If you replace some number for "p", you can calculate the sum. Otherwise, just write it as "p+9" - you can't simplify this.
If you replace some number for "p", you can calculate the sum. Otherwise, just write it as "p+9" - you can't simplify this.
If you replace some number for "p", you can calculate the sum. Otherwise, just write it as "p+9" - you can't simplify this.
If you replace some number for "p", you can calculate the sum. Otherwise, just write it as "p+9" - you can't simplify this.
S.I. = (P x R x T)/100 where R is rate, T is time, P is Original sum and S.I. is simple interest. 800x100 = P x R x T P = 80000/(5x7) = 80000/35 = 2285.71 So, the original sum is Rs 2285.71
The sum of 1 + 8 is 9.
the sum is the answer to an addition problem e.g.4+5=9 9 is the sum
The sum of 8, 11 and 9 is 28.
4 and 5 have the sum of 9 and the product of 20.
1/2(9+p)=p-3
P(9) ≈ 11.11%When two dice are rolled there are 36 equiprobable events.The number of events that produce a sum of 9 are 4: (6,3), (5,4), (4,5) and (3,6).So the probability of getting a sum of 9 is, P(9) = 4/36 = 1/9 = 0.1111.... ≈ 0.1111≈ 11.11%
P(x≥9) = 0.27777... ≈ 28%Let x equal to the sum of the numbers showing on dice when you roll a pair of dice.P(x=12) = 1/36; P(x=11) = 2/36; P(x=10) = 3/36; P(x=9) = 4/36.So, P(x≥9) = P(x=12) + P(x=11) + P(x=10) + P(x=9) = 1/36 [1+2+3+4] = 10/36P(x≥9) = 0.2777... ≈ 28%
The sum of p and q means (p+q). The difference of p and q means (p-q).
Aces and 9s are disjoint events, so the probability of either is the sum of the probabilities of each. P(A or 9) = P(A) + P(9) = 1/13 + 1/13 = 2/13
p + 21.
q - p
The sum of 19 and 9 is 28.
S.I. = (P x R x T)/100 where R is rate, T is time, P is Original sum and S.I. is simple interest. 800x100 = P x R x T P = 80000/(5x7) = 80000/35 = 2285.71 So, the original sum is Rs 2285.71
If ( p ) is an integer between 1000 and 1030, it can be expressed as ( p = 1000 + n ), where ( n ) ranges from 0 to 30. The sum of the digits of ( p ) is given by ( 1 + \text{(sum of the digits of } n) ). Since 1 is odd, for the total sum of the digits to be odd, the sum of the digits of ( n ) must be even. As a result, if ( p ) is odd, ( n ) must be odd (e.g., 1, 3, 5, etc.), confirming that ( p ) is indeed odd. Thus, the statement is true: if the sum of the digits of ( p ) is odd, then ( p ) must be odd.
p+n=x
p + 7