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Annette [7]
3 years ago
13

Suppose two dice (one red, one green) are rolled. Consider the following events. A: the red die shows 1; B: the numbers add to 4

; C: at least one of the numbers is 3; and D: the numbers do not add to 9. Express the given event in symbolic form. HINT [See Example 5.]
Either the numbers add to 9 or the red die shows a 1.

D ∩ B

D ∩ A

D' ∪ A

D' ∩ A

D' ∪ B

How many elements does it contain?
Mathematics
1 answer:
kirill115 [55]3 years ago
5 0

Answer: Third option is correct.

Step-by-step explanation:

Since we have given by

Event A: the red die shows 1

Event B :  the numbers add to 4

Event C: at least one of the numbers is 3

Event D : the numbers do not add to 9.

D' means the number add to 9.

We need to write the following statement in symbolic form:

Either the numbers add to 9 or the red die shows a 1.

So, it becomes,

D'\cup A

Number of elements in D'∪A = 1.

Hence, Third option is correct.

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Answer:

Volume of the cone: 32π cubed unit

Step-by-step explanation:

Volume of cone: V=1/3hπr²

1/3 x 6 x π x 4^2 = 32π

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If the length of the longest side of a right triangle is 10 inches is 60 degrees, what is the length (in inches) of the shortest
Scorpion4ik [409]
Yep 5 is correct! Good job
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3 years ago
George has a mass of 65kg. What would George weigh on the Moon? (The gravitational field strength on the Moon is 1.6N/kg.)
maks197457 [2]

George would weigh

65 kg × 1.6N /kg = 104 Newtons

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3 years ago
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What is 5.316 - 1.942 btw (show ur work) :)
erastovalidia [21]

Answer:

3.374

Step-by-step explanation:

\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}

\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}

\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}

\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}

\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4\\

\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}

\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}

\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&10&\:\:&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&3&1&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{\:\:}&\:\:&\:\:&\:\:&4\end{matrix}}\\

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{7}&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&11&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&\linethrough{1}&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{3}&7&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}

\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}

=3.374

6 0
2 years ago
The magnitudes of the moon and sirius are -12.5 and -1.44, respectively. approximately how many times is the moon brighter than
Lelu [443]

Solution:

we are given that

The magnitudes of the moon and Sirius are -12.5 and -1.44 respectively.

we have been asked to find approximately how many times is the moon brighter than Sirius?

The word magnitude is used by the scientist to describe the brightness of moon/star/object etc.

So we can find the brightness by dividing the given values. so we get

\frac{-12.5}{-1.44} =8.6805

Hence we can say

The moon is 8.68 times brighter than Sirius.

3 0
3 years ago
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