Michael has 3 quarters, 2 dimes, and 3 nickels in his pocket. He randomly draws two coins from his pocket, one at a time, and they are both dimes. He says the probability of that occurring is 1 4 because 2 of the 8 coins are dimes. Is he correct

No, he is wrong

Step-by-step explanation:

To find that probability, the first thing we need to know is the total number of coins present.

Mathematically, that would be 3 + 2 + 3 = 8

Since there are 8 coins, the probability of selecting a dime is number of dimes/total number of coins = 2/8 = 1/4

The probability we want to work with is that the two selections are dimes, let’s say with replacement.

That means; first selection is a dime and second selection is also a dime

Mathematically in probability, the term and means that we have to multiply our results.

So the probability that both are dimes would be 1/4 * 1/4 = 1/16

And this makes his answer wrong

answer:  

step-by-step explanation:

odd numbers are only divisible by 1 and itself.

even numbers are divisible by more than 1 and itself.

for example 6 is even because it is divisible by more than 1 and its self. 1 x 6= 6 ; 2 x 3= 6

and 7 is odd because only 7 x 1 equals seveb

x-150=475   x=625

step-by-step explanation:

if the bike after the sale ($150) is $475 then just simply add 150 to 475.