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2 years ago

Suppose you roll a fair, six-sided number cube.

Mathematics

1 answer:

Serga [27]2 years ago

5 0

Answer:

Sample space= S= {1, 2, 3, 4, 5, 6}

One possible event is rolling an even number

probability (even number) $=P\{2, 4, 6\}$ $=\frac{3}{6}=\frac{1}{2}$

Step-by-step explanation:

Sample Space

The simple space is the combination of all possible outcomes

so,

Sample space= S= {1, 2, 3, 4, 5, 6}

One possible event

One possible event is rolling an even number

Probability of that event

to find the probability of the event i.e. rolling an even number, we first find the sample for the event i.e. event space

event space={2, 4, 6}

Now,

probability (even number) $=P\{2, 4, 6\}$ $=\frac{3}{6}=\frac{1}{2}$

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What is y=3/8x-5 in a table of values

iogann1982 [59]

It is 8x-5 you replace the variable y with 3 in the expression

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3 years ago

Which of the following is equivalent to x^-1/2 1) -1/2x 2) -√x 3) 1/√x 4) -1/2x

irina [24]

Answer:

Option (3) is correct.

Step-by-step explanation:

We need to find the expression that is equivalent to $x^{-1/2}$ .

The property is : $x^{-a}=\dfrac{1}{x^a}$

Here, a = 1/2

So,

$x^{-1/2}=\dfrac{1}{x^{1/2}}$

We know that, $x^{1/2}=\sqrt{x}$

$\dfrac{1}{x^{1/2}}=\dfrac{1}{\sqrt{x} }$

So, $x^{-1/2}$ is equivalent to $\dfrac{1}{\sqrt{x} }$

Hence, the correct option is (3).

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2 years ago

(05.04 MC) At a school carnival, the diameter of the mat of a trampoline is 16 feet and the diameter of its metal frame is 18 fe

xenn [34]

Answer:

Step-by-step explanation:

You're looking for the circumference here, which is the distance around the outside of a circle. The formula is

C = πd or C = 2πr

Since we are given the diameter, we will use that one:

C = (3.14)(18) so

C = 56.5 feet

3 0

2 years ago

Graph the image of this figure after a dilation with a scale factor of 3 centered at (-7,-6)

Yuri [45]

Given that the triangle is dilated by factor 3, the image will be found as follows;
The object is at:
A(-7,-3), B(-3,-2), C(-4,-5)
when the image was enlarged the new coordinates will be:
A'=3(-7,-3)=(-21,-9)
B'=3(-3,-2)=(-9,-6)
C'=3(-4,-5)=(-12,-15)
since the image is centered at the point (-7,-6), the final point will be at:
A"=[(-21+-7),(-9+-6)=(-28,-15)
B''=[(-9+-7),(-6+-6)]=(-16,-12)
C''=[(-12+-7),(-15+-6)]=(-19,-17)
thus the coordinates of the final image are:
A''(-28,-15),B''(-16,-12),C''(-19,-17)

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2 years ago

If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for y with respect to x, i.e.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}$

Compute the first derivative. By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta$

$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta$

and so the second derivative is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta$

4 0

2 years ago