<p>How would you do this problem?</p>
<p>How many four-digit numbers are possible if the first digit cannot be zero and the number is divisible by 5.</p>
<p>A. 900
B. 1458
C. 1800
D. 2000
E. 4500</p>
<p>Look at how many numbers can be in each “place” of the four-digit number:</p>
<p>Thousands place: 9 possible digits (1-9)
Hundreds place: 10 possible digits
Tens place: 10 possible digits
Ones place: 2 possible digits (0 or 5, since it has to be divisible by 5)</p>
<p>Now you can multiply the the number of possible digits in each place to find the total number of possible combinations:
9x10x10x2=1800</p>
<p>hmm… I see, thanks tanman.</p>
<p>How about this question?</p>
<p>Hakim and Chris began running a 50-yard race at the same time. When Hakim finished the race, chris was 4 yards behind him. If hakim ran the race in 7 seconds, what was the difference in their rates in yards per second for those 7 seconds?</p>
<p>if hakim ran 50yds/7sec and chris ran 46yds/7sec, then 50/7 - 46/7 = 4/7 yrds/sec…i hope im not just being dumb</p>
<p>8888, where did u get this question?
“Hakim and Chris began running a 50-yard race at the same time. When Hakim finished the race, chris was 4 yards behind him. If hakim ran the race in 7 seconds, what was the difference in their rates in yards per second for those 7 seconds?”</p>
<p>i’ve seen it somewhere before…</p>
<p>it’s from one of those practice test from pr, college board, barrons,etc… I’m not sure though.</p>
<p>^^^ that’s how you do it borwn?? … I don’t know lol why did you divide 46 by 7? :/</p>
<p>the problem said hakim finished the 50 yard race in 7 seconds and that chris wsa 4 yards behind him when he finished so he had to have ran (50-4)yds in 7 secs or 46yds/7sec. so, the difference in their rates wud be 50/7 - 46-7…</p>