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Masja [62]
3 years ago
9

-6/10×-1/8=3/40? is this correct

Mathematics
1 answer:
Leokris [45]3 years ago
6 0

your answer is correct !


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30 points: Answer the questions in the picture provided
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1).  Enrico Fermi directed construction of the the thing
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5 0
2 years ago
Expand 4(z+6) into the simplest form.
Ratling [72]
4z + 24 is the answer
4 0
2 years ago
A loaf of bread that is baked today costs $7. All of the bread baked yesterday is 40% off.
vovikov84 [41]

Answer:

The price of today's bread is $7.

40% of $7 is $2.80.

The price of yesterdays bread is $7-$2.80.

7-2.80=4.20

The price of yesterday's bread is $4.20

Tobin has enough money to purchase a loaf of yesterday's bread.

Hope this helps!

4 0
3 years ago
7) PG & E have 12 linemen working Tuesdays in Placer County. They work in groups of 8. How many
BabaBlast [244]

Part A

Since order matters, we use the nPr permutation formula

We use n = 12 and r = 8

_{n}P_{r} = \frac{n!}{(n-r)!}\\\\_{12}P_{8} = \frac{12!}{(12-8)!}\\\\_{12}P_{8} = \frac{12!}{4!}\\\\_{12}P_{8} = \frac{12*11*10*9*8*7*6*5*4*3*2*1}{4*3*2*1}\\\\_{12}P_{8} = \frac{479,001,600}{24}\\\\_{12}P_{8} = 19,958,400\\\\

There are a little under 20 million different permutations.

<h3>Answer: 19,958,400</h3>

Side note: your teacher may not want you to type in the commas

============================================================

Part B

In this case, order doesn't matter. We could use the nCr combination formula like so.

_{n}C_{r} = \frac{n!}{r!(n-r)!}\\\\_{12}C_{8} = \frac{12!}{8!(12-8)!}\\\\_{12}C_{8} = \frac{12!}{4!}\\\\_{12}C_{8} = \frac{12*11*10*9*8!}{8!*4!}\\\\_{12}C_{8} = \frac{12*11*10*9}{4!} \ \text{ ... pair of 8! terms cancel}\\\\_{12}C_{8} = \frac{12*11*10*9}{4*3*2*1}\\\\_{12}C_{8} = \frac{11880}{24}\\\\_{12}C_{8} = 495\\\\

We have a much smaller number compared to last time because order isn't important. Consider a group of 3 people {A,B,C} and this group is identical to {C,B,A}. This idea applies to groups of any number.

-----------------

Another way we can compute the answer is to use the result from part A.

Recall that:

nCr = (nPr)/(r!)

If we know the permutation value, we simply divide by r! to get the combination value. In this case, we divide by r! = 8! = 8*7*6*5*4*3*2*1 = 40,320

So,

_{n}C_{r} = \frac{_{n}P_{r}}{r!}\\\\_{12}C_{8} = \frac{_{12}P_{8}}{8!}\\\\_{12}C_{8} = \frac{19,958,400}{40,320}\\\\_{12}C_{8} = 495\\\\

Not only is this shortcut fairly handy, but it's also interesting to see how the concepts of combinations and permutations connect to one another.

-----------------

<h3>Answer: 495</h3>
5 0
2 years ago
Please help I'm really confused I need have with these 6 questions
Lelechka [254]
Ahh, basic shapes. Split up the weird shapes into easier ones. #1 can be truned into 2 trapezoids. #2 can be turned into 2 circles. #3 is a triangle and a trapezoid. #4 is 2 right triangles. #5 is a rectangel and a traingle. Finally, #6 is 3 traigles and a rectangle. Do you see how we broke the hard shapes into easier shapes?
6 0
2 years ago
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