Answer:
- 3 (die)
- 4 (slips)
- 6 (spinner)
- 5 (ace)
Step-by-step explanation:
Josie rolls a six-sided die 18 times. What is the estimated number of times she rolls a two? 3 = (1/6)(18)
Slips of paper are numbered 1 through 10. If one slip is drawn and replaced 40 times, how many times should the slip with number 10 appear? 4 = (1/10)(40)
A spinner consists of 10 equal- sized spaces: 2 red, 3 black, and 5 white. If the spinner is spun 30 times, how many times should it land on a red space? 6 = (2/10)(30)
A card is picked from a standard deck of playing cards 65 times and replaced each time. About how many times would the card drawn be an ace? 5 = (4/52)(65)
_____
The probability of a given event is the number of ways it can occur divided by the number of possibilities. For example, a 2 is one of 6 numbers on a die, so we expect its probability of showing up to be 1/6. The expected number of times it will show up in 18 rolls of the die is (1/6)(18) = 3.
Answer:
x=82/81
Step-by-step explanation:
The lengths of the first, second and third piece are respectively 8 ft, 16 ft and 32 ft
<h3>How to solve Algebra word Problems?</h3>
Let the length of the first piece be x.
Let the length of the second piece be 2x
Let the length of the thrid piece be 4x
Since the total length of the piece is 56 ft, then we can say that;
x + 2x + 4x = 56
7x = 56
x = 56/7
x = 8 ft
Thus, length of first piece = 8 ft
Length of second piece = 2 * 8 = 16 ft
Length of third piece = 4 * 8 = 32 ft
Thus, the lengths of the first, second and third piece are respectively 8 ft, 16 ft and 32 ft
Read more about Algebra Word Problems at; brainly.com/question/13818690
#SPJ1
2.0076* 10^4
move the decimal 4 places to the right
20076.
20076 < 26970
Answer:
<em>m</em>
perpendicular
=
−
3
/4
