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3 years ago

You roll a standard number cube. What is the probability that you will roll an even number?

Mathematics

1 answer:

Nadusha1986 [10]3 years ago

7 0

Answer:

The probability of rolling an even number would be $\frac{1}{2}$ or 50%.

Step-by-step explanation:

A standard number cube has 6 sides that features numbers 1 to 6 on each side. Since there are three even numbers (2, 4, 6) total on the sides of the number cube, that means the theoretical probability of rolling an even number would be $\frac{3}{6}$ , which simplifies to $\frac{1}{2}$ or 50%.

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Plsss help mee quickkk !!

Andreyy89

Answer:

2.5 overtime hours

Step-by-step explanation:

90$+195$=285

390-285=105$

105/42=2.5 overtime hours

5 0

3 years ago

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How do I graph this? In picture below<br> please include the graph

Savatey [412]

Answer:

See the attached figure.

Step-by-step explanation:

The given function is called piecewise function which is the function that can be in pieces, i.e: defined by multiple sub-functions.

$f(x)=\left \{ {{2x} \ , \ x\geq3 \atop {-\frac{1}{3}x+7 \ , \ x\leq 3 }} \right.$

So, need to graph 2x in the interval [3,∞)

And graph -(1/3) x + 7 in the interval (-∞,3]

We will find that f(3) at the function 2x will be equal f(3) at the function -(1/3) x + 7

Which mean the function is continuous.

The attached figure represents the graph of function, it was graphed using the tables on the graph.

3 0

3 years ago

Would he earn more money with simple interest or with compound interest

zaharov [31]

Answer:

The more often your interest compounds, the more interest you'll earn on your investment. It's easy to see that money grows more quickly when it's earning compound interest than when it's earning simple interest

Step-by-step explanation:

6 0

2 years ago

If Li goes on 32 out of 100 rides at an amusement park, what percentage of

lyudmila [28]

Answer:

A. 68%

Step-by-step explanation:

68 + 32 = 100, therefore the answer is 68%.

5 0

3 years ago

Read 2 more answers

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

3 years ago