1) We can find the magnitude of a vector(a.k.a. the norm) of a vector and the direction, by making use of the following formulas:
2) In this question, the magnitude and the direction of that vector have been given to us. So, let's do the other way around to identify which one has this magnitude and direction.
Note that since we want a positive value, we need to add 360 degrees.
To correctly solve x from the given equation,
3x - 3 = 6x + 1
We first subtract 3x from both sides of the equation and this will give us,
-3 = (6x - 3x) + 1
Simplifying,
-3 = 3x + 1
Then, subtract 1 again from both sides of the equation to give,
-4 = 3x
Lastly, divide both sides by 3 giving us a final answer for x which is x = -4/3.
Answer:
As shown in picture, this circle has radius 1.5 and passes (0, 1.5)
=> According to the general form of equation of circle that has radius r and passes (a, b): (x - a)^2 + (y - b)^2 = r^2, we have:
x^2 + (y - 1.5)^2 = 1.5^2
<=>
x^2 + (y - 1.5)^2 = 2.25
Hope this helps!
:)
Answer:
The third choice.
Step-by-step explanation:
This is a linear function.
The equation is y = (x - 5)(5x + 4) - 5x^2.
y = 5x^2 - 25x + 4x - 20 - 5x^2
y = 5x^2 - 5x^2 - 21x - 20
y = -21x - 20
That means this is a linear function, with a linear term of -21x and constant term of -20.
Hope this helps!
Answer:
The 99% confidence interval for the mean number of toys purchased each year is between 7.6 toys and 7.8 toys.
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
So it is z with a pvalue of , so
Now, find M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 7.7 - 0.1 = 7.6 toys
The upper end of the interval is the sample mean added to M. So it is 7.7 + 0.1 = 7.8 toys
The 99% confidence interval for the mean number of toys purchased each year is between 7.6 toys and 7.8 toys.