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3 years ago

Copy and complete the table for the sequence 3n-1

1 answer:

defon3 years ago

8 0

3n-1

where n = 1 ; n = 2 ; n = 3

3(n) - 1 = 3(1) - 1 = 3 - 1 = 2
3(n) - 1 = 3(2) - 1 = 6 - 1 = 5
3(n) - 1 = 3(3) - 1 = 9 - 1 = 8

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5a-2<18a<

Margarita [4]

The first one is (4) The second one is (-6) I'm not 101% sure but yeah.

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3 years ago

Which problem is molded by the equation 4x-5=45?

kolezko [41]

Answer: B.

Step-by-step explanation:

See attached. We examine the equation and decide which situation, or problem, models it correctly.

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2 years ago

The dimensions of a rectangular solid are 4 ft by 6 ft by 10 ft. What is the volume?

mrs_skeptik [129]

Answer:

Step-by-step explanation:

Volume = length x breadth x height

= 4 x 6 x 10

= 240 cubic feet

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3 years ago

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

Graph the linear equation y = 2

sleet_krkn [62]

Answer: Please look at the steps to see the answers.

Step-by-step explanation:

The line y= 2 has a slope of 0 meaning that for any input of x the results will always be 2.

This means the the graph will be a horizontal line through the point, (0,2)

You can also use this table to graph it;

x y

0 2

1 2

2 2

3 2

4 2

5 2

6 2

-1 2

-2 2.

8 0

3 years ago