DOMAIN:
- Set of allowable inputs (set of 1st elements of the ordered pairs in relation)
- X point
- AKA the input
DEFINE INPUT:
A number that'll be used in a function rule to determine the value of the output
RANGE:
- Set of possible outputs (set of 2nd elements of the ordered pairs in relation)
- Y point
- AKA the output
DEFINE OUTPUT:
Result of applying a function rule to the value of ani input
ORDERED PAIRS:
(-1, 7): -1 is the domain & 7 is the range
(3, 5): 3 is the domain & 5 is the range
(4, 9): 4 is the domain & 9 is the range
(-6, -10): -6 is the domain & -10 is the range
(-2, 8): -2 is the domain & 8 is the range
Hope this helps you!!! :)
Answer:
Neither binomial nor normal distribution
Step-by-step explanation:
In binomial distribution Sumner of trials are fixed and there is only two outcomes either success or failure
But in this question there are no fixed trials and outcomes is not proper so this is not a binomial distribution.
In normal distribution there is information of mean and variance which is also not give in the question so it is also nit a normal distribution
So it is neither binomial nor normal distribution
Answer:
d≤48/2.15
Step-by-step explanation:
Step one:
Given data
Claudia spends $2.15 a day for lunch.
We are told that her balance is $48.00
let the number of days be d
Step two:
The maximum number of days she can eat is given as
d≤48/2.15
d≤22.33 days
Answer:
201
Step-by-step explanation:
Given that Erica and AAron,are using lottery system to decide who will wash dishes every night.
They put some red and blue power chips and draw each one. If same colour, Aaron will wash and if not same colours Erica will wash
If the game is to be fair, then both should have equal chances of opportunity for washing.
i.e. Probability for Erica washing = Prob of Aaron washing
i.e. P(different chips) = P(same colour chips)
Say there are m red colours and n blue colours.
Both are drawing at the same time.
Hence Prob (getting same colour) = (mC2+nC2)/(m+n)C2
Probfor different colour = mC1+nC1/(m+n)C2
The two would be equal is mC2 +nC2 = m+n
This is possible if mC2 =m and nC2 = n.
Or m = 2+1 =3 and n =3
That for a fair game we must have both colours to be 3.