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3 years ago

Find the solution for this system of equations.

2 answers:

8 0

Let,
12x + 15y =34--------(1)
-6x + 5y =3-------(2)

(2) x 2 = -12x + 10y =6-------(3)

(3) + (1)

20y =40
y= 2

substitute y=2 into (1)
12x + 15(2) =34
12x =4
x= 1/3

belka [17]3 years ago

3 0

Answer:

$x=\frac{5}{6}$

$y=\frac{8}{5}$

Step-by-step explanation:

$12x + 15=34$

$-6x + 5y=3$

In elimination method we try to make the coefficient of one variable same

LEts multiply the second equation by 2

$-6x + 5y=3$ * 2

$-12x + 10y =6$

Now add it with first equation

$12x + 15y=34$

$-12x + 10y =6$

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

$25y=40$

Divide by 25 on both sides

$y=\frac{8}{5}$

Now plug it in first equation and find out x

$12x + 15y=34$

$12x + 15(\frac{8}{5})=34$

$12x + 24=34$ , subtract 24

$12x =10$

Divide by 12 on both sides

$x=\frac{5}{6}$

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There are 15 students in a kindergarten class. If each kindergartner can have only one task, in how many ways can the teacher as

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Answer: 75 possible ways

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Multiply the number of kindergarten students by the number of tasks to get how many possible ways the tasks could be assigned.

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4 years ago

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Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen

k0ka [10]

Let $a,b,c$ be the randomly selected lengths. Without loss of generality, suppose $a[tex]P(A + B \ge C) = P(A + B - C \ge 0)$

where $A,B,C$ are independent random variables with the same uniform distribution on [0, 1].

By their mutual independence, we have

$P(A=a,B=b,C=c) = P(A=a) \times P(B=b) \times P(C=c)$

so that the joint density function is

$P(A=a,B=b,C=c) = \begin{cases}1 & \text{if }(a,b,c)\in[0,1]^3 \\ 0 & \text{otherwise}\end{cases}$

where $[0,1]^3=[0,1]\times[0,1]\times[0,1]$ is the cube with vertices at (0, 0, 0) and (1, 1, 1).

Consider the plane

$a + b - c = 0$

with $(a,b,c)\in\Bbb R^3$ . This plane passes through (0, 0, 0), (1, 0, 1), and (0, 1, 1), and thus splits up the cube into one tetrahedral region above the plane and the rest of the cube under it. (see attached plot)

The point (0, 0, 1) (the vertex of the cube above the plane) does not belong the region $a+b-c\ge0$ , since $0+0-1=-1$ . So the probability we want is the volume of the bottom "half" of the cube. We could integrate the joint density over this set, but integrating over the complement is simpler since it's a tetrahedron.

Then we have

$\displaystyle P(A+B-C\ge0) = 1 - P(A+B-C < 0) \\\\ ~~~~~~~~ = 1 - \int_0^1\int_0^{1-a}\int_{a+b}^1 P(A=a,B=b,C=c) \, dc\,db\,da \\\\ ~~~~~~~~ = 1 - \int_0^1 \int_0^{1-a} (1 - a - b) \, db \, da \\\\ ~~~~~~~~ = 1 - \int_0^1 \frac{(1-a)^2}2\,da \\\\ ~~~~~~~~ = 1 - \frac16 = \boxed{\frac56}$

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2 years ago

Calculate the value of each of the following

Ghella [55]

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104.7664068

Step-by-step explanation:

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3 years ago

If we list all the natural numbers below 20 that are multiples of 7 or 11, we get 7, 11 and 14. The sum of these multiples is 32

velikii [3]

Number of multiples of 7 up to 1337: $\left\lfloor\dfrac{1337}7\right\rfloor=191$
Number of multiples of 11 up to 1337: $\left\lfloor\dfrac{1337}{11}\right\rfloor=121$
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This means there are $191+121-17=295$ distinct multiples of 7 *or* 11 up to 1337.

The sum of these multiples is

$\displaystyle\sum_{k=1}^{191}7k+\sum_{k=1}^{121}11k-\sum_{k=1}^{17}77k$

which can be computed using the well-known formula,

$\displaystyle\sum_{k=1}^nk=\dfrac{n(n+1)}2$

So you have

$\displaystyle\sum_{k=1}^{191}7k+\sum_{k=1}^{121}11k-\sum_{k=1}^{17}77k=7\dfrac{191\times192}2+11\dfrac{121\times122}2-77\dfrac{17\times18}2=197762$

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3 years ago

What is the approximate volume of the cylinder? Use 3.14 for п.

Veseljchak [2.6K]

<u>Given</u>:

Given that the radius of the cylinder is 4 cm.

The height of the cylinder is 9 cm.

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<u>Volume of the cylinder:</u>

The volume of the cylinder can be determined using the formula,

$V=\pi r^2 h$