Answer:
(1) .20 (2) .40 (3) .12 (4) Less than
Step-by-step explanation:
You have to look at the table. There are 5 columns with 10 rows. 5x10=50
Then simply count the boxes that have the correct number of currency for instance, if they are asking for EXACTLY 1 dime then you rule out the ones that have 2 or 3 dimes and only the count the ones that have a single dime. So you count PDN but you would not count PDD. There are 20 boxes that have a single dime in them. 20 out of the 50 boxes. 20/50=.40 (answer 2)
The estimated probability that exactly two of the three coins Avery randomly picked are nickels is .
20
The estimated probability that exactly one of the three coins Avery randomly picked is a dime is .
40
The estimated probability that all three coins Avery randomly picked are pennies is .
12
The answer to #1 is .20 or 20% and the answer to #2 is .40 or 40%. 20% is less than 40% so...
The estimated probability that exactly two of the three coins Avery randomly picked are nickels is LESS THAN the estimated probability that exactly one of the three coins Avery randomly picked is a dime.
Answer:
3 slices
Step-by-step explanation:
12-75%=3
Answer:
12 - p
Step-by-step explanation:
12 - p
Answer:
Kindly check explanation
Step-by-step explanation:
Given that :
Upper level seat = $25
Mid level seat = $40
Number of tickets to give away is at least 25
Budget constraint = $1000
part A: write a system of two inequalities that describe this situation
Number of tickets constraint:
Upper level + mid level ≥ 25
u + m ≥ 25
Cost constraint :
$25u + $40m ≤ $1000
Part B:
Give away 10 upper level seat tickets and 15 mid level seat tickets
