Answer:
The answer should be the range of the 7th grade data is greater than the range of the 8th grade data
Step-by-step explanation:
You can find the range of data by subtracting the lowest value from the largest value, looking at the chart:
7th Graders: Lowest is 10, Highest is 80
Range for 7th Graders is 70
8th Graders: Lowest is 15, Highest is 75
Range for 8th Graders is 60
Answer:
y = 5x-7
Step-by-step explanation:
y-y/x-x = slope
-12-3/-1-2 = -15/-3 = 5
next plug in slope and coordinates
y=mx+b
3=5(2)+b
solve for b
3=10+b
-7=b
Answer:
i think it's
x is greater than or equal to -8
Step-by-step explanation:
Answer:
[See Below]
Step-by-step explanation:
C.
The solution to this is x ≤ 4.
So then we look at the graphs and choose the correct one.
When the sign is ≤ it's like an arrow whatever way it's pointing it goes that way. In this case it'd go the other way even though the sign is ≤, because the solution is stating 4 is greater then or equal to so it can't be less than 4. And the only option of +4 and higher is C.
Answer:
a-bi
Step-by-step explanation:
If a quadratic equation lx^2+mx+n=0 has one imaginary root as a+bi then the other root is the conjugate of a+bi = a-bi
Because we have l, m and n are real numbers and they are the coefficients.
Sum of roots = a+bi + second root = -m/l
When -m/l is real because the ratio of two real numbers, left side also has to be real.
Since bi is one imaginary term already there other root should have -bi in it so that the sum becomes real.
i.e. other root will be of the form c-bi for some real c.
Now product of roots = (a+bi)(c-bi) = n/l
Since right side is real, left side also must be real.
i.e.imaginary part =0
bi(a-c) =0
Or a =c
i.e. other root c-bi = a-bi
Hence proved.