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Alekssandra [29.7K]

3 years ago

8

There are 17 appetizers available at a restaurant. From these, Frank is to choose 13 for his party. How many groups of 13 appeti

zers are possible

1 answer:

Andrej [43]3 years ago

8 0

Answer:

2,380

Step-by-step explanation:

This is a selection problem, hence it is combination. It is simply telling us in how many ways can we select a group of 13 from 17.

This is straightforward, we only need to apply the combination formulae. Hence this goes thus:

17C13 = 2,380

There are 2,380 ways to do this task

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Step-by-step explanation:

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A tractor, starting from a stationary position, drives down the road. It accelerates at 4 m/s2 for 10 seconds, then travels at a

chubhunter [2.5K]

Answer:

B. 1400 meters.

Step-by-step explanation:

According to the statement, there are two stages for the motion of the tractor, in which we need to find the travelled distance:

(i) <em>Uniform accelerated motion.</em>

<em /> $s_{i} = s_{o,i} +v_{o,i}\cdot t_{i} + \frac{1}{2}\cdot a\cdot t_{i}^{2}$ <em> </em>(1)

$v_{i} = v_{o,i}+a\cdot t_{i}$ (2)

Where:

$s_{o,i}$ - Initial position of the tractor, measured in meters.

$s_{i}$ - Final position of the tractor, measured in meters.

$v_{o,i}$ - Initial velocity of the tractor, measured in meters per second.

$t_{i}$ - Time, measured in seconds.

$a$ - Acceleration, measured in meters per square second.

$v_{i}$ - Final velocity of the tractor, measured in meters per second.

(ii) <em>Uniform motion.</em>

$s_{ii} = s_{i} + v_{i}\cdot t_{ii}$ (3)

Where:

$s_{ii}$ - Final position of the tractor, measured in meters.

$t_{ii}$ - Time, measured in seconds.

The distance covered by the tractor ( $\Delta s$ ), measured in meters, is:

$\Delta s = s_{ii}$ (4)

If we know that $a = 4\,\frac{m}{s^{2}}$ , $t_{i} = 10\,s$ , $v_{o,i} = 0\,\frac{m}{s}$ , $s_{o,i} = 0\,m$ and $t_{ii} = 30\,s$ , then distance covered by the tractor is:

By (1) and (2):

$s_{i} = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (10\,s)+ \frac{1}{2}\cdot \left(4\,\frac{m}{s^{2}} \right) \cdot (10\,s)^{2}$

$s_{i} = 200\,m$

$v_{i} = \left(0\,\frac{m}{s} \right)+\left(4\,\frac{m}{s^{2}} \right)\cdot (10\,s)$

$v_{i} = 40\,\frac{m}{s}$

By (3):

$s_{ii} = 200\,m + \left(40\,\frac{m}{s} \right)\cdot (30\,s)$

$s_{ii} = 1400\,m$

By (4):

$\Delta s = 1400\,m$

The correct answer is B.

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Step-by-step explanation:

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Step-by-step explanation:

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