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marysya [2.9K]
3 years ago
10

What is the least multiple of 1/2+1/6

Mathematics
2 answers:
vredina [299]3 years ago
6 0
The leat common multiple is 12
multiply your denominator 2×6=12
next multiply the rest
(6×1)+(2×1)/(12)

finally solve
(6+2)/(12)

is (8/12) now we can reduce by dividing both terms by 4

(8÷4)/(12÷4)
gives you (2)/(3) as your answer
eimsori [14]3 years ago
5 0
2/3.  1/2 is equal to 3/6 and 3/6 + 1/6 is 4/6 and 4/6 simplified is 2/3.
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The amount of money in a bank account increased by 15.8% over the last year. If the amount of money at the beginning of the year
Lostsunrise [7]

Answer:

The expression represents the amount of money in the bank account after the increase is: x=1.158n.

Step-by-step explanation:

With the information provided, you can say that the amount of money in the bank after the increase would be equal to multiplying the amount at the beginning of the year for the result of adding up 1 plus the percentage of the increase, which would be:

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5 0
3 years ago
We are again studying the times required to solve two elementary math problems. Suppose we ask four students to attempt both Pro
KiRa [710]

Answer:

(-16.494 ; -3.506)

Step-by-step explanation:

student Prob A Prob B difference, d (A-B)

1 20 35____ - 15

2 30 40 ___ - 10

3 15 20 ___ - 5

4 40 50 __ - 10

Difference, d = -15, -10, -5, -10

Xd = Σd/ n = - 40 / 4 = - 10

Standard deviation of d ; Sd = 4.082

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Xd ± Tcritical*(Sd/√n)

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6 0
3 years ago
A sailboat is sitting at rest near its dock. A rope attached to the bow of the boat is drawn in over a pulley that stands on a p
irina [24]

Answer:

The boat is approaching the dock at rate of 2.14 ft/s

Step-by-step explanation:

The situation given in the question can be modeled as a triangle, please refer to the attached diagram.

A rope attached to the bow of the boat is drawn in over a pulley that stands on a post on the end of the dock that is 5 feet higher than the bow that means x = 5 ft.

The length of rope from bow to pulley is 13 feet that means y = 13 ft.

We know that Pythagorean theorem is given by

x^{2} + y^{2} = z^{2}

Differentiating the above equation with respect to time yields,

2x\frac{dx}{dt}  + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}

x\frac{dx}{dt}  + y\frac{dy}{dt} = z\frac{dz}{dt}

dx/dt = 0  since dock height doesn't change

y\frac{dy}{dt} = z\frac{dz}{dt}

\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}

The rope is being pulled in at a rate of 2 feet per second that is dz/dt = 2 ft/s

First we need to find z

z² = (5)² + (13)²

z² = 194

z = √194

z = 13.93 ft

So,

\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}

\frac{dy}{dt} = \frac{13.93}{13}(2)

\frac{dy}{dt} = 2.14 ft/s

Therefore, the boat is approaching the dock at rate of 2.14 ft/s

8 0
3 years ago
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