All categories

3 years ago

HELP!!

Mathematics

2 answers:

coldgirl [10]3 years ago

4 0

Answer:

Step-by-step explanation:

Sample Response: Alex is not correct. Rolling the two number cubes are independent events and the probability of getting an odd or an even number on either roll will be ½.

Zinaida [17]3 years ago

3 0

As long as they have the same number of sides, no. The outcome of the first roll does not affect the outcome of the second roll in any way for this situation.

You might be interested in

Write the equation of the line in standard form that has a slope of 2/3 and y-intercept of -7.

trasher [3.6K]

Answer:

-2/3x + y = -7

Step-by-step explanation:

So, y = 2/3x - 7 is the slope-intercept equation. Now, we need to turn it into the standard equation of a slope.

Standard equation --> Ax + By = C

y = 2/3x - 7

-2/3x -2/3x

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

-2/3x + y = -7 --> this is your equation in standard slope form.

3 0

2 years ago

Read 2 more answers

Help me ASAP giving brianliest!!!!! Find the X

ioda

Hey! So I’m pretty sure you multiply 56 and 2, then you swap the sides of the equation. Answer: X=112

6 0

3 years ago

PLSSSSS HELP MEEEEEE DUE TODAY!!!!!

BartSMP [9]

Answer:

4/5 or .8

Step-by-step explanation:

4 0

2 years ago

Use the table of integrals, or a computer or calculator with symbolic integration capabilities, to find the indefinite integral.

andriy [413]

Answer:

$\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C$

Step-by-step explanation:

We have been given a indefinite integral $\int \frac{2}{3x\left(3x-5\right)}dx$ . We are asked to find the indefinite integral.

We will use partial fraction formula to solve our given problem.

$\frac{2}{3x\left(3x-5\right)}=\frac{3}{5(3x-5)}-\frac{1}{5x}$

$\int \frac{2}{3x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx$

$\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{3}{5(3x-5)}-\frac{1}{5x}dx$

Using difference rule of integrals, we will get:

$\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)$

Now, we need to use u-substitution as:

Let $u=3x-5$ .

$\frac{du}{dx}=3$

$dx=\frac{1}{3}du$

$\int \frac{3}{5(3x-5)}dx= \frac{3}{5}\int \frac{1}{(u)}*\frac{1}{3}du=\frac{3}{5}*\frac{1}{3}\int \frac{1}{(u)}du=\frac{1}{5}ln|u|=\frac{1}{5}ln|3x-5|$

$\int \frac{1}{5x}dx=\frac{1}{5}\int \frac{1}{x}dx=\frac{1}{5}ln|x|$

Substitute back these values:

$\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)=\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)$

Let us add a constant C.

$\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C$

Therefore, our required integral would be $\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C$ .

5 0

3 years ago

What is a minimum - pre cal

UkoKoshka [18]

Answer: y-value of the vertex of a ∪-shaped (positive) parabola

Step-by-step explanation:

For a quadratic: it is the y-value of the vertex of a positive parabola.

If it is a negative parabola ( ∩-shaped), it is the maximum.

For a cubic: it is the y-value of all of the vertices of the ∪-shaped sections of the graph. These are typically referred to as the "relative minima" when given an interval.

4 0

2 years ago