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baherus [9]
2 years ago
8

Assume that two fair dice are rolled. First compute P(F) and then P(FIE). Explain why one would expect the

Mathematics
1 answer:
saveliy_v [14]2 years ago
5 0

The probability that a three shows on at least one of the dice is 11/36.

<h3>How to calculate the probability?</h3>

It should be noted that of probability means the likelihood of the occurence of an event.

The probability that a three shows on at least one of the dice will be:

= 11/36

This can be seen from the table that is attached.

The probability that the total is less than 8 will be:

= 21/36

= 7/12

Here, there are 21 places where the total is less than 8.

Learn more about probability on:

brainly.com/question/24756209

#SPJ1

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Step-by-step explanation:jol

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Find the two square roots of each complex number by creating and solving polynomial equations.
son4ous [18]

Answer:

1) w₁=4 - i w₂= -4 + i

2) w₁= 3 - i w₂= -3  + i

3) w₁= 1 + 2i w₂= - 1 - 2i

4) w₁= 2- 3i w₂= -2 + 3i

5) w₁= 5 - 2i w₂= -5 + 2i

6) w₁= 5 - 3i w₂= -5 + 3i

Step-by-step explanation:

The root of a complex number is given by:

\sqrt[n]{z}=\sqrt[n]{r}(Cos(\frac{\theta+2k\pi}{n}) + i Sin(\frac{\theta+2k\pi}{n}))

where:

r: is the module of the complex number

θ: is the angle of the complex number to the positive axis x

n: index of the root

1) z = 15 − 8i  ⇒ r=17 θ= -0.4899 rad

w₁=\sqrt{17}(Cos(\frac{-0.4899}{2}) + i Sin(\frac{-0.4899}{2}))=4-i

w₂=\sqrt{17}(Cos(\frac{-0.4899+2\pi}{2}) + i Sin(\frac{-0.4899+2\pi}{2}))=-1+i

2) z = 8 − 6i  ⇒ r=10 θ= -0.6435 rad

w₁=\sqrt{10}(Cos(\frac{ -0.6435}{2}) + i Sin(\frac{ -0.6435}{2}))= 3 - i

w₂=\sqrt{10}(Cos(\frac{ -0.6435+2\pi}{2}) + i Sin(\frac{ -0.6435+2\pi}{2}))= -3  + i

3) z = −3 + 4i  ⇒ r=5 θ= -0.9316 rad

w₁=\sqrt{5}(Cos(\frac{-0.9316}{2}) + i Sin(\frac{-0.9316}{2}))= 1 + 2i

w₂=\sqrt{5}(Cos(\frac{-0.9316+2\pi}{2}) + i Sin(\frac{-0.9316+2\pi}{2}))= -1 - 2i

4) z = −5 − 12i  ⇒ r=13 θ= 0.4426 rad

w₁=\sqrt{13}(Cos(\frac{0.4426}{2}) + i Sin(\frac{0.4426}{2}))= 2- 3i

w₂=\sqrt{13}(Cos(\frac{0.4426+2\pi}{2}) + i Sin(\frac{0.4426+2\pi}{2}))= -2 + 3i

5) z = 21 − 20i  ⇒ r=29 θ= -0.8098 rad

w₁=\sqrt{29}(Cos(\frac{-0.8098}{2}) + i Sin(\frac{-0.8098}{2}))= 5 - 2i

w₂=\sqrt{29}(Cos(\frac{-0.8098+2\pi}{2}) + i Sin(\frac{-0.8098+2\pi}{2}))= -5 + 2i

6) z = 16 − 30i ⇒ r=34 θ= -1.0808 rad

w₁=\sqrt{34}(Cos(\frac{-1.0808}{2}) + i Sin(\frac{-1.0808}{2}))= 5 - 3i

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