Compute how many 7-digit numbers can be made from the digits 1,2,3,4,5,6,7 if there is no repetition and the odd digits must appear in an unbroken sequence

There are four odd digits, and the odd digits sequence begin on the 1st empty spot, 2nd, 3r  or 4rth.
So basically we have four choices, where the first digit is odd.
We get
Four ways (4!) to arrange odd digits.
Three ways (3!) to arrange even digits.
Total arrangements = 4 x 4! x 3!
Simplify
Total arrangements= 4 x 4 x 3 x 2 x 1 x 3 x 2 x 1
Total arrangements= 576

576

One shot. if 60% of 10 is 6, then what is 70%? we know it would be seven (since 60% is mathematically 60/100, or 6/10, then 70% is 70/100, or 7/10) but we can check anyway using ratios: 6 : 60% = x : 70%*inverse multiplicationx * 60% = 6 * 70%*divide both sides by 60%x = (6 * 70%) / 60%x = 7and since he has made 6 shots already, he would need just one more to get 7 shots or 70%. if the explanation is too confusing, comment and i will try to