A cashier will randomly choose two coins one at a time without replacement. What is the probabilty that the cashier will chose a penny first, followed by a dime?

Complete question is;

A cash register contains coins. 10 pennies, 11 nickels, 13 dimes, and 6 quarters. A cashier will randomly choose two coins one at a time without replacement. What is the probability that the cashier will choose a penny first, followed by a dime?

1/12

Step-by-step explanation:

We are told there are 10 pennies, 11 nickels, 13 dimes, and 6 quarters.

Thus total number of coins = 10 + 11 + 13 + 6 = 40 coins

Probability of picking a penny first is;

10/40 = 1/4

After which we will have 39 coins left.

Probability of a dime after that will be 13/39 = 1/3

Thus,probability of picking a penny first then a dime will be; ¼ × ⅓ = 1/12

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5

answer: 1722

step-by-step explanation:

given: the mass of saturn=

the mass of mercury=

the number of times the mass of saturn is greater than the mass of mercury

hence, the mass of saturn is 1722 times greater than the mass of mercury.