Language Arts 7 B Unit 6: Language and Style Help
I go to connexus and and literaly 2 years and 4 days later Nobody is still correct
do the word problem on your own
The spinner has a total of 8 sections, out of which 2 are red. The experimental probability of the spinner landing on red is the number of times red was spun divided by the total number of spins:
Experimental probability of landing on red = 2 / (6 + 4 + 3 + 5) = 2 / 18 = 1 / 9
The fraction is 1 over 9. None of the given answer choices matches with this, so we need to select the closest option.
The nearest option to 1/9 is "the fraction is 1 over 4." However, this is not accurate. The experimental probability is less than one-fourth. The best option is therefore:
Answer: B. one-third.
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
The spinner has a total of 8 sections, out of which 2 are blue. The experimental probability of the spinner landing on blue is the number of times blue was spun divided by the total number of spins:
Experimental probability of landing on blue = 2 / (6 + 4 + 3 + 5) = 2 / 18 = 1 / 9
Answer: C. start fraction 2 over 9 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
3 / 8
2 of 8 Answered
In each coin toss, there are two possible outcomes - heads (H) or tails (T). So, the probability of getting tails in one toss is 1/2, and the probability of getting heads is also 1/2.
To find the probability that at least one of the coins landed on heads, we can find the complement of the probability that both coins landed on tails (i.e., both coins landed on tails in all 10 tosses).
The probability that both coins landed on tails in one toss is 1/2 x 1/2 = 1/4. So, the probability that both coins landed on tails in all 10 tosses is (1/4)^10 = 1/2,56,00,000.
Therefore, the probability that at least one of the coins landed on heads is 1 - probability that both coins landed on tails in all 10 tosses = 1 - 1/2,56,00,000 ≈ 1. Hence, the answer is:
Answer: A. 0.8 (expressed as a decimal).
Two coins were tossed 10 times. The results are shown in the table below.
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
4 / 8
3 of 8 Answered
In each coin toss, there are two possible outcomes - heads (H) or tails (T). So, the probability of getting tails in one toss is 1/2, and the probability of getting heads is also 1/2.
To find the probability that both coins landed on tails in one toss, we need to multiply the probability of getting tails on the first coin toss with the probability of getting tails on the second coin toss.
The probability that both coins landed on tails in one toss is 1/2 x 1/2 = 1/4.
Since the coins were tossed 10 times, we need to multiply this probability with itself 10 times to find the probability of both coins landing on tails in all 10 tosses:
Probability of both coins landing on tails in all 10 tosses = (1/4)^10 = 1/2,56,00,000 (approximately).
Therefore, the experimental probability that both coins landed on tails is:
Answer: B. 0.2 (expressed as a decimal).
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
We can set up a proportion to solve the problem. If 8 out of 20 people purchased popcorn, then the proportion of people who purchased popcorn can be expressed as:
8/20 = x/80
To solve for x, we can cross-multiply the proportion:
20x = 8 * 80
20x = 640
x = 640/20
x = 32
Therefore, we should expect 32 people to have purchased popcorn in a theater with 80 people.
Answer: C. 32 people.
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
We can set up a proportion to solve the problem. If 3 cartons out of 50 cartons contain cracked eggs, then the proportion of cartons that contain cracked eggs can be expressed as:
3/50 = x/400
To solve for x, we can cross-multiply the proportion:
50x = 3 * 400
50x = 1200
x = 1200/50
x = 24
Therefore, the supermarket can expect 24 cartons out of 400 to contain cracked eggs.
Answer: A. 24 cartons.
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
We can use a proportion to estimate the number of flawed jeans in a shipment of 20,000. If 5 out of 80 jeans have a flaw, then the proportion of flawed jeans can be expressed as:
5/80 = x/20,000
To solve for x, we can cross-multiply the proportion:
80x = 5 * 20,000
80x = 100,000
x = 100,000/80
x ≈ 1250
Therefore, we can predict that there will be around 1250 flawed jeans in a shipment of 20,000.
Answer: B. 1250 jeans.
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
Answer: B. randomly guessing the answers on a true/false test.
The table shows the results of rolling a number cube labeled one through six 50 times.
