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EleoNora [17]
3 years ago
13

There are 40 GT students going on a field trip. Each one pays their teacher $8.75 to cover admission to the museum and lunch. Ad

mission for the students is $120 and each one gets an equal amount to spend on lunch. How much will each GT student be able to spend on lunch?
Mathematics
1 answer:
Norma-Jean [14]3 years ago
3 0

Step-by-step explanation:

There are 40 DuBois students going on a field trip.

Each one pays their teacher $8.75 to cover admission to the museum and lunch.

Admission for the students is $120 and each one gets an equal amount to spend on lunch.

To Find :

How much will each DuBois student be able to spend on lunch.

Solution :

Amount they paid for admission :

\begin{gathered}P =\$ \dfrac{120}{40}\\\\P=\$3\end{gathered}P=$40120P=$3

Now, amount of money to spend on lunch is 8.75 -8.75−3 = $5.75 .

Hence, this is the required solution.

You might be interested in
A flat rectangular piece of aluminum has a perimeter of 70 inches. The length is 11 inches longer than the width. Find the width
rodikova [14]

perimeter = 2L +2w

L = w+11

70 = 2(w+11) +2w

70 = 2w+22+2w

70= 4w + 22

48 = 4w

w=48/4 = 12

width = 12

length = 12+11 = 23

2x12 = 24

2x23 = 46

46+24 = 40

 length = 23 inches, width = 12 inches

Answer is D

5 0
3 years ago
Read 2 more answers
A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021
egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

  • A is approximately <u>2020.022</u>

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

  • \displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

  • \displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}

Therefore, for the last term we have;

  • \displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

\displaystyle \frac{A}{2}  = \mathbf{ 1011 \cdot  \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460}  \right)}

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2} +\frac{1}{2} -  \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022}  \right)

Which gives;

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2022}  \right)

\displaystyle  A = 2 \times 1011 \cdot  \left(1 - \frac{1}{2022}  \right) = \frac{1032231}{511} \approx \mathbf{2020.022}

  • A ≈ <u>2020.022</u>

Learn more about the sum of a series here:

brainly.com/question/190295

8 0
2 years ago
Read 2 more answers
Can an absolute value equal a negative?
Montano1993 [528]
The distance of it away from zero cannot be a negative distance, distance is always positive therefor the absolute value will always be positive.

Short answer: no, absolute value is always positive

I hope this helps :)
7 0
3 years ago
In 2000, the total population of the U.S. was 281.4 million people. In 2010, it was 308.7 million people. What is the average ra
kifflom [539]

Answer:

2.73 million people per year

Step-by-step explanation:

Turn sentences into coordinates/points:

In 2000, 281.4 million people

  • (x₁, y₁) = (2000, 281.4)

In 2010, 308.7 million people

  • (x₂, y₂) = (2010, 308.7)

Rate of population per year:

\sf \dfrac{y_2 - y_1}{x_2- x_1} \ \  where \ (x_1 , \ y_1), ( x_2 , \ y_2) \ are \ points

\rightarrow \sf \dfrac{308.7-281.4}{2010-2000}

\rightarrow \sf 2.73 \ million \ people

5 0
2 years ago
A jewelry store is having a 50% off sale for all necklaces. During this sale, what is the cost of a necklace that regulary costs
Ymorist [56]
50% off means half price, so half of 49.98 is 24.99
the necklace now costs 24.99 dollars
3 0
3 years ago
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