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SVETLANKA909090 [29]

3 years ago

7

8. How many ways can a committee of 5 students be chosen from a student council of 30 students? Is the order in which the member

s of the committee are chosen important?

1 answer:

tankabanditka [31]3 years ago

4 0

Answer: A committee of 5 students can be chosen from a student council of 30 students in 142506 ways.

No , the order in which the members of the committee are chosen is not important.

Step-by-step explanation:

Given : The total number of students in the council = 30

The number of students needed to be chosen = 5

The order in which the members of the committee are chosen does not matter.

So we Combinations (If order matters then we use permutations.)

The number of combinations of to select r things of n things = $^nC_r=\dfrac{n!}{r!(n-r)!}$

So the number of ways a committee of 5 students can be chosen from a student council of 30 students= $^{30}C_5=\dfrac{30!}{5!(30-5)!}$

$=\dfrac{30\times29\times28\times27\times26\times25!}{(120)\times25!}=142506$

Therefore , a committee of 5 students can be chosen from a student council of 30 students in 142506 ways.

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The height of a free falling object at time t can be found using the function, h(t) = - 12t2 + 36t. Where h(t) is the height in

never [62]

For this case we have the following equation:
h (t) = - 12t2 + 36t
When the object hits the ground we have:
- 12t2 + 36t = 0
We look for the roots of the polynomial:
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t2 = 3
Therefore, the time it takes the object to hit the ground is:
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Answer:
the time when the object hits the ground is:
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3 years ago

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Ymorist [56]

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3 years ago

When an amount of heat Q (in kcal) is added to a unit mass (kg) of a

lisov135 [29]

Answer:

The change in temperature per minute for the sample, dT/dt is 71. $\overline {6}$ °C/min

Step-by-step explanation:

The given parameters of the question are;

The specific heat capacity for glass, dQ/dT = 0.18 (kcal/°C)

The heat transfer rate for 1 kg of glass at 20.0 °C, dQ/dt = 12.9 kcal/min

Given that both dQ/dT and dQ/dt are known, we have;

$\dfrac{dQ}{dT} = 0.18 \, (kcal/ ^{\circ} C)$

$\dfrac{dQ}{dt} = 12.9 \, (kcal/ min)$

Therefore, we get;

$\dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = {\dfrac{dQ}{dt} } \times \dfrac{dT}{dQ} = \dfrac{dT}{dt}$

$\dfrac{dT}{dt} = \dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = \dfrac{12.9 \, kcal / min }{0.18 \, kcal/ ^{\circ} C } = 71.\overline 6 \, ^{\circ } C/min$

For the sample, we have the change in temperature per minute, dT/dt, presented as follows;

$\dfrac{dT}{dt} = 71.\overline 6 \, ^{\circ } C/min$

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2 years ago

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GrogVix [38]

Answer:

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Step-by-step explanation:

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3 years ago

The equation h(t) = -16t² + 80t + 64 represented the height, in feet, of a potato t seconds after it has been launched.

Lyrx [107]

Answer:

Part A) The potato hit the ground at t=5.70 seconds (see the explanation)

Part B) The potato is 40 feet off the ground at the time t=5.28 seconds (see the explanation)

Step-by-step explanation:

we have

$h(t)=-16t^2+80t+64$

where

h(t) is the height of a potato in feet

t is the time in seconds

Part A) Write an equation that can be solved to find when the potato hits the ground. Then solve the equation

we know that

When the potato hit the ground, the value of h(t) must be equal to zero

so

For h(t)=0

$-16t^2+80t+64=0$

Solve the quadratic equation

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-16t^2+80t+64=0$

so

$a=-16\\b=80\\c=64$

substitute in the formula

$t=\frac{-80\pm\sqrt{80^{2}-4(-16)(64)}} {2(-16)}$

$t=\frac{-80\pm\sqrt{10,496}} {-32}$

$t=\frac{-80+\sqrt{10,496}} {-32}=-0.70$

$t=\frac{-80-\sqrt{10,496}} {-32}=5.70$

therefore

The potato hit the ground at t=5.70 seconds

Part B) Write an equation that can be solved to find when the potato is 40 feet off the ground. Then solve the equation

For h(t)=40 ft

substitute in the quadratic equation

$-16t^2+80t+64=40$

$-16t^2+80t+24=0$

Solve the quadratic equation

we have

$a=-16\\b=80\\c=24$

substitute in the formula

$t=\frac{-80\pm\sqrt{80^{2}-4(-16)(24)}} {2(-16)}$

$t=\frac{-80\pm\sqrt{7,936}} {-32}$

$t=\frac{-80+\sqrt{7,936}} {-32}=-0.28$

$t=\frac{-80-\sqrt{7,936}} {-32}=5.28$

therefore

The potato is 40 feet off the ground at the time t=5.28 seconds

3 0

3 years ago