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3 years ago

An experiment consists of flipping a single coin followed by rolling a six-sided cube (a die).

Mathematics

2 answers:

Over [174]3 years ago

8 0

Answer: $\frac{1}{12}$

Step-by-step explanation:

Okay, first, you want to find the probability of getting a head, which would be 1/2 because there are two options and we are only looking for one out of the two. Then, the probability of getting a 6 would be 1/6 because there is only one chance of getting a 6 and there are 6 options. Next, you would have to multiply both quantities to find out what the probability would be of getting both. So it would be $\frac{1}{2}$ × $\frac{1}{6}$ = $\frac{1}{12}$

Reptile [31]3 years ago

5 0

Probability of getting a head followed by a 6 is 1/12

Step-by-step explanation:

There are 2 possible outcomes of tossing a coin a head and a tail.

The probability is 1 out of 2 each for head and tail.

Probability of getting a 6 when a dice is rolled is 1 out of 6 for each face i.e 1,2,3,4,5,6

So the probability of getting a head followed by 6 is 1/2*1/6 = 1/12

It can also be understood as below.

There can be 12 possible combinations once a coin is tossed and a die is rolled. Head-1, Head-2. Head-3, Head-4, Head-5, Head-6, Tail-1, Tail-2, Tail-3, Tail-4, Tail-5, Tail-6

In the list there is 1 outcome that we need. Head-6. This again gives us the chance of getting Head-6 as 1 out of 12.

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Help me out thanks !!!!!!

dexar [7]

Answer:

a = 12

Step-by-step explanation:

as C^2 = A^2 + B^2

by Pythagoras theorem

A^2 = C^2 - B^2

= 13^2 - 5^2

= 169- 25

144

as 12^2 is equal to 144

hence side a is equal to 12

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olya-2409 [2.1K]

50 - 12 = 38. X = 12 is the answer

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3 years ago

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The equations that must be solved for maximum or minimum values of a differentiable function w=f(x,y,z) subject to two constrai

sammy [17]

The Lagrangian is

$L(x,y,z,\lambda,\mu)=x^2+y^2+z^2+\lambda(4x^2+4y^2-z^2)+\mu(2x+4z-2)$

with partial derivatives (set equal to 0)

$L_x=2x+8\lambda x+2\mu=0\implies x(1+4\lambda)+\mu=0$

$L_y=2y+8\lambda y=0\implies y(1+4\lambda)=0$

$L_z=2z-2\lambda z+4\mu=0\implies z(1-\lambda)+2\mu=0$

$L_\lambda=4x^2+4y^2-z^2=0$

$L_\mu=2x+4z-2=0\implies x+2z=1$

Case 1: If $y=0$ , then

$4x^2-z^2=0\implies4x^2=z^2\implies2|x|=|z|$

Then

$x+2z=1\implies x=1-2z\implies2|1-2z|=|z|\implies z=\dfrac25\text{ or }z=\dfrac23$

$\implies x=\dfrac15\text{ or }x=-\dfrac13$

So we have two critical points, $\left(\dfrac15,0,\dfrac25\right)$ and $\left(-\dfrac13,0,\dfrac23\right)$

Case 2: If $\lambda=-\dfrac14$ , then in the first equation we get

$x(1+4\lambda)+\mu=\mu=0$

and from the third equation,

$z(1-\lambda)+2\mu=\dfrac54z=0\implies z=0$

Then

$x+2z=1\implies x=1$

$4x^2+4y^2-z^2=0\implies1+y^2=0$

but there are no real solutions for $y$ , so this case yields no additional critical points.

So at the two critical points we've found, we get extreme values of

$f\left(\dfrac15,0,\dfrac25\right)=\dfrac15$ (min)

and

$f\left(-\dfrac13,0,\dfrac23\right)=\dfrac59$ (max)

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3 years ago

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11111nata11111 [884]

Answer:

x-intercept = 9 y-intercept = 5

Step-by-step explanation:

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3 years ago

8TH GRADE GEOMETRY:100 POINTS+BRAINLIEST: Please answer the question and explain how you did it. Also, please tell me what a "ra

tatyana61 [14]

Answer:

Option B 1 : 3

Step-by-step explanation:

Volume formulas & definitions:

To solve this problem, we must remember the formulas for volume of each shape:

$V_{cylinder}=\pi r^2h\\V_{cone}=\frac{1}{3} \pi r^2 h$ where r represents the radius of the circular part of the shape, and h represents the height of the shape (the plural of height is "heights", and the plural of radius is "radii")

So, when the question states that the shapes "have congruent heights and radii" it means that the height of both objects is the same, and the radius of both objects is the same.

Ratios

To find the ratio of the volume of the cone to the volume of the cylinder, we must setup the ratio. Ratios can be setup as a fraction where the first quantity is the numerator (top of fraction), and the second quantity is the denominator (bottom of the fraction):

$\dfrac{V_{cone}}{V_{cylinder}}=\dfrac{\frac{1}{3} \pi r^2 h} {\pi r^2 h}$

Since the pi, the "r", and the "h" are the same in both the numerator and denominator (and since there is only multiplication & division in the fraction) these common factors can cancel, and reduce the fraction to the following:

$\dfrac{V_{cone}}{V_{cylinder}}=\dfrac{\frac{1}{3}} {1}$

Any number divided by 1 doesn't change, so one-third divided by 1 is still one-third.

$\dfrac{V_{cone}}{V_{cylinder}}=\frac{1}{3}$

Other representations of ratios

Lastly, we must remember that a ratio an also be written with a colon symbol, where the numerator is written first, and the denominator is written second.

So, the ratio of the volume of the cone to the volume of the cylinder is:

1 : 3

7 0

2 years ago

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