In this situation, it is similar to a coin flip, so think about flipping a 2 sided chip, one side colored yellow the other blue, flipping the chip will result in a 50/50 chance for it to land on either side, so no matter how many times you flip it it will always be a 1/2 chance of it being on one of the sides
There are two numbers whose sum is 64. The larger number subtracted from 4 times the smaller number gives 31. Then the numbers are 45 and 19
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Given that, There are two numbers whose sum is 64.
Let the number be a and b in which a is bigger.
Then, a + b = 64 ------ eqn (1)
The larger number subtracted from 4 times the smaller number gives 31.
4 x b – a = 31
4b – a = 31 ----- eqn (2)
We have to find the numbers.
So, from eqn (2)
a = 4b – 31
Subatitute a in (1)
4b – 31 + b = 64
On solving we get
5b = 64 + 31
5b = 95
b = 19
So, b = 19, then eqn 1
a + 19 = 64
On simplification,
a = 64 – 19
a = 45
Hence, the two numbers are 45 and 19
Answer: The given logical equivalence is proved below.
Step-by-step explanation: We are given to use truth tables to show the following logical equivalence :
P ⇔ Q ≡ (∼P ∨ Q)∧(∼Q ∨ P)
We know that
two compound propositions are said to be logically equivalent if they have same corresponding truth values in the truth table.
The truth table is as follows :
P Q ∼P ∼Q P⇔ Q ∼P ∨ Q ∼Q ∨ P (∼P ∨ Q)∧(∼Q ∨ P)
T T F F T T T T
T F F T F F T F
F T T F F T F F
F F T T T T T T
Since the corresponding truth vales for P ⇔ Q and (∼P ∨ Q)∧(∼Q ∨ P) are same, so the given propositions are logically equivalent.
Thus, P ⇔ Q ≡ (∼P ∨ Q)∧(∼Q ∨ P).
Answer:
A, B, C
Step-by-step explanation:
whole number: the set of counting numbers, 1, 2, 3, ..., and includes the number 0.
integer: all whole numbers, zero, and all the negatives of the whole numbers: ..., -3, -2, -1, 0, 1, 2, 3, ...
rational: any number that can be written as a fraction of integers
irrational number: a number that cannot be written as a fraction of integers
1.00 is the same as 1
It is:
A whole number
B integer
C rational
