The given statement is:
An integer is divisible by 100 if and only if its last two digits are zeros
The two conditional statements that can be made are:
1) If an integer is divisible by 100 its last two digits are zeros.
This is a true statement. If a number is divisible by 100, it means 100 must be a factor of that number. When 100 will be multiplied by the remaining factors, the number will have last two digits zeros.
2) If the last two digits of an integer are zeros, it is divisible by 100.
This is also true. If last two digits are zeros, this means 100 is a factor of the integer. So the number will be divisible by 100.
Therefore, the two conditional statements that are formed are both true.
So, the option A is the correct answer.
Yes, it is. When the definition is separated into two conditional statements, both of the statements are true
The events A and B are not independent.
Given that,
P(A|B) = 0.4
P(A) = 0.5
P(B) = 0.8
Two events A and B are said to be independent, if P(A∩B) = P(A) P(B)
We have, P(A|B) = P(A∩B)/P(B)
Therefore, P(A∩B) = P(A|B) P(B) = 0.4 x 0.8 = 0.32
Now P(A) P(B) = 0.5 x 0.8 = 0.4
So P(A∩B) ≠ P(A)P(B)
Therefore, the events A and B are not independent.
Learn more about independent events at brainly.com/question/22881926
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Answer:
B
Step-by-step explanation:
cause it's got a 5.....5 x 5 = 25.......make sense
When taking square roots, you can't take square roots of negative roots of negative numbers. So, what will work for the domain of u(x) is what makes u(x) zero or more. We can make an inequality for that.
u(x) ≥ 0.
9x + 27 ≥ 0 by squaring both sides
9x ≥ -27
x ≥ -3
So the domain of the function is when x ≥ -3 is true.
Answer:
the cost of running the boarding
house for 600 students is N61,000
Step-by-step explanation:
Let C represents cost
K1 represents first constant
K2 represents second constant
C= k1+k2n
3500 = k1 + 25 k2............. Eqn(1)
6000= k1 + 50 k2 .............. Eqn(2)
Subtract eqn(1) from eqn(2)
2500= 25k2
K2= 2500/25
K2= 100
To get k1 from eqn(1)
3500 = k1 + 25 k2
Substitute the value of k2
3500 = k1 + 25 (100)
3500= k1 +2500
K1= 3500- 2500
K1= 1000
The equation connecting them;
C= 1000+ 100n
The cost of running the boarding
house for 600 students is
n= 600
C= 1000+ 100(600)
C= 1000+60000
C= N 61,0000
