Answer:
a) Frequency table:
Category Frequency
Dog 4
Cat 3
Fish 2
Hamster 1
b) Relative frequencies of each animal type
- Dog: 4/10 = 0.4
- Cat: 3/10 = 0.3
- Fish: 2/10 = 0.2
- Hamster: 1/10 = 0.1
c) Popularity
- Dog is the most popular because it has the highest relative frequency.
- Hamster is the least popular because it has the lowest relative frequency.
Explanation:
<u>a. Make a frequency table for the results.</u>
There are four kind of pets: dog, cat, hamster, and fish.
A <em>frequency table </em>shows the number of items for each category (kind of pets).
Count the number of each kind of pet:
- Dog: 4
- Cat: 3
- Hamster: 1
- Fish: 2
- Total: 10
With that you build your frequency table.
Frequency table:
Category Frequency
Dog 4
Cat 3
Fish 2
Hamster 1
<u>b. Relative frequencies of each animal type.</u>
The relative frequency is how often an outcome appears divided by the total number of outmcomes.
Here the total number of outcomes is 10 (the ten pets).
So, calculate each relative frequency:
- Dog: 4/10 = 0.4
- Cat: 3/10 = 0.3
- Fish: 2/10 = 0.2
- Hamster: 1/10 = 0.1
An important feature of the relative frequency is that they must add up 1. Check:
- 0.4 + 0.3 + 0.2 + 0.1 = 1.
<u>c. Using the relative frequencies explain which animals are most and least popular.</u>
Popularity is determined by the frequency with each outcome is repeated. The most popular is the most repeated. The least popular is the least repeated.
- Dog is the most popular because it has the highest relative frequency.
- Hamster is the least popular because it has the lowest relative frequency.
Answer:
about 314.16 units squared
Step-by-step explanation:
The formula for area is A = πr^2. Plugging in r, the radius, we get A = π10^2. Evaluating π*10, we get A = [about] 314.16^2.
If the problem asks you to plug in 3.14 for pi, the answer is 314 units squared.
Answer: C. 150 km²
<u>Step-by-step explanation:</u>
Use similar triangle proportion to find x:
![\dfrac{x}{4.5}=\dfrac{15.3+7.4}{7.4}\\\\\\x=\dfrac{4.5(22.7)}{7.4}\\\\\\x=13.8\\\\\text{Notice that x is the diameter of the circle}\implies r=\dfrac{13.8}{2}=6.9\\\\\\\text{Find the area of the circle using }A=\pi r^2\\A=\pi (6.9)^2\\\\.\ =\pi (47.6)\\\\.\ =149.6\\\\.\ \approx150](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%7D%7B4.5%7D%3D%5Cdfrac%7B15.3%2B7.4%7D%7B7.4%7D%5C%5C%5C%5C%5C%5Cx%3D%5Cdfrac%7B4.5%2822.7%29%7D%7B7.4%7D%5C%5C%5C%5C%5C%5Cx%3D13.8%5C%5C%5C%5C%5Ctext%7BNotice%20that%20x%20is%20the%20diameter%20of%20the%20circle%7D%5Cimplies%20r%3D%5Cdfrac%7B13.8%7D%7B2%7D%3D6.9%5C%5C%5C%5C%5C%5C%5Ctext%7BFind%20the%20area%20of%20the%20circle%20using%20%7DA%3D%5Cpi%20r%5E2%5C%5CA%3D%5Cpi%20%286.9%29%5E2%5C%5C%5C%5C.%5C%20%3D%5Cpi%20%2847.6%29%5C%5C%5C%5C.%5C%20%3D149.6%5C%5C%5C%5C.%5C%20%5Capprox150)
The probability you're asked to find is
![\mathbb P(A\cup B)-\mathbb P(A\cap B)](https://tex.z-dn.net/?f=%5Cmathbb%20P%28A%5Ccup%20B%29-%5Cmathbb%20P%28A%5Ccap%20B%29)
where
![A\cup B](https://tex.z-dn.net/?f=A%5Ccup%20B)
is the event that either event occurs (A, B, or both), and
![A\cap B](https://tex.z-dn.net/?f=A%5Ccap%20B)
is the event that both events occur.
Recall that
![\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)-\mathbb P(A\cap B)](https://tex.z-dn.net/?f=%5Cmathbb%20P%28A%5Ccup%20B%29%3D%5Cmathbb%20P%28A%29%2B%5Cmathbb%20P%28B%29-%5Cmathbb%20P%28A%5Ccap%20B%29)
which means
![\mathbb P(A\cup B)-\mathbb P(A\cap B)=\mathbb P(A)+\mathbb P(B)-2\mathbb P(A\cap B)](https://tex.z-dn.net/?f=%5Cmathbb%20P%28A%5Ccup%20B%29-%5Cmathbb%20P%28A%5Ccap%20B%29%3D%5Cmathbb%20P%28A%29%2B%5Cmathbb%20P%28B%29-2%5Cmathbb%20P%28A%5Ccap%20B%29)
You're told that
![\mathbb P(A)=0.5](https://tex.z-dn.net/?f=%5Cmathbb%20P%28A%29%3D0.5)
,
![\mathbb P(B)=0.7](https://tex.z-dn.net/?f=%5Cmathbb%20P%28B%29%3D0.7)
, and (if I'm reading the diagram correctly)
![\mathbb P(A\cap B)=0.3](https://tex.z-dn.net/?f=%5Cmathbb%20P%28A%5Ccap%20B%29%3D0.3)
.
So,
![\mathbb P(A\cup B)-\mathbb P(A\cap B)=0.5+0.7-2\times0.3=0.6](https://tex.z-dn.net/?f=%5Cmathbb%20P%28A%5Ccup%20B%29-%5Cmathbb%20P%28A%5Ccap%20B%29%3D0.5%2B0.7-2%5Ctimes0.3%3D0.6)
Another way of seeing this is that the event A consists of the regions "A not B" and "A and B". So the probability that "A not B" occurs is
![\mathbb P(A\setminus B)=\mathbb P(A)-\mathbb P(A\cap B)=0.5-0.3=0.2](https://tex.z-dn.net/?f=%5Cmathbb%20P%28A%5Csetminus%20B%29%3D%5Cmathbb%20P%28A%29-%5Cmathbb%20P%28A%5Ccap%20B%29%3D0.5-0.3%3D0.2)
Similarly, B consists of "B not A" and "A and B", so you have
![\mathbb P(B\setminus A)=\mathbb P(B)-\mathbb P(A\cap B)=0.7-0.3=0.4](https://tex.z-dn.net/?f=%5Cmathbb%20P%28B%5Csetminus%20A%29%3D%5Cmathbb%20P%28B%29-%5Cmathbb%20P%28A%5Ccap%20B%29%3D0.7-0.3%3D0.4)
So the probability that A or B, but not both, occur is