The answer is no, the side lengths are not persevered. <span />
The answer for the given above is letter "a, -6r2s4t3". This is obtained by dividing 18 by -3 and subtracting the exponent below with that from above of the same variable. For r, subtract 2 from 4 to get 2. For s, subtract 1 from 5 to get 4 and lastly for t, subtract 3 from 6 to get 3.
Answer:
p ≥ 2
Step-by-step explanation:
10p – 8 ≥ 12
10p ≥ 20
p ≥ 2
Answer:
Look below.
Step-by-step explanation:
K(x)=x^2 isnt graphed but P(x)=x^2+n is. We could use this information to figure out what n is. In a quadratic function we always have the same points unless we are transforming it.
X Y
2 4
1 1
0 0
-1 1
-2 4
If we compare what the points were in the parent function to what is shown now we could see that the graph went down 6. From this information we can conclude that n must be -6.
Answer:
Step-by-step explanation:
We know that half of the students has two pets. The rest of the students make up the other half. So, we have 3 students + 2 students + 8 students = 13 students that make half of the sample population
That means total number of students being surveyed is 13+13=26 students
Then we work out the probability
P(One pet) = 8/26 = 4/13
P(Two pets) = 1/2
P(Three pets) = 3/26
P( Four pets) = 2/26 = 1/13
The probability distribution is shown in the table below. Let be the number of pets and is the probability of owning the number of pets