Slope=m=y/x=rise/run=(y1-y2)/(x1-x2)
Labeling the question; 0=x1, 32=y1, 100=x2, 212=y2
so just go ahead and fill the numbers in;
(32-212)/(0-100)=(-180/-100)=1.8
so therefore, the slope is at 1.8
Answer:
y = 1/12 (x − 5)²
Step-by-step explanation:
We can solve this graphically without doing calculations.
The y component of the focus is y = 3. Since this is above the directrix, we know this is an upward facing parabola, so it must have a positive coefficient. That narrows the possible answers to A and C.
The x component of the focus is x = 5. Since this is above the vertex, we know the x component of the vertex is also x = 5.
So the answer is A. y = 1/12 (x−5)².
But let's say this wasn't a multiple choice question and we needed to do calculations. The equation of a parabola is:
y = 1/(4p) (x − h)² + k
where (h, k) is the vertex and p is the distance from the vertex to the focus.
The vertex is halfway between the focus and the directrix. So p is half the difference of the y components:
p = (3 − (-3)) / 2
p = 3
k, the y component of the vertex, is the average:
k = (3 + (-3)) / 2
k = 0
And h, the x component of the vertex, is the same as the focus:
h = 5
So:
y = 1/(4×3) (x − 5)² + 0
y = 1/12 (x − 5)²
Answer:
-0.06 left-middle
0.28 right-middle
0.03 right-top
0.1 middle-top
0.12 middle-bottom
Step-by-step explanation:
column-row
The number of tests that it would take for the probability of committing at least one type I error to be at least 0.7 is 118 .
In the question ,
it is given that ,
the probability of committing at least , type I error is = 0.7
we have to find the number of tests ,
let the number of test be n ,
the above mentioned situation can be written as
1 - P(no type I error is committed) ≥ P(at least type I error is committed)
which is written as ,
1 - (1 - 0.01)ⁿ ≥ 0.7
-(0.99)ⁿ ≥ 0.7 - 1
(0.99)ⁿ ≤ 0.3
On further simplification ,
we get ,
n ≈ 118 .
Therefore , the number of tests are 118 .
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