The equation of the ellipse is required \frac{(x)^2}{16} +\frac{(y+4)^2}{25}=1
The required equation is
It can be seen that the major axis is parallel to the y axis.
The major axis points are
Subtracting the equations
The foci are
Subtracting the equations
2c=6
c=3
k+c=-1 implies that k=-1-c
k=-1-3
k=-4
<h3>What is the equation of the ellipse?</h3>So we get,
Therefore the equation is, .
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Answer:
A
Step-by-step explanation:
Since the triangle is right use Pythagoras' identity to find EF
The square on the hypotenuse is equal to the sum of the squares on the other two sides, that is
EF² + DF² = DE² ← substitute values
EF² + 12² = 18², that is
EF² + 144 = 324 ( subtract 144 from both sides )
EF² = 180 ( take the square root of both sides )
EF = → A
Answer:
For this case we know this:
We can find the standard error like this:
So then our random variable can be described as:
Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:
We can use the z score given by:
And using this we got this:
And we can find this probability using the Ti 84 on this way:
2nd> VARS> DISTR > normalcdf
And the code that we need to use for this case would be:
1-normalcdf(-1000, 3.04; 0;1)
Or equivalently we can use:
1-normalcdf(-1000, 0.4; 0.2;0.0658)
Step-by-step explanation:
We need to check if we can use the normal approximation:
We assume independence on each event and a random sampling method so we can conclude that we can use the normal approximation and then ,the population proportion have the following distribution :
For this case we know this:
We can find the standard error like this:
So then our random variable can be described as:
Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:
We can use the z score given by:
And using this we got this:
And we can find this probability using the Ti 84 on this way:
2nd> VARS> DISTR > normalcdf
And the code that we need to use for this case would be:
1-normalcdf(-1000, 3.04; 0;1)
Or equivalently we can use:
1-normalcdf(-1000, 0.4; 0.2;0.0658)