Answer:
(a) false
(b) true
(c) true
(d) true
(e) false
(f) true
(g) false
(h) true
(i) true
Step-by-step explanation:
(a) 15 ⊂ A, since 15 is not a set, but an element, we cannot say of an element to be subset of a set. False
(b) {15} ⊂ A The subset {15} is a subset of A, since every element of {15}, that is 15, belongs to A.
15 ∈ {15} and 15 ∈ { x ∈ Z: x is an integer multiple of 3 } 15 is an integer multiple of 3. since 15/3=5. True
(c)∅ ⊂ A
∅ is a subset of any set. True
(d) A ⊆ A
A is a subset of itself. True
(e)∅ ∈ B
∅ is not an element, it is a subset, so it does not belong to any set. False
(f)A is an infinite set.
Yes, there are infinite integers multiple of 3. True
(g)B is a finite set.
No, there are infinite integers that are perfect squares. False
(h)|E| = 3
The number of elements that belong to E are 3. True
(i)|E| = |F|
The number of elements that belong to F are 3. So is the number of elements of E. True
Answer:
The answer is 150
Step-by-step explanation:
It's 150 because ye evening sales are three times of his afternoon sales.
SO that means you have to times 50x3=150 or you can add 50+50+50=150. there is your answer.
Step-by-step explanation:
every flip has 2 possible outcomes.
1 flip has 2.
2 flips have 2×2 = 2² = 4
3 flips have 2×2×2 = 2³ = 8
and so on.
the number of possible combinations flipping a coin n times is
C(n) = 2^n
for 10 flips
C(10) = 2¹⁰ = 1024
possible outcomes or combinations.
Answer:
<em>Given </em><em>x </em><em>=</em><em> </em><em>4</em><em> </em><em>and </em><em>y </em><em>=</em><em> </em><em>3</em>
Step-by-step explanation:
5x +2y -6 = <em>5</em><em>*</em><em>4</em><em> </em><em>+</em><em> </em><em>2</em><em>*</em><em>3</em><em> </em><em>-</em><em>6</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>2</em><em>0</em><em> </em><em>+</em><em> </em><em>6</em><em> </em><em>-</em><em> </em><em>6</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>2</em><em>0</em>
So I'm hoping you mean 6, or else the probability is 0. So the possible ways of getting a sum of 6 is 1&5, 2&4, 3&3, 4&2, 5&1. There are 36 different ways of rolling the dice, so divide the 5 possible ways by 36. Then, we need to add the possibility of getting at least one 6 in the roll. It doesn't matter what the other die is, so for each die, the probability is 1/6, and for both, the probability is 2/6. Add this to the probability of getting a sum of 6, and you get your answer.
