Answer:
Step-by-step explanation:
The imaginary number 'i' is equal to the square root of -1. If you were to square the square root, they both cancel and you get -1.
Please help real quick, least to greatst, math
2 answers:
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Complete Question:
300 students in a high school freshman class are surveyed about what kinds of pets they have. Of the 300 students, 200 have a dog, 180 have a cat, and 150 have a cat and a dog. Using this information, answer each of the following questions, Let D be the event that a randomly selected student has a dog and C be the event that a randomly selected student has a cat.
a . What is P(D), the probability that a student in the class has a dog?
b. What is P(C), the probability that a student in the class has a cat?
c. What is P(D and C), the probability that a student in the class has a dog and a cat?
d. What is P(D or C), the probability that a student in the class has a dog or a cat?
Answer:
a. 2/3
b) 3/5
c) 1/2
d) 19/ 15 = 1 4/15
Step-by-step explanation:
The formula for probability =
Number of possible outcomes ÷ Number of events.
Total number of Students = 300
Number of Students that have a dog = 200
Number of Students that have a cat = 180
Number of Students that have a dog and cat = 150
a. What is P(D), the probability that a student in the class has a dog?
In the question, the given data is interpreted as :
Number of possible outcomes = 200
Number of events = 300
P(D) = 200/300
= 2/3
Therefore, the probability that a student has a dog = 2/3
b. What is P(C), the probability that a student in the class has a cat?
In the question, the given data is interpreted as :
Number of possible outcomes = 180
Number of events = 300
P(D) = 180/300
= 3/5
Therefore, the probability that a student has a cat = 3/5
c. What is P(D and C), the probability that a student in the class has a dog and a cat?
In the question, the given data is interpreted as :
Number of possible outcomes = 150
Number of events = 300
P(D) = 150/300
= 1/2
Therefore, the probability that a student has a dog and cat = 1/2
d. What is P(D or C), the probability that a student in the class has a dog or a cat?
The is calculated as
P(D) + P(C)
In the above question,
P(D) is calculated as 2/3
P(C) is calculated as 3/5
Therfore, P(D or C), the probability that a student in the class has a dog or a cat is calculated as:
= 2/3 +3/5
= 19/15 = 1 4/15
<h2>Answer:</h2>
The correct answer is: Option: A
A. 85, 78, 80, 108, 46, 66, 68, 82, 72, 68
<h2>Step-by-step explanation:</h2>It is given that:
The whisker ranges from 66 to 85.
The box ranges from 68 to 82 with the vertical bar inside the box at 75.
- Hence, the minimum value of the box plot is: 66
- Maximum value of box plot is: 85
Also,
- The first quartile or the lower quartile i.e.
is: 68
- The middle quartile or median i.e.
is: 75
- The upper quartile or third quartile i.e.
is: 82
Also, it is given that:
One dot mark above forty-six.
One dot mark above one hundred and eight.
A)
On arranging the data in increasing order after removing outliers 46 and 108 we have:
66 68 68 72 78 80 82 85
Hence, from this data we have:
Minimum value=66
maximum value=85
As all the values matches the value defined.
Hence, this option is correct.
B)
On arranging the data in increasing order after removing 46 and 108 we have:
66 68 68 70 70 80 82 85
Hence, from this data we have:
Minimum value=66
maximum value=85
As the value of .
Hence, this option is incorrect.
C)
On arranging the data in increasing order after removing 46 and 108 we have:
66 68 72 78 80 80 82 85
Hence, from this data we have:
Minimum value=66
maximum value=85
As the value of do not match.
Hence, this option is incorrect.
D)
On arranging the data in increasing order after removing 46 and 108 we have:
66 68 68 72 78 80 80 85
Hence, from this data we have:
Minimum value=66
maximum value=85
As the value of do not match.
Hence, this option is incorrect.