A spinner is shown with two blue, two green, two yellow, and two red sections. The arrows is pointing to a red section. Color
red
blue
green
yellow
# of times spun
6
4
3
5
Use the table to answer the question.
What is the experimental probability of the spinner landing on blue?
A. Start Fraction 2 over 7 End Fraction
B. one-fourth
C. start fraction 2 over 9 end fraction
12 answers
A. Start Fraction 2 over 7 End Fraction
Two coins were tossed 10 times. The results are shown in the table.
Toss 1 2 3 4 5 6 7 8 9 10
Result HH TT HT TH HT HH TH TT TH HT
Use the table and information to answer the question.
What is the experimental probability that at least one of the coins landed on heads? Express the probability as a decimal.
A. 0.8
B. 0.2
C. 0.6
Toss 1 2 3 4 5 6 7 8 9 10
Result HH TT HT TH HT HH TH TT TH HT
Use the table and information to answer the question.
What is the experimental probability that at least one of the coins landed on heads? Express the probability as a decimal.
A. 0.8
B. 0.2
C. 0.6
C. 0.6
Out of the 10 tosses, there were 6 times when at least one coin landed on heads (HH, HT, TH, HT, HH, TH). Therefore, the experimental probability that at least one of the coins landed on heads is 6/10, which simplifies to 0.6.
Out of the 10 tosses, there were 6 times when at least one coin landed on heads (HH, HT, TH, HT, HH, TH). Therefore, the experimental probability that at least one of the coins landed on heads is 6/10, which simplifies to 0.6.
Toss 1 2 3 4 5 6 7 8 9 10
Result HH TT HT TH HT HH TH TT TH HT
Use the table and information to answer the question.
What is the experimental probability that both of the coins landed on tails? Express the probability as a decimal.
A. 0.8
B. 0.2
C. 0.6
Result HH TT HT TH HT HH TH TT TH HT
Use the table and information to answer the question.
What is the experimental probability that both of the coins landed on tails? Express the probability as a decimal.
A. 0.8
B. 0.2
C. 0.6
B. 0.2
Out of the 10 tosses, there was only 1 time when both coins landed on tails (TT). Therefore, the experimental probability that both of the coins landed on tails is 1/10, which simplifies to 0.1.
Out of the 10 tosses, there was only 1 time when both coins landed on tails (TT). Therefore, the experimental probability that both of the coins landed on tails is 1/10, which simplifies to 0.1.
A survey of a random sample of 20 moviegoers shows that 8 people purchased popcorn to eat while watching the movie. If a theater has 80 people in it, how many should you expect to have purchased popcorn?
A. 40 people
B. 36 people
C. 32 people
A. 40 people
B. 36 people
C. 32 people
B. 36 people
If 8 out of 20 people (or 2 out of 5 people) purchased popcorn, we can estimate that 2/5 of the total number of people in the theater will purchase popcorn. So, we can set up a proportion:
2/5 = x/80
where x is the number of people we expect to have purchased popcorn. To solve for x:
2(80) = 5x
160 = 5x
x = 32
Therefore, we expect 32 people in the theater to have purchased popcorn.
If 8 out of 20 people (or 2 out of 5 people) purchased popcorn, we can estimate that 2/5 of the total number of people in the theater will purchase popcorn. So, we can set up a proportion:
2/5 = x/80
where x is the number of people we expect to have purchased popcorn. To solve for x:
2(80) = 5x
160 = 5x
x = 32
Therefore, we expect 32 people in the theater to have purchased popcorn.
In a random sample of egg cartons, a supermarket finds that 3 cartons in every 50 cartons contain cracked eggs. If the supermarket receives a shipment of 400 cartons, how many cartons can they expect will contain cracked eggs?
A. 24 cartons
B. 30 cartons
C. 14 cartons
A. 24 cartons
B. 30 cartons
C. 14 cartons
B. 30 cartons
If 3 out of 50 cartons contain cracked eggs, we can set up a proportion to find out how many cartons out of 400 we can expect to contain cracked eggs:
3/50 = x/400
where x is the number of cartons we expect to contain cracked eggs. To solve for x:
50x = 3(400)
50x = 1200
x = 24
Therefore, we can expect 24 out of 400 cartons to contain cracked eggs.
If 3 out of 50 cartons contain cracked eggs, we can set up a proportion to find out how many cartons out of 400 we can expect to contain cracked eggs:
3/50 = x/400
where x is the number of cartons we expect to contain cracked eggs. To solve for x:
50x = 3(400)
50x = 1200
x = 24
Therefore, we can expect 24 out of 400 cartons to contain cracked eggs.
A manufacturer checks a random sample of jeans and finds that 5 out of 80 jeans have a flaw. Which is the best prediction for the number of flawed jeans in a shipment of 20,000?
A. 125 jeans
B. 1,250 jeans
C. 12,500 jeans
A. 125 jeans
B. 1,250 jeans
C. 12,500 jeans
B. 1,250 jeans
If 5 out of 80 jeans have a flaw, we can set up a proportion to find out how many flawed jeans we can expect out of 20,000:
5/80 = x/20,000
where x is the number of flawed jeans we can expect in a shipment of 20,000. To solve for x:
80x = 5(20,000)
80x = 100,000
x = 1,250
Therefore, we can predict that there will be 1,250 flawed jeans in a shipment of 20,000.
If 5 out of 80 jeans have a flaw, we can set up a proportion to find out how many flawed jeans we can expect out of 20,000:
5/80 = x/20,000
where x is the number of flawed jeans we can expect in a shipment of 20,000. To solve for x:
80x = 5(20,000)
80x = 100,000
x = 1,250
Therefore, we can predict that there will be 1,250 flawed jeans in a shipment of 20,000.
Which of the following situations could be modeled by flipping a coin?
A. randomly picking one of four students to be a group leader
B. randomly guessing the answers on a true/false test
C.
randomly selecting a prize from a bag containing six different prizes
A. randomly picking one of four students to be a group leader
B. randomly guessing the answers on a true/false test
C.
randomly selecting a prize from a bag containing six different prizes