To gather data involving their classmates, 50 fourth graders each filled out a personal information survey sheet. In the category of pets, 22 students had a cat, 17 students had a dog, and 9 students had both a dog and a cat. If a student is randomly selected from this class, what is the probability the student will have a dog or a cat

0.60

Step-by-step explanation:

In the category of pets

22 students had a cat = P(C)

17 students had a dog = P(D)

9 students had both a dog and a cat = P(C and D)

The formula to find the Probability of two events =

P(C or D) = P(C) + P(D) - P(C and D)

P(cat or dog) = P(cat) + P( dog) - P(cat and dog)

P(cat or dog) = 22/50 + 17/50 - 9/50

P(cat or dog) = 0.44 + 0.34 - 0.18

P(cat or dog) = 0.60

The probability the student will have a dog or a cat is 0.60

80π/3 or about 83.775

step-by-step explanation:

since a cone is 1/3 of a cylinder, we need to find the base's area. since area = πr^2, we get 16π. then, multiplying this by 5, we get 80π. dividing by 3, since the cone's volume is 1/3 of that of a cylinder's, we get 80π/3.

hope this !

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.