Given that the hyperbola has a center at (0,0), and its vertices and foci are on y-axis. This, the equation of the hyperbola is of the form
x²/a²-y²/b²=-1 (a>0, b>0)
In the equation, vertices are (0, +/-b) .
Thus,
b=60
Foci (0,+/-√(a²+b²))
thus
√(a²+60²)=65
hence solving for a²
a²=65²-60²
a²=625
a²=25²
hence the equation is:
x²/25²-y²/60²=-1
Answer:
The data is skewed to the bottom and contains an outlier.
Step-by-step explanation:
1. Test for outlier
An outlier is a point that is more than 1.5IQR below Q1 or above Q3.
IQR = Q3 - Q1 = 74 - 51 = 23
1.5 IQR = 1.5 × 23 = 34.5
51 - 15 = 36 > 1.5IQR
The point at 15 is an outlier.
2. Test for normal distribution
The median is not in the middle of the box.
Rather, it cuts the box into two unequal parts, so the data does not have a normal distribution.
3. Test for skewness
The longer part is to the left of the median, so the data is skewed left.
Answer:
{3,6,9,0}
Step-by-step explanation:
The range is the set of all second elements of ordered pairs (y-coordinates).
Use the the double angle formula:
sin(2A)=2sin(A)cos(A)
substitute 2x for A, then
20sin(2x)cos(2x)=10(sin(2(2x))cos(2(2x))=10sin(4x)
Answer:
the right triangles are B,C,D