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Pavel [41]
3 years ago
15

The fifth-grade classes at Brookfield School used five identical buses to go on a field trip. •There were a total of 40 seats on

each bus. •All of the seats on four buses were filled. •The fifth bus had 4/5 of the seats filled. •1/8 of all the passengers on the buses were adults. How many adults went on the field trip.
Mathematics
1 answer:
Eva8 [605]3 years ago
6 0
First things first, you need to figure out the amount of passengers on the fifth bus.

Solution:
40 x 4
--      -  = 32 passengers
 1  x 5
*The fifth bus has 32 people.
You can now figure out the total number of passengers in all of the buses.

(40 x 4) + 32 = 192
*Overall, all 5 buses have a total of 192 passengers.
Since you have the total number of passengers, you can now figure out the numbers of adults who went on the field trip, which is 1/8 of all the passengers.

Solution:
192 x 1
-----    - = 24 passengers
  1   x 8

The final answer is: 24 adults went on the field trip.
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PLEASE HELP I WILL MARK YOU BRAINIEST PLEASE
cupoosta [38]

Answer:

Zeniab makes the most money by $0.25, and would make $2.50 more.

Step-by-step explanation:

Okay, the good news is, this problem is about finding how much you make in an hour.

Zainab makes $44 for 8 hours

and

Keziah makes $47.25 in 9 hours.

in each problem, we can set them up in an equation of

$44=8x

and

$47.25=9x

In the first equation, divide both sides by 8 to isolate the x.

that would be 44/8 = x which is $5.50 per hour for Zainab.

In the second equation, divide both sides by 9 to isolate the x.

That would be 47.25/9 = x which is $5.25 per hour for Keziah.

This means that Zainab would make more per hour, and to find the second part of the question for a 10 hour job,

(<u>You can multiply any number by 10 easily by moving each number up one place. Hundreds go to the thousands when you multiply by 10 as an example. In this problem, 5.50 times 10 would be $55.00 because the 5 behind the decimal would be moved up infront of the decimal)</u>

<u>Zainab would make $55 for 10 hours</u>

<u>Keziah would make $52.50 for 10 hours</u>

<u>Then, take the difference of those final totals, which is $2.50, or multiply $0.25 by 10 to get $2.50</u>

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3 years ago
PLEASE HELP WILL MARK BRAINLIEST!!(plz show work too)
trapecia [35]

\huge \bf༆ Answer ༄

Let the capacity of bus be x students

And van be y students, now ;

From the given statements we get two equations ~

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2x + 10y = 260 \:   \:  \:  \: \:  (1)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:13x + 5y = 670 \:  \:  \:  \:  \: (2)

multiply the equation (2) with 2 [ it won't change the values ]

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2x + 10y = 260 \: \:  \:  \:   \:  \:  \:  \:  \: (1)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:26x + 10y = 1340 \:  \:  \:  \:  \: (3)

Now, deduct equation (1) from equation (3)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:26x + 10y - 2x - 10y = 1340 - 260

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:24x = 1080

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:x = 1080 \div 24

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:x = 45

Therefore each bus can carry (x) = 45 students

Now, plug the value of x in equation (1) to find y ~

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2x + 10y = 260

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:(2 \times 45) + 10y = 260

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:90 + 10y = 260

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:10y = 260 - 90

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:10y = 170

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:y = 170 \div 10

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:y = 17

Hence, each van can carry (y) = 17 students in total.

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What is the approximate circumference of a circle with a diameter of 14 inches, rounded to the nearest inch? (Use 3.14 as an app
soldi70 [24.7K]

Answer:

44 in.

Step-by-step explanation:

pi = 3.14 (in our case)

circumference = diameter x pi

circ. = 14 x 3.14, which equals to 43.96

round that and you get 44 in.

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2 years ago
-3(p-7) &gt; 21 solve for p​
Goryan [66]

Answer:

p < 0

Step-by-step explanation:

Given

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3 years ago
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choli [55]

Answer:

a = 2^\alpha.14 \ ,\alpha\geq 8\\b = 7^\delta.14 \ , \delta\geq 3

Step-by-step explanation:

As 2  and 7 are the only prime divisors of both a and b we know that both can be written as:

a = 2^\alpha . 7^\beta \ , \alpha ,\beta\ \in \mathbb{N}_{0}\\b = 2^\gamma . 7^\delta \ , \gamma ,\delta\ \in \mathbb{N}_{0}\\

Where \mathbb{N}_{0} is the set of all natural numbers adding the zero (careful because this part is important as I'll explain next).

We also know that 14 divides both numbers and that is actually the greatest common divisor between them. So we can rewrite a and b as follows:

a = 2^\alpha . 7^\beta.14 \ , \alpha ,\beta\ \in \mathbb{N}_{0}\\b = 2^\gamma . 7^\delta.14 \ , \gamma ,\delta\ \in \mathbb{N}_{0}\\

Why do I write them like this? Because this way is easier to observe that if \alpha and \gamma were both greater than zero, then 28 would divide both hence 14 wouldn't be their g.c.d.. Likewise, if \beta and \delta were both greater than zero, then 98 would divide both and once again, 14 wouldn't be their g.c.d.

So either of them has to be equal to zero. And then we have that

a = 2^\alpha.14 \\b = 7^\delta.14

All we have left to do is find the possible values for \alpha and \delta so that a>2000 \ , b>2000 and that only happens if \alpha\geq 8 and \delta\geq 3

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3 years ago
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