Number Rolled
Frequency
1
7
2
9
3
11
4
6
5
9
6
8
Use the table and information to answer the question.
What is the experimental probability of rolling a 3?
A. 0.11
B. 0.22
C. 0.30
D. 0.27
The experimental probability of rolling a 3 is the number of times a 3 was rolled divided by the total number of rolls:
Experimental probability of rolling a 3 = 11/50
Answer: A. 0.11.
A quality control expert randomly samples 60 pairs of sunglasses and finds 5 defective pairs. Predict how many defective pairs will be in a shipment of 420 sunglasses.
A. 30
B. 35
C. 40
D. 70
We can use a proportion to estimate the number of defective pairs in a shipment of 420 sunglasses. If 5 out of 60 sunglasses are defective, then the proportion of defective sunglasses can be expressed as:
5/60 = x/420
To solve for x, we can cross-multiply the proportion:
60x = 5 * 420
60x = 2100
x = 2100/60
x = 35
Therefore, we can predict that there will be 35 defective pairs in a shipment of 420 sunglasses.
Answer: B. 35.
Use the following information for the question.
A school has an equal number of boys and girls. You use a coin to simulate the first three students to arrive at school each day, where “heads” represents a boy and “tails” represents a girl. The table below shows a sample of 20 coin tosses
T H T
T T T
T H T
H T H
H H H
T T T
H T T
H H T
T H T
T T T
T T H
T H T
H H T
H H H
H T H
T H T
H H T
T T T
H H H
T H T
Find the experimental probability that the first three students to arrive at school are boys.
A. Start Fraction 3 over 20 End Fraction
B. one-fifth
C. start fraction 3 over 10 end fraction
D. one-fourth
We need to count the number of times "heads" appears three times consecutively (representing three boys) in the 20 coin tosses.
From the sample, we can see that "heads" appears three times consecutively in 3 out of 20 tosses. Therefore, the experimental probability that the first three students to arrive at school are boys is:
Experimental probability = 3/20
Answer: A. Start Fraction 3 over 20 End Fraction
Use the following information for the question.
A school has an equal number of boys and girls. You use a coin to simulate the first three students to arrive at school each day, where “heads” represents a boy and “tails” represents a girl. The table below shows a sample of 20 coin tosses
T H T
T T T
T H T
H T H
H H H
T T T
H T T
H H T
T H T
T T T
T T H
T H T
H H T
H H H
H T H
T H T
H H T
T T T
H H H
T H T
Find the experimental probability that exactly two out of the first three students to arrive at school are girls.
A. one-fifth
B. one-fourth
C. two-fifths
D. one-eighth
We need to count the number of times "tails" appears exactly twice (representing two girls) in the first three coin tosses.
From the sample, we can see that "tails" appears exactly twice in the first three tosses in 5 out of 20 trials - THT, TTT, TTH, TTT, and TTH. Therefore, the experimental probability that exactly two out of the first three students to arrive at school are girls is:
Experimental probability = 5/20 = 1/4
Answer: B. one-fourth.
You mix the letters S, E, M, I, T, R, O, P, I, C, A, and L thoroughly. Without looking, you draw one letter. Find the probability that you select a vowel. Write your answer as a fraction in simplest form.
A. twelve-fifths
B. Start fraction 5 over 12 End fraction
C. start fraction 1 over 3 end fraction
D. start fraction 7 over 12 end fraction
There are 12 letters in total, including 5 vowels (E, I, O, U, and A) and 7 consonants (S, M, T, R, P, C, and L).
The probability of selecting a vowel can be found by dividing the number of vowels by the total number of letters:
Probability of selecting a vowel = Number of vowels / Total number of letters
= 5/12
Therefore, the probability of selecting a vowel is:
Answer: B. Start fraction 5 over 12 End fraction.
You roll a standard number cube once. Find P(0).
A. The term shows 7 over 6.
B. 1
C. one-half
D. 0
The probability of rolling a 0 on a standard number cube is zero because the numbers on a standard number cube range from 1 to 6, and there is no such number as 0.
Answer: D. 0.
