

Divide both sides by 600.

\dfrac{600(x+2y)}{600}\geq \dfrac{720000}{600}

x+2y\geq 1200

Hence proved.

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Enter the missing numbers in the boxes to complete the table of equivalent ratios of time to distance.
andre [41]
Answer:  Here is the complete table, with the filled in values:
______________________________________________________________
Time (h)    Distance (mi)
    3                2 
    9                6
   12                8
   18               12
___________________________________________________

Explanation:
___________________________________________________
Let us begin by obtaining the "?" value; that is, the "distance" (in "mi.") ;
         when the time (in "h") is "18" ; 
___________________________________________________

12/8 = 18/?  

Note: "12/8 = (12÷4) / (8÷4) = 3/2 ;

Rewrite:  3/2 = 18/? ;  cross-multiply:  3*? = 2 * 18 ; 
                                                                        3*? = 36 ;

                                                                      Divide each side by "3" ;
                                                                            The "?" = 36/3 = 12 ;

So, 12/8 = 18/12 ;

The value: "12" takes the place for the "?" in the table for "distance (in "mi.);
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__________________________________________________________
Now, let us obtain the "? " value for the "distance" (in "mi.");
                when the "time" (in "h") is:  "9" .

12/8 = 9/? ;  Solve for "?" ;

We know (see aforementioned) that "12/8 = 3/2" ; 

So, we can rewrite:  3/2 = 9/? ;  Solve for "?" ; 

Cross-multiply:  3* ? = 2* 9 ;    3* ? = 18 ;  
                                                           Divide each side by "3" ;
                                                    to  get:  "6" for the "?" value.

When the time (in "h") is "9", the distance (in "mi.") is "6" .
____________________________________________________
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9/6 = ?/2 ;  Note:  We get the "6" from our "calculated value" (see above problem). 

9/6 = (9÷3) / (6÷3) = 3/2 ;  

So, we know that the "?" value is:  "3" .

Alternately:   9/6 = ?/2 ; 

Cross-multiply:  6*? = 2*9 ;   6 * ? = 18 ;  Divide each side by "6" ;
                                                               to find the value for the "?" ; 
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When the "distance" (in "mi.") is:  "2" ;  the time (in "h") is:  "3" .
____________________________________________________
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____________________________________________________

<span>Time (h)    Distance (mi)
____________________________________________________
    3                2 
    9                6
   12                8
   18               12
____________________________________________________
</span>
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