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4vir4ik [10]
3 years ago
10

The first class had a kids in it, the second had b kids in it, and the third class had c kids in it. Kids from all three classes

were equally divided between two buses. How many kids were on each bus?
Mathematics
2 answers:
astra-53 [7]3 years ago
5 0
There are no numbers and only variables so I can't tell you the amount of kids on each bus but here is the equation:
(a*b*c)/2
kramer3 years ago
3 0
(a+b+c)/2. this is your answer. if it had the numbers i would solve it fully, but it only has variables.
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sergejj [24]

Answer:

16 units

Step-by-step explanation:

6.1 + 5.8 + 4.1 = 16

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3 years ago
Please help. A trapezoid has base measures of 10 and 14 inches and a height of 6 inches. What is the area?
yaroslaw [1]

Answer:

84 in^2

Step-by-step explanation:

Find the area by first finding the average of 10 and 14 (it is 12) and then multiplying this result (12) by 6:  area of trapezoid = 84 in^2.



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3 years ago
Problem 2-11 (Algorithmic) Let A be an event that a person's primary method of transportation to and from work is an automobile
dangina [55]

The events are mutually exclusive, because your primary method can't be both a driving a car or taking a bus.

Since they are exclusive, the probability of the "or" clause is the sum of the probabilities:

P(A\lor B)=P(A)+P(B)=0.48+0.32=0.8

Since a person primarily takes a bus with probability 0.32, the opposite happens with probability

1-P(B) = 1-0.32=0.68

7 0
3 years ago
the height h(t) of a trianle is increasing at 2.5 cm/min, while it's area A(t) is also increasing at 4.7 cm2/min. at what rate i
nekit [7.7K]

Answer:

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

Step-by-step explanation:

From Geometry we understand that area of triangle is determined by the following expression:

A = \frac{1}{2}\cdot b\cdot h (Eq. 1)

Where:

A - Area of the triangle, measured in square centimeters.

b - Base of the triangle, measured in centimeters.

h - Height of the triangle, measured in centimeters.

By Differential Calculus we deduce an expression for the rate of change of the area in time:

\frac{dA}{dt} = \frac{1}{2}\cdot \frac{db}{dt}\cdot h + \frac{1}{2}\cdot b \cdot \frac{dh}{dt} (Eq. 2)

Where:

\frac{dA}{dt} - Rate of change of area in time, measured in square centimeters per minute.

\frac{db}{dt} - Rate of change of base in time, measured in centimeters per minute.

\frac{dh}{dt} - Rate of change of height in time, measured in centimeters per minute.

Now we clear the rate of change of base in time within (Eq, 2):

\frac{1}{2}\cdot\frac{db}{dt}\cdot h =  \frac{dA}{dt}-\frac{1}{2}\cdot b\cdot \frac{dh}{dt}

\frac{db}{dt} = \frac{2}{h}\cdot \frac{dA}{dt} -\frac{b}{h}\cdot \frac{dh}{dt} (Eq. 3)

The base of the triangle can be found clearing respective variable within (Eq. 1):

b = \frac{2\cdot A}{h}

If we know that A = 130\,cm^{2}, h = 15\,cm, \frac{dh}{dt} = 2.5\,\frac{cm}{min} and \frac{dA}{dt} = 4.7\,\frac{cm^{2}}{min}, the rate of change of the base of the triangle in time is:

b = \frac{2\cdot (130\,cm^{2})}{15\,cm}

b = 17.333\,cm

\frac{db}{dt} = \left(\frac{2}{15\,cm}\right)\cdot \left(4.7\,\frac{cm^{2}}{min} \right) -\left(\frac{17.333\,cm}{15\,cm} \right)\cdot \left(2.5\,\frac{cm}{min} \right)

\frac{db}{dt} = -2.262\,\frac{cm}{min}

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

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2 years ago
Describe a real-life multiplication situation for which an estimate makes sense
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