The answer is:
B.
rational numbers
Because if you perform subtraction in a set of rational numbers, the result you get would still be a rational number.
This does not apply to the other sets.
The answer is 21. 29^2 - 20^2 = 21^2.
Answer:
Only one solution.
Step-by-step explanation:
We have to solve for x from a linear single variable equation of x.
We have to determine the number of the solution of the given equation.
The equation is 
⇒ x + 12 = 8x - 2
⇒ 7x = 14
⇒ x = 2
So, there is only one solution of the given equation of x. (Answer)
Solve for x over the real numbers:
5 x^2 - 21 = 10 x
Subtract 10 x from both sides:
5 x^2 - 10 x - 21 = 0
Divide both sides by 5:
x^2 - 2 x - 21/5 = 0
Add 21/5 to both sides:
x^2 - 2 x = 21/5
Add 1 to both sides:
x^2 - 2 x + 1 = 26/5
Write the left hand side as a square:
(x - 1)^2 = 26/5
Take the square root of both sides:
x - 1 = sqrt(26/5) or x - 1 = -sqrt(26/5)
Add 1 to both sides:
x = 1 + sqrt(26/5) or x - 1 = -sqrt(26/5)
Add 1 to both sides:
Answer: x = 1 + sqrt(26/5) or x = 1 - sqrt(26/5)
Answer:
a) 0.06
b) 0.778
Step-by-step explanation:
Let's suppose a community of 100 families just to facilitate the calculation.
30% of the families own a dog
Dog = 30% of 100 = 30
20% of the families that own a dog also own a cat = 20% of 30 = 6
27% of all the families own a cat = 27% of 100 = 27
So, 6 families own a dog and a cat.
As 30 families own a dog, [30 - 6 =] 24 families own only dogs
As 27 families own a cat, [27 - 6 = ] 21 families own only cats
See picture attached.
a) What is the probability that a randomly selected family owns both a dog and a cat?
P(dog and cat) = dog ∩ cat/total = 6/100 = 0.06
b) What is the conditional probability that a randomly selected family doesn't own a dog given that it owns a cat?
So, only cat/total cat
P (not dog|cat) = 21/27 = 0.778
