7/50 = ??
Multiply that answer times 100o to find about how many are defective each day.
Tony works in a factory that produces 1,000 computers each day. When 50 computers were sampled, it was fount that 7 were defective. Estimate how many defective computers are made each day.
10 answers
I got 140.
Tony works in a factory that produces 1,000 computers each day. When 50 computers were sampled, it was found that 7 were defective. Estimate how many defective computers are made each day.
A. 140 computers
B. 7 computers
C. 169 computers
D. 265 computers
A. 140 computers
B. 7 computers
C. 169 computers
D. 265 computers
To estimate how many defective computers are made each day, we can set up a proportion:
defective/total = defective in sample/total in sample
Let d be the number of defective computers made each day. Then:
d/1000 = 7/50
We can cross-multiply to solve for d:
50d = 7000
d = 140
Therefore, the estimated number of defective computers made each day is 140, which is option A.
defective/total = defective in sample/total in sample
Let d be the number of defective computers made each day. Then:
d/1000 = 7/50
We can cross-multiply to solve for d:
50d = 7000
d = 140
Therefore, the estimated number of defective computers made each day is 140, which is option A.
The spinner is divided into equal parts. Use a tree diagram to find the probability that the spinner will land on a consonant both times if the spinner is spun twice.
A circle is divided equally into three sections.· One of the sections is labeled with an upper E.
· One of the sections is labeled with an upper U.
· One of the sections is labeled with an upper N.
· An arrow originating from the center of the circle is pointing at the section labeled with upper U.
A. one-ninth
B. one-third
C. start fraction 5 over 9 end fraction
D. three-fourths
A circle is divided equally into three sections.· One of the sections is labeled with an upper E.
· One of the sections is labeled with an upper U.
· One of the sections is labeled with an upper N.
· An arrow originating from the center of the circle is pointing at the section labeled with upper U.
A. one-ninth
B. one-third
C. start fraction 5 over 9 end fraction
D. three-fourths
Here is a tree diagram showing all the possible outcomes of spinning the spinner twice:
```
C V
/ \ / \
C V C V
/ \ / \ / \ / \
C CC VC VV CV VC
/ \ / \
C V C V
```
Each branch represents a possible outcome of spinning the spinner twice. The letters "C" and "V" represent consonant and vowel, respectively. To find the probability that the spinner will land on a consonant both times, we can multiply the probabilities along the branches that lead to two consonants:
P(consonant twice) = P(CC) + P(CV) + P(VC)
Note that we also include the possibility of getting a consonant-vowel or vowel-consonant combination, as long as one of the letters is a consonant.
From the tree diagram, we can see that:
- P(CC) = (2/3)(2/3) = 4/9, since the probability of getting a consonant on the first spin is 2/3, and the probability of getting another consonant on the second spin, given that the first spin was a consonant, is also 2/3.
- P(CV) = (2/3)(1/3) = 2/9, since the probability of getting a consonant on the first spin is 2/3, and the probability of getting a vowel on the second spin, given that the first spin was a consonant, is 1/3.
- P(VC) = (1/3)(2/3) = 2/9, since the probability of getting a vowel on the first spin is 1/3, and the probability of getting a consonant on the second spin, given that the first spin was a vowel, is 2/3.
So:
P(consonant twice) = 4/9 + 2/9 + 2/9 = 8/9
Therefore, the probability that the spinner will land on a consonant both times if the spinner is spun twice is 8/9, which is not one of the answer choices provided.
```
C V
/ \ / \
C V C V
/ \ / \ / \ / \
C CC VC VV CV VC
/ \ / \
C V C V
```
Each branch represents a possible outcome of spinning the spinner twice. The letters "C" and "V" represent consonant and vowel, respectively. To find the probability that the spinner will land on a consonant both times, we can multiply the probabilities along the branches that lead to two consonants:
P(consonant twice) = P(CC) + P(CV) + P(VC)
Note that we also include the possibility of getting a consonant-vowel or vowel-consonant combination, as long as one of the letters is a consonant.
From the tree diagram, we can see that:
- P(CC) = (2/3)(2/3) = 4/9, since the probability of getting a consonant on the first spin is 2/3, and the probability of getting another consonant on the second spin, given that the first spin was a consonant, is also 2/3.
- P(CV) = (2/3)(1/3) = 2/9, since the probability of getting a consonant on the first spin is 2/3, and the probability of getting a vowel on the second spin, given that the first spin was a consonant, is 1/3.
- P(VC) = (1/3)(2/3) = 2/9, since the probability of getting a vowel on the first spin is 1/3, and the probability of getting a consonant on the second spin, given that the first spin was a vowel, is 2/3.
So:
P(consonant twice) = 4/9 + 2/9 + 2/9 = 8/9
Therefore, the probability that the spinner will land on a consonant both times if the spinner is spun twice is 8/9, which is not one of the answer choices provided.
A yogurt shop offers 5 different flavors of frozen yogurt and 11 different toppings. How many choices are possible for a single serving of frozen yogurt with one topping?
A. 16 choices
B. 53 choices
C. 55 choices
D. 58 choices
A. 16 choices
B. 53 choices
C. 55 choices
D. 58 choices
There are 5 choices for the flavor of frozen yogurt and 11 choices for the topping. To find the total number of choices possible for a single serving with one topping, we can use the multiplication rule of counting:
total choices = (number of choices for frozen yogurt) x (number of choices for topping)
total choices = 5 x 11
total choices = 55
Therefore, there are 55 choices possible for a single serving of frozen yogurt with one topping, which is option C.
total choices = (number of choices for frozen yogurt) x (number of choices for topping)
total choices = 5 x 11
total choices = 55
Therefore, there are 55 choices possible for a single serving of frozen yogurt with one topping, which is option C.
Sammy likes to mix and match her 4 necklaces, 2 bracelets, and 3 hats. The colors are listed in the table. On Monday, she randomly picks a bracelet, a necklace, and a hat. What is the probability of Sammy choosing a red necklace and yellow bracelet?
Necklace Bracelet Hat
Red Red Silver
Green Yellow Yellow
Gold Green
Silver
A. one-half
B. one-fifth
C. one-ninth
D. one-eighth
Necklace Bracelet Hat
Red Red Silver
Green Yellow Yellow
Gold Green
Silver
A. one-half
B. one-fifth
C. one-ninth
D. one-eighth
Sammy has 4 choices for her necklace, 2 choices for her bracelet, and 3 choices for her hat. The total number of ways she could choose one of each item is:
4 x 2 x 3 = 24
Out of these 24 possible outcomes, there is only one outcome where Sammy chooses a red necklace and a yellow bracelet, since there is only one red necklace and one yellow bracelet. Therefore, the probability of her choosing a red necklace and yellow bracelet is:
1/24
Therefore, the answer is not one of the answer choices given.
4 x 2 x 3 = 24
Out of these 24 possible outcomes, there is only one outcome where Sammy chooses a red necklace and a yellow bracelet, since there is only one red necklace and one yellow bracelet. Therefore, the probability of her choosing a red necklace and yellow bracelet is:
1/24
Therefore, the answer is not one of the answer choices given.