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3 years ago

14

You have to answer 3 essay questions for an exam. There are 6 essays to choose from. How many different groups of 3 essays could

you possibly choose?

1 answer:

Paladinen [302]3 years ago

3 0

Answer:

The answer is 120

Step-by-step explanation:

3 places. 6 options in the first place, 5 options in the second place, and then 4 in the last place. Because one option is used every time you pick one.

6x5x4=120

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Write the standard equation for the hyperbola with the following conditions: vertices: (-2, -3) and (6, -3) foci: (-4, -3) and (

STALIN [3.7K]

Hello,

Vertices are on a line parallele at ox (y=-3)

The hyperbola is horizontal.

Equation is (x-h)²/a²- (y-k)²/b²=1

Center =middle of the vertices=((-2+6)/2,-3)=(2,-3)
(h+a,k) = (6,-3)
(h-a,k)=(-2,-3)
==>k=-3 and 2h=4 ==>h=2
==>a=6-h=6-2=4 (semi-transverse axis)

Foci: (h+c,k) ,(h-c,k)
h=2 ==>c=8-2=6

c²=a²+b²==>b²=36-4²=20

Equation is:
$\boxed{ \dfrac{(x-2)^2}{16} - \dfrac{(y+3)^2}{20} =1}

$

8 0

3 years ago

The time to assemble the first unit on a production line is 8 hours. The learning rate is 0.81. Approximately how long will it t

pochemuha

Answer:

4.428 hours

Step-by-step explanation:

If the learning rate is 0.81, the slope of the learning curve is:

$b=\frac{ln(0.81)}{ln(2)} \\b=-0.304$

The time it takes to produce the n-th unit is:

$T_n=T_1*n^b$

If T1 = 8 hours, the time required to produce the seventh unit will be:

$T_n=8*7^{-0.304}\\T_n=4.428\ hours$

It will take roughly 4.428 hours.

3 0

3 years ago

If (5^2)p = 5^12, what is the value of p?

mr Goodwill [35]

Divide both sides by 5^2 (which is 25). So p = (5^12)/25. I can't be bothered to plug in into a calculator but be my guest.

3 0

3 years ago

Please help me<br><br><br> Simplify.<br><br> 3(x + 4)

valentinak56 [21]

Distribute the 3.

3*x + 3*4
3x + 12

Hope this helps :)

8 0

3 years ago

For the figures below, assume they are made of semicircles, quarter circles and squares. For each shape, find the area and perim

ICE Princess25 [194]

Answer:

Part a) The area of the figure is $\frac{9}{2}(4+\pi )\ cm^{2}$

Part b) The perimeter of the figure is $3(2+2\sqrt{2}+ \pi)\ cm$

Step-by-step explanation:

Step 1

Find the area of the figure

In this problem we have that

The figure ABC is the half of a square and the other figure is a semicircle

<u>Find the area of the figure ABC</u>

we have

$AB=6\ cm, BC=6\ cm$

The area of the half square ABC is equal to find the area of triangle ABC

so

$A1=\frac{1}{2}*6*6=18\ cm^{2}$

<u>Find the area of the semicircle</u>

The area of the semicircle is equal to

$A2=\pi r^{2}/2$

we have that

$r=6/2=3\ cm$

substitute

$A2=\pi (3)^{2}/2$

$A2=(9/2) \pi\ cm^{2}$

The area of the figure is equal to

$18\ cm^{2}+(9/2) \pi\ cm^{2}= \frac{9}{2}(4+\pi )\ cm^{2}$

Step 2

Find the perimeter of the figure

The perimeter of the figure is equal to

$P=AB+AC+length\ CB$

we have

$AB=6\ cm$

Applying Pythagoras theorem

$AC=\sqrt{6^{2}+6^{2}}\\AC=6\sqrt{2}\ cm$

Remember that

the circumference of a semicircle is equal to

$C=\frac{1}{2}2\pi r=\pi r$

$r=6/2=3\ cm$

$C=\pi(3)$

$C=3 \pi\ cm$

The perimeter of the figure is equal to

$P=6\ cm+6\sqrt{2}\ cm+3 \pi\ cm$

Simplify

$P=3(2+2\sqrt{2}+ \pi)\ cm$

5 0

3 years ago