Answer:
Let us say the domain in the first case, has the numbers. And the co-domain has the students, .
Now for a relation to be a function, the input should have exactly one output, which is true in this case because each number is associated (picked up by) with only one student.
The second condition is that no element in the domain should be left without an output. This is taken care by the equal number of students and the cards. 25 cards and 25 students. And they pick exactly one card. So all the cards get picked.
Note that this function is one-one and onto in the sense that each input has different outputs and no element in the co domain is left without an image in the domain. Since this is an one-one onto function inverse should exist. If the inverse exists, then the domain and co domain can be interchanged. i.e., Students become the domain and the cards co-domain, exactly like Mario claimed. So, both are correct!
99-21=78
Answer:The floor lamp costs $78
Y = 2 ( X - 3 ) ^ 2 - 4 =-4
Answer:
The probability that a household has at least one of these appliances is 0.95
Step-by-step explanation:
Percentage of households having radios P(R) = 75% = 0.75
Percentage of households having electric irons P(I) = 65% = 0.65
Percentage of households having electric toasters P(T) = 55% = 0.55
Percentage of household having iron and radio P(I∩R) = 50% = 0.5
Percentage of household having radios and toasters P(R∩T) = 40% = 0.40
Percentage of household having iron and toasters P(I∩T) = 30% = 0.30
Percentage of household having all three P(I∩R∩T) = 20% = 0.20
Probability of households having at least one of the appliance can be calculated using the rule:
P(at least one of the three) = P(R) +P(I) + P(T) - P(I∩R) - P(R∩T) - P(I∩T) + P(I∩R∩T)
P(at least one of the three)=0.75 + 0.65 + 0.55 - 0.50 - 0.40 - 0.30 + 0.20 P(at least one of the three) = 0.95
The probability that a household has at least one of these appliances is 0.95
You write it as 3.16 I am pretty sure