A Lights-A-Lot quality inspector examines a sample of 25 strings of lights and finds that 6 are defective.
a. What is the experimental probability that a string of lights is defective?
b. What is the best prediction of the number of defective strings of lights in a delivery of 1,000 strings of lights?
A. three over five hundred; 6 lights
B. six over twenty-five; 24 lights
C. one over forty; 25 lights
D. six over twenty-five; 240 lights
a. The experimental probability that a string of lights is defective is the number of defective strings divided by the total number of strings in the sample:
Experimental probability = Number of defective strings / Total number of strings
= 6/25
Answer: B. six over twenty-five.
b. We can use a proportion to estimate the number of defective strings of lights in a delivery of 1,000 strings. If 6 out of 25 strings are defective, then the proportion of defective strings can be expressed as:
6/25 = x/1000
To solve for x, we can cross-multiply the proportion:
25x = 6 * 1000
25x = 6000
x = 6000/25
x = 240
Therefore, the best prediction of the number of defective strings of lights in a delivery of 1,000 strings is:
Answer: D. 240 lights.
Coin Toss Results
H = Heads T = Tails
T H T H T H T T T T H T H T T
Find the experimental probability of tossing heads.
A. five-eighths
B. one-third
C. two-thirds
D. 2
To find the experimental probability of tossing heads, we need to count the number of heads that were tossed and divide it by the total number of coin tosses:
Number of heads = 6
Total number of tosses = 15
Experimental probability of tossing heads = Number of heads / Total number of tosses
Experimental probability of tossing heads = 6/15
Answer: C. two-thirds.
Clarissa is having lunch at a sandwich shop. She can choose white bread or pumpernickel bread. Her options for fillings are turkey, tuna, ham, or egg salad. Her choices for condiments are mayonnaise, salad dressing, or mustard. How many different sandwich choices does Clarissa have?
A. 36
B. 6
C. 24
D. 12
Clarissa has two options for the bread: white or pumpernickel.
She has four options for the filling: turkey, tuna, ham, or egg salad.
She has three options for the condiment: mayonnaise, salad dressing, or mustard.
The total number of different sandwich choices that Clarissa can make is the product of the number of choices for each category:
Number of sandwich choices = Number of bread choices x Number of filling choices x Number of condiment choices
Number of sandwich choices = 2 x 4 x 3
Number of sandwich choices = 24
Therefore, Clarissa has 24 different sandwich choices.
Answer: C. 24.
Janelle wants to buy a shirt for a friend. Her choices of material are polyester and cotton. Shirts are available in yellow, orange, and blue. Make an organized list showing the different choices Janelle has.
A. clothing options
B. clothing options
C. clothing options
D. clothing options
Question 7 of 8
Shoe Bargin Corner
Types Colors
Walking
Basketball
Cross-Trainer
Badminton
Baseball Black
Blue
Red
White
Silver
You notice the sign above at a mall.
a. How many different type/color combinations does the store offer?
b. Suppose the store has equal numbers of each shoe. You select a shoe at random. What is the probability of selecting a baseball shoe?
A. 5; one over twenty-five
B. 25; one-fifth
C. 10; one over twenty-five
D. 10; one-fifth
Evidence for Common Characteristics Quick Check
1 of 41 of 4 Items
Question
Which of these statements about DNA is true?
(1 point)
Responses
All animals have DNA.
All animals have DNA.
Proteins signal the cells to make DNA.
Proteins signal the cells to make DNA.
DNA is a protein.
DNA is a protein.
DNA is too similar to use for grouping organisms.
DNA is too similar to use for grouping organisms.
Which characteristic is common among vertebrates?(1 point)
Responses
supported by an exoskeleton
supported by an exoskeleton
eat with teeth
eat with teeth
breathe with lungs or gills
breathe with lungs or gills
make own food
1.Sarah worked like Cinderella until all of her chores were done.
2." shape" and "affect"
3." tolerate" and "bear
4. an artist needs a canvas just like a pottor needs clay
5. antonyms
6.confidence
7.force and course
8. the night is able to move, has eyes, and evil nature
your welcome :)
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 red?
A. one-half
B. one-third
C. The fraction is 1 over 4.