First two rolls have to be 1-4 that is 2/3 chance twice and the third can be 4or 5
2/3*2/3*1/3 + the chance that the fourth is the 5 or 6.
2/3*2/3*2/3*1/3
So the solution is : P=2/3*2/3*1/3 + 2/3*2/3*2/3*1/3
Answer:
→<u> </u><u>First</u><u> </u><u>value</u><u> </u><u>is</u><u> </u><u>1</u>
→<u> </u><u>Second</u><u> </u><u>value</u><u> </u><u>is</u><u> </u><u>2</u>
→<u> </u><u>Third</u><u> </u><u>value</u><u> </u><u>is</u><u> </u><u>5</u>
Step-by-step explanation:
• let numbers be x, y and z
• from eqn 2, make x the subject:
• substitute all variables in eqn 3:
• find z
• find x:
Rounding to nearest value:
Answer:
6+_3√2/2
Step-by-step explanation:
y=2x²-12x+9
y=6+3√2/2
Answer:
53
Step-by-step explanation:
37+90=127
180-127=53