Can someone help? At least solve one that you might know, explain if you can.

Are the expressions equivalent 7y + 21 and 3(2y+7)+y

DochEvi [55]

Answer:

Yes

Step-by-step explanation:

3(2y + 7) + y

6y + 21 + y

7y + 21

4 0

1 year ago

Read 2 more answers

Each day, X arrives at point A between 8:00 and 9:00 a.m., his times of arrival being uniformly distributed. Y arrives independe

astraxan [27]

Answer:

Y will arrive earlier than X one fourth of times.

Step-by-step explanation:

To solve this, we might notice that given that both events are independent of each other, the joint probability density function is the product of X and Y's probability density functions. For an uniformly distributed density function, we have that:

$f_X(x) = \frac{1}{L}$

Where L stands for the length of the interval over which the variable is distributed.

Now, as X is distributed over a 1 hour interval, and Y is distributed over a 0.5 hour interval, we have:

$f_X(x) = 1\\\\f_Y(y)=2$ .

Now, the probability of an event is equal to the integral of the density probability function:

$\iint_A f_{X,Y} (x,y) dx\, dy$

Where A is the in which the event happens, in this case, the region in which Y<X (Y arrives before X)

It's useful to draw a diagram here, I have attached one in which you can see the integration region.

You can see there a box, that represents all possible outcomes for Y and X. There's a diagonal coming from the box's upper right corner, that diagonal represents the cases in which both X and Y arrive at the same time, under that line we have that Y arrives before X, that is our integration region.

Let's set up the integration:

$\iint_A f_{X,Y} (x,y) dx\, dy\\\\\iint_A f_{X} (x) \, f_{Y} (y) dx\, dy\\\\2 \iint_A dx\, dy$

We have used here both the independence of the events and the uniformity of distributions, we take the 2 out because it's just a constant and now we just need to integrate. But the function we are integrating is just a 1! So we can take the integral as just the area of the integration region. From the diagram we can see that the region is a triangle of height 0.5 and base 0.5. thus the integral becomes:

$2 \iint_A dx\, dy= 2 \times \frac{0.5 \times 0.5 }{2} \\\\2 \iint_A dx\, dy= \frac{1}{4}$

That means that one in four times Y will arrive earlier than X. This result can also be seen clearly on the diagram, where we can see that the triangle is a fourth of the rectangle.

6 0

2 years ago

Kyle is raising funds for his company by selling preferred stock. The preferred stock has a par value of $122 and a dividend ra

kirza4 [7]

no i think your answer is 9082

6 0

2 years ago

Prove algebraically that (m + 2)2 – m 2 – 12 is always a multiple of 4

Julli [10]

Answer:

$(m + 2)^{2} - m^{2} - 12 = 4(m - 2)$

Step-by-step explanation:

Step 1:

Write the expression

$(m+2)^{2} - m^{2} - 12$

Step 2: Expand $(m + 2)^{2}$

$(m+2)^{2} - m^{2} - 12\\(m+2)(m+2) - m^{2} - 12\\m^{2} + 2m + 2m + 4 - m^{2} - 12$

Step 3: Collect similar terms

$m^{2} - m^{2} + 4m + 4 - 12\\4m - 8$

Step 4: Factor 4 out of the expression to prove that the expression is a multiple of 4.

$Therefore\\4m - 8 = 4(m - 2)\\Hence,\\(m+2)^{2} - m^{2} - 12 is a multiply of 4 because the expression is equal to 4(m-2)$

3 0

2 years ago

The ordered pair R(-2, 7) is translated 5 units to the right and 2 units down.

Usimov [2.4K]

Left and right affect domain. (x)

Up and down effects range. (y)

So,

-2+5 (because the right of the number line is an increase)

7-2 (because you are decreasing by going down)

You are left with:

R'(3,5)

I hope this helps!
~cupcake

8 0

2 years ago