The result of rolling a number cube 7 times is a 7-digit number composed of digits 1,2,3,4,5 and 6 so that digits can repeat. The total number of possibilities is 6^7.
The number of possibilities where 4 appears exactly two times is 5^5*(7!-6!/2).
5^5 is the number of 5-digits numbers composed of digits 1,2,3,5 and 6 so that digits can repeat.
7! is the number of permutations of digits 1,2,3,4,4,5 and 6.
6! is the number of permutations of digits 1,2,3,{4,4},5 and 6.
We don't want to subtract all numbers where digits 4 appear side by side. That's why we must divide 6! by 2.
Finally, the probability is P=5^5(7!-6!/2)/7^7
Answer:
840
Step-by-step explanation:
(−30)(−7) − (210)(−3)
=210 − (210)(−3)
=210 − (−630)
=840
Hope this helps!
brainliest?
:)
The answer to this one is A
Answer:
<h2><em><u>Pythagorean </u></em><em><u>theorem </u></em><em><u>reads </u></em><em><u>as:</u></em></h2>
<h2><em><u>H²</u></em><em><u>=</u></em><em><u>P²</u></em><em><u>+</u></em><em><u>B</u></em><em><u>²</u></em></h2>
<h2><em><u>in </u></em><em><u>which </u></em><em><u>p </u></em><em><u>reads </u></em><em><u>as </u></em><em><u>perpendicular </u></em><em><u>so </u></em></h2>
<h2><em><u>P²</u></em><em><u>=</u></em><em><u>H²</u></em><em><u>-</u></em><em><u>B²</u></em></h2>
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