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alisha [4.7K]
3 years ago
12

The teachers at West Elementary are planning a field trip for 385 students. It will cost $1,750 for transportation and $3,080 fo

r tickets to the museum. If the teachers charge each student $12, will they have enough money to cover the costs of the field trip?
Mathematics
1 answer:
melomori [17]3 years ago
5 0
The cost needed for the trip is <span>$1,750 for transportation and $3,080 for tickets to the museum.
This makes a total of: 1750 + 3080 = $4830 needed

Now, we have 385 students. Teachers will collect $12 from each. This means that the amount collected would be: 12 * 385 = $4620

Based on the above calculations, the amount needed for the trip is $4830 and the amount collected is $4620.
This means that collecting $12 from each student would not be enough for the trip.</span>
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What % of 40 is 15? Please put how u do it
andriy [413]

Answer:

37.5

Step-by-step explanation:

so x% of 40 is 15

the equation for that is

x/100 * 40 = 15

40x/100 = 15

multiply 100 on both sides

40x = 1500

divide 40 on both sides

37.5

37.5% of 40 is 15

6 0
3 years ago
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Five more than a half number is 6. find the number.
Artyom0805 [142]
First write the equation:
1/2x + 5 = 6

Then subtract five from both sides:

1/2x = 1

Multiply both sides by 2:

x = 2
7 0
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Using a calculator, work out:<br> V2704 x 22 - 149
goldfiish [28.3K]

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7 0
3 years ago
If you create a matrix, C, to show the inventory at the end of July, the value of the entry represented by C22 is ?
Inga [223]

Answer:

C_{22} = 389

Max(A_{31}) =376

Step-by-step explanation:

Given

See attachment for complete question

Solving (a): The entry C22

First, matrix C represents the inventory at the end of July.

The entry of C is calculated as:

C = Inventory - Sales

i.e.

C = \left[\begin{array}{ccc}543&356&643\\364&476&419\\376&903&409\end{array}\right]  -  \left[\begin{array}{ccc}102&78&97\\98&87&59\\54&89&79\end{array}\right]

C = \left[\begin{array}{ccc}543-102&356-78&643-97\\364-98&476-87&419-59\\376-54&903-89&409-79\end{array}\right]

C = \left[\begin{array}{ccc}441&278&546\\266&389&360\\322&814&330\end{array}\right]

Item C22 means the entry at the second row and the second column.

From the matrix

C_{22} = 389

Solving (b): The maximum A31 possible.

From the given data, we have:

Inventory = \left[\begin{array}{ccc}543&356&643\\364&476&419\\376&903&409\end{array}\right]

Unit\ Sales =   \left[\begin{array}{ccc}102&78&97\\98&87&59\\54&89&79\end{array}\right]

From the matrices above.

A31 means entry at the 3rd row and 1st column.

So, the possible values of A31 are:

A_{31} = 376

and

A_{31} = 54

By comparison, 376 > 54

So:

Max(A_{31}) =376  

8 0
2 years ago
two trains leave the station at the same time, one heading west and the other east. the westbound train travels at 95 miles per
Alinara [238K]
Remark
They are travelling in opposite directions. Their distances will be added until that distance = 462 miles.

You need only add the r*t values together. You do not need to find the distance.

Givens 
d = 462 miles
r1 = 115 mph
r2 = 95 mph
t = the time each was traveling.

Formula
r1 * t   +  r2*t = d

Substitute and solve
95 * t   +   115 * t = 462   Add the like terms on the left.
210 * t  = 462    Divide by 210
t = 462 / 210
t = 2.2 hours.

t = 2 1/5 hours

t = 2 hours  and 12 minutes 
All three answers are the same.

4 0
3 years ago
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