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3 years ago

Solve the inequality.

Mathematics

1 answer:

klemol [59]3 years ago

5 0

Answer:

-24.4 ≥ w

Step-by-step explanation:

-13.4 ≥ 6.7 + 4.3 + w

-13.4 - 6.7 - 4.3 ≥ w

-24.4 ≥ w

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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

3 years ago

Un triángulo rectángulo tiene un área de 84 pies2 y una hipotenusa de 25 pies de largo. ¿Cuáles son las longitudes de sus otros

jolli1 [7]

Answer:

7 feet and 24 feet

Step-by-step explanation:

In the right triangle, the hypotenuse is 25 feet long. The area of this triangle is 84 square feet.

Let x feet and y feet be the lengths of triangle's legs.

By the Pythagorean theorem,

$x^2+y^2=25^2\\ \\x^2+y^2=625$

The area of the right triangle is half the product of its legs, thus

$84=\dfrac{1}{2}xy\\ \\xy=168$

Solve the system of two equations:

$\left\{\begin{array}{l}x^2+y^2=625\\ \\xy=168\end{array}\right.$

From the second equation:

$x=\dfrac{168}{y}$

Substitute it into the first equation:

$\left(\dfrac{168}{y}\right)^2+y^2=625\\ \\168^2+y^4=625y^2\\ \\y^4-625y^2+168^2=0\\ \\D=(-625)^2-4\cdot 168^2=(625-2\cdot 168)(625+2\cdot 168)=289\cdot 961=17^2\cdot 31\\ \\\sqrt{D}=17\cdot 31=527\\ \\y^2_{1,2}=\dfrac{-(-625)\pm 527}{2}=49,\ 576\\ \\y_1=7,\ y_2=-7,\ y_3=24,\ y_4=-24$