The value of C is 0.03030 if the f(x, y) = c(x 2y), x = 1, 2 and y =1, 2, 3 because the sum of the probability is always 1
<h3>What is a function?</h3>
It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.
We have:
f(x, y) = c(x+2y),
x = 1, 2 and y =1, 2, 3
f(1, 1) = C(1 + 2) = 3C
f(1, 2) = C(1 + 4) = 5C
f(1, 3) = C(1 + 6) = 7C
f(2, 1) = C(2 + 2) = 4C
f(2, 2) = C(2+ 4) = 6C
f(2, 3) = C(2 + 6) = 8C
We know the sum of the all probability is 1
3C+5C+7C+4C+6C+8C=1
33C =1
C = 0.03030
Thus, the value of C is 0.03030 if the f(x, y) = c(x 2y), x = 1, 2 and y =1, 2, 3 because the sum of the probability is always 1
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Answer:
Step-by-step explanation:
hi im new im not sure what to do but yea
Answer:
Step-by-step explanation:
You would find your answer by dividing
<em>19.50/3.25 = 6 family members over 3 years old</em>
<em />
A:amount of money invested in the account with 3% simple interest
B: amount of money invested in the account with 4.5% simple interest
a+b=25000 ——>a=25000-b
0.03a+0.045b=900
30a+45b=900000
2a+3b=60000
2(25000-b)+3b=60000
50000-2b+3b=60000
B=10000
A=25000-10000=15000
When choosing an interval to use for a frequency table, the low value and the high value of the dat is considered so that the interval refrects a true data size for all the intervals.
Given that a<span>
data set has values ranging from a low of 10 to a high of 52. Using the class limits of 10-19, 20-29, 30-39, 40-49 for a
frequency table will make the last interval not to refrect the true size of the other intervals. i.e. the last interval will be 50 - 52 which has a size of 3, different from the size of 10 the other intervals have.</span>