The possible digits are:
5, 6, 7, 8 and
9. Let's mark the case when the locker code begins with a prime number as
A and the case when <span>the locker code is an odd number as
B. We have
5 different digits in total,
2 of which are prime (
5 and
7).
First propability:
</span>
<span>
By knowing that digits don't repeat we can say that code is an odd number in case it ends with
5, 7 or
9 (three of five digits).
Second probability:
</span>
1/23 = 0.043
I hope this helps :)
Step-by-step explanation:
<h2>
<em><u>3</u></em><em><u>x</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>1</u></em><em><u>0</u></em><em><u> </u></em><em><u>≤</u></em><em><u> </u></em><em><u>3</u></em><em><u>0</u></em><em><u> </u></em></h2><h2>
<em><u>3</u></em><em><u>x</u></em><em><u> </u></em><em><u>≤</u></em><em><u> </u></em><em><u>3</u></em><em><u>0</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>1</u></em><em><u>0</u></em><em><u> </u></em></h2><h2>
<em><u>3</u></em><em><u>x</u></em><em><u> </u></em><em><u>≤</u></em><em><u> </u></em><em><u>2</u></em><em><u>0</u></em></h2><h2>
<em><u>x </u></em><em><u>≤</u></em><em><u> </u></em><em><u>2</u></em><em><u>0</u></em><em><u>/</u></em><em><u>3</u></em></h2>
Answer:
An equation is c = 15m
Step-by-step explanation:
Answer: of each shape? 3 & 4?
Step-by-step explanation:
