Ask question

Ask question

All categories

2 years ago

7

Twelve girls try out for the hockey team. Two play goal, four play defence, and the rest play forward. How many different starti

ng lineups could the coach choose if she needs 1 goalie, 2 defence, and 3 forwards?

1 answer:

Leto [7]2 years ago

4 0

There are 2 choices for 1 goalie, 4C2 possible combinations for the 2 defence and 6C3 combinations for the 3 forward. Therefore the total number of different lineups is found from:
$2\times4C2\times6C3=240$
There are 240 possible different lineups.

You might be interested in

If 2 and 4 each divide q without remainder, which of the following must q divide

Marrrta [24]

Any multiple of 4 would work.

7 0

2 years ago

Help!<br> Change 2/5<br> to a percent.

nalin [4]

Answer:

40%

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Which method of gathering data is described by the statement "these are run only when probabilities are known as fact

HACTEHA [7]

It seems that you have missed the necessary options for us to answer this question. But anyway, here is the answer. The method <span>of gathering data that is described by the statement "these are run only when probabilities are known as fact" is called SIMULATION. Hope this answers your question.</span>

3 0

3 years ago

-64=(2cos(pi/2)+2i sin(pi/2))^n..... n=??

zvonat [6]

Answer:

The real solution is $n=6$ .

Step-by-step explanation:

$\cos(\frac{\pi}{2})=0$ while $\sin(\frac{\pi}{2})=1$

So the equation becomes:

$-64=(2(0)+2i(1))^n$

$-64=(0+2i)^n$

$-64=(2i)^n$

We know that $2^6=64$ . So let's see what $n=6$ gives us:

$(2i)^6=64i^6=64i^4i^2=64(1)(-1)=-64$ .

$-64$ is the result we wanted.

$n=6$ is therefore a solution.

4 0

2 years ago

What are the solutions to the following system? StartLayout Enlarged left-brace 1st row negative 2 x squared + y = negative 5 2n

Yakvenalex [24]

Answer:

$(\sqrt{2},-1),(-\sqrt{2},-1)$

(StartRoot 2 EndRoot, negative 1) and (negative StartRoot 2 EndRoot, negative 1)

Step-by-step explanation:

we have

$-2x^{2} +y=-5$ ----> equation A

$y=-3x^{2} +5$ -----> equation B

solve by substitution

substitute equation B in equation A

$-2x^{2} +(-3x^{2} +5)=-5$

solve for x

$-5x^{2} +5=-5$

$-5x^{2}=-10$

$x^{2}=2$

$x=\pm\sqrt{2}$

<em>Find the value of y</em>

$y=-3x^{2} +5$

For $x=\sqrt{2}$ ----> $y=-3(\sqrt{2})^{2} +5=-1$

For $x=-\sqrt{2}$ ----> $y=-3(-\sqrt{2})^{2} +5=-1$

therefore

The solutions are

$(\sqrt{2},-1),(-\sqrt{2},-1)$

(StartRoot 2 EndRoot, negative 1) and (negative StartRoot 2 EndRoot, negative 1)

8 0

3 years ago

Read 2 more answers