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muminat
3 years ago
5

You roll a six-sided number cube. What is

Mathematics
2 answers:
givi [52]3 years ago
7 0

Answer: 1, 2, 3, 4, 5, 6

p greater than 2 = 4/6 = 2/3

p 2 or smaller and prime = 2/6 = 1/3

2/3 + 1/3 = 1

every single side of the cube is either greater than 2 or prime (two and one) so you simply will get the desired result in 100% of your rolls

Zepler [3.9K]3 years ago
6 0

Step-by-step explanation:

no of favourable cases, n(E) = 4

no of sample space, n(S) = 6

Probability that it will be rolled on the number greater than two, P(E) = n(E) / n(S) = 4/6 = 2/3

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Martha has widow's peak and attached earlobes. Martha's dad had a straight hairline and unattached earlobes. If Martha marries a
krok68 [10]

Answer:

the probability of having a widow peak and attached earlobe = 0.6667*0.66667 = 0.4445 = 44.5%

Step-by-step explanation:

Martha = widow's peak + attached earlobes.

Martha's dad = straight hairline + unattached earlobes.

Martha husband =straight hairline+ widow's peak + attached earlobes

what is the probability that Martha child will have a widow's peak + attached earlobes?

since,

Martha + Martha husband =straight hairline+ widow's peak + attached earlobes

the probability of the child having widow peak is 2/3 = 0.6667

the probability of the child having attached earlobes is 2/3 = 0.66667

the probability of having both = 0.6667*0.66667 = 0.4445

6 0
3 years ago
Solve each equation.<br> 9. m - 17=-8<br> 10. k - 55=67<br> 11.-44 + n =36
Alex
9) m= 9
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4 0
4 years ago
Read 2 more answers
Do these ratios form a proportion?
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Answer:

no

Step-by-step explanation:

8 0
3 years ago
How many times greater is 700,000 + 70,000
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3 years ago
find an equation of the tangent plane to the given parametric surface at the specified point. x = u v, y = 2u2, z = u − v; (2, 2
Alexxandr [17]

The surface is parameterized by

\vec s(u,v) = x(u, v) \, \vec\imath + y(u, v) \, \vec\jmath + z(u, v)) \, \vec k

and the normal to the surface is given by the cross product of the partial derivatives of \vec s :

\vec n = \dfrac{\partial \vec s}{\partial u} \times \dfrac{\partial \vec s}{\partial v}

It looks like you're given

\begin{cases}x(u, v) = u + v\\y(u, v) = 2u^2\\z(u, v) = u - v\end{cases}

Then the normal vector is

\vec n = \left(\vec\imath + 4u \, \vec\jmath + \vec k\right) \times \left(\vec \imath - \vec k\right) = -4u\,\vec\imath + 2 \,\vec\jmath - 4u\,\vec k

Now, the point (2, 2, 0) corresponds to u and v such that

\begin{cases}u + v = 2\\2u^2 = 2\\u - v = 0\end{cases}

and solving gives u = v = 1, so the normal vector at the point we care about is

\vec n = -4\,\vec\imath+2\,\vec\jmath-4\,\vec k

Then the equation of the tangent plane is

\left(-4\,\vec\imath + 2\,\vec\jmath - 4\,\vec k\right) \cdot \left((x-2)\,\vec\imath + (y-2)\,\vec\jmath +  (z-0)\,\vec k\right) = 0

-4(x-2) + 2(y-2) - 4z = 0

\boxed{2x - y + 2z = 2}

6 0
2 years ago
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