Answer:
510
Step-by-step explanation:
divide into chunks. I divided it like how you would draw the equal sign.
for top chunk, 20*8=160
for 2nd chunk, 8*25=200
for the last chunk, 15*10=150
add up 160, 200, and 150
Order the list from least to greatest:
-12, -9, 0, 8, 15, 16, 17
subtract the least from the greatest: 17 - (-12)
17 + 12 = 29
The range is 29
hope this helps
Answer:
Number of times she will hit the ball the next time she plays softball if she is at bat 20 times = 9 times
Step-by-step explanation:
Percentage of times Sierra hit the ball when she was at bat playing softball= 45%
So, we can find:
Number of times she will hit the ball the next time she plays softball if she is at bat 20 times = 
At at least one die come up a 3?We can do this two ways:) The straightforward way is as follows. To get at least one 3, would be consistent with the following three mutually exclusive outcomes:the 1st die is a 3 and the 2nd is not: prob = (1/6)x(5/6)=5/36the 1st die is not a 3 and the 2nd is: prob = (5/6)x((1/6)=5/36both the 1st and 2nd come up 3: prob = (1/6)x(1/6)=1/36sum of the above three cases is prob for at least one 3, p = 11/36ii) A faster way is as follows: prob at least one 3 = 1 - (prob no 3's)The probability to get no 3's is (5/6)x(5/6) = 25/36.So the probability to get at least one 3 is, p = 1 - (25/36) = 11/362) What is the probability that a card drawn at random from an ordinary 52 deck of playing cards is a queen or a heart?There are 4 queens and 13 hearts, so the probability to draw a queen is4/52 and the probability to draw a heart is 13/52. But the probability to draw a queen or a heart is NOT the sum 4/52 + 13/52. This is because drawing a queen and drawing a heart are not mutually exclusive outcomes - the queen of hearts can meet both criteria! The number of cards which meet the criteria of being either a queen or a heart is only 16 - the 4 queens and the 12 remaining hearts which are not a queen. So the probability to draw a queen or a heart is 16/52 = 4/13.3) Five coins are tossed. What is the probability that the number of heads exceeds the number of tails?We can divide
