Answer:a) L.C.M OF 2,3,5 is 30
THEREFORE=60/30+15/30÷10/30-6/30
=60/30+15/30×30/10-6/30(DMAS)
=60/30+3/2-6/30
=60+45/30 -6/30(ADDITION FIRST THEN SUBTRACTION.
=105/30-6/30
=105-6/30
=99/30
=33/10
b)(3+1/2)÷(2-1/2)=L.C.M IS 2,=(6+1/2)÷(4-1/2)=7/2÷3/2=7/3
The answer would be: 0.0065
If all were children, revenue would be 700*$7 = $4900. Revenue is actually $6400 -4900 = $1500 more than that. Each adult admission that replaces a child's admission adds $10 -7 = $3 to the revenue, so there must have been
$1500/$3 = 500 . . . . adult admissions
There were 200 children at the show.
There were 500 adults at ths show.
Explanation:
For the purpose of filling in the table, the BINOMPDF function is more appropriate. The table is asking for p(x)--not p(n≤x), which is what the CDF function gives you.
If you want to use the binomcdf function, the lower and upper limits should probably be the same: 0,0 or 1,1 or 2,2 and so on up to 5,5.
The binomcdf function on my TI-84 calculator only has the upper limit, so I would need to subtract the previous value to find the table entry for p(x).
3x+6 I believe because you can’t add a number with a variable to a number without one
