796,000/2388=10,000/x
796,000x= (2388*10,000)
796,000x/796,000=23,880,000/796,000
x=30 births
Digit numbers.
Let the numbers 00 to 89 represent that the train is on time.
Let the numbers between 90 and 99 represent that the train is late.
Randomly select 6 numbers, with repetition allowed.
Count the number of times the train is late.
Repeat this simulation multiple times.
You will most likely obtain a result of between 0 and 2 times that the train is late.
For the house A we have:
f (x) = 124270 (1.04) ^ x
Evaluating for 7, 8, 9 and 10 we have:
f (7) = 124270 (1.04) ^ 7 = 163530.8422
f (8) = 124270 (1.04) ^ 8 = 170072.0759
f (9) = 124270 (1.04) ^ 9 = 176874.9589
f (10) = 124270 (1.04) ^ 10 = 183949.9573
For house B we have:
f (x) = 114270 (1.05) ^ x
Evaluating for 7, 8, 9 and 10 we have:
f (7) = 114270 (1.05) ^ 7 = 160789.3653
f (8) = 114270 (1.05) ^ 8 = 168828.8336
f (9) = 114270 (1.05) ^ 9 = 177270.2752
f (10) = 114270 (1.05) ^ 10 = 186133.789
We observe that for years 7 and 8 the value of house A is greater than the value of house B.
Answer:
7 and 8
It is definitely D because on part 1 n=24n+20 and if n equals 0 than 24 (0) +20= 20 and that is true. on part 2 if n =0 than 24 (0)+ 20 (0-1)=20 than 20 (-1)=20 -20= 20. that statement is false because-20 does not equal 20. AND for part 3 n=1 and f (1-1) + 24= 44 so f (0) + 24 =44 and keep in mind f (0) equals 20 so 44=44 because 24+20 is 44. so that is a true statement too