Abox contains a 3 nickel, 2 pennies, and 5dimes. find the probability of choosing first a dime, and then, without replacing the dime, choosing a penny?

Finding a dime, 5/10

Finding a penny, 2/9

Step-by-step explanation:

add up the total number of coins, 10, and put the number of dimes over it

5/10

add up the total number of coins, 10, and subtract a dime, so there is 9 coins, but 2 pennies so 2/9

d

step-by-step explanation:

given

y + 3 = - 4(x - 2) ← distribute the parenthesis

y + 3 = - 4x + 8 ( subtract 3 from both sides )

y = - 4x + 5 → d

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)