Answer:
There are (63) combinations. The notation means "six choose three". Out of six items (flavors) choose three.
(nk)=n!k!(n−k)!.
(63)=6!3!3!.
Think of it this way. There are 6 ways to choose a flavor. Once you choose, there are 5 ways to choose the next. After that, there are 4 flavors left. which is 6!/3!=6⋅5⋅4⋅3⋅2⋅13⋅2⋅1=6⋅5⋅4=120.
But, you could have chosen {chocolate,vanilla,strawberry} and you get the same combination as {vanilla, strawberry, chocolate} so we have to divide by 3!=3⋅2⋅1=6 to account for the order of choosing.
So the number of combinations of flavors is (63)=1206=20.
<h3>Mark me a brainlist</h3>
You do the implcit differentation, then solve for y' and check where this is defined.
In your case: Differentiate implicitly: 2xy + x²y' - y² - x*2yy' = 0
Solve for y': y'(x²-2xy) +2xy - y² = 0
y' = (2xy-y²) / (x²-2xy)
Check where defined: y' is not defined if the denominator becomes zero, i.e.
x² - 2xy = 0 x(x - 2y) = 0
This has formal solutions x=0 and y=x/2. Now we check whether these values are possible for the initially given definition of y:
0^2*y - 0*y^2 =? 4 0 =? 4
This is impossible, hence the function is not defined for 0, and we can disregard this.
x^2*(x/2) - x(x/2)^2 =? 4 x^3/2 - x^3/4 = 4 x^3/4 = 4 x^3=16 x^3 = 16 x = cubicroot(16)
This is a possible value for y, so we have a point where y is defined, but not y'.
The solution to all of it is hence D - { cubicroot(16) }, where D is the domain of y (which nobody has asked for in this example :-).
(Actually, the check whether 0 is in D is superfluous: If you write as solution D - { 0, cubicroot(16) }, this is also correct - only it so happens that 0 is not in D, so the set difference cannot take it out of there ...).
If someone asks for that D, you have to solve the definition for y and find that domain - I don't know of any [general] way to find the domain without solving for the explicit function).
Answer:
<u>y = -3x + 8</u>
Step-by-step explanation:
<u>Given :-</u>
Point on line = (5, -7)
Slope (m) = -3
<u>To find :-</u>
Equation of line
<u>Solving :-</u>
Using point-slope form of equation,
y - y₁ = m(x - x₁)
y + 7 = -3(x - 5)
y + 7 = -3x + 15
<u>Solution :-</u>
<u>y = -3x + 8</u>
Answer:
(a) Circle Q is 9.4 units to the center of circle P
(b) Circle Q has a smaller radius
Step-by-step explanation:
Given


Solving (a): The distance between both
The equation of a circle is:

Where


P and Q can be rewritten as:


So, for P:


For Q:


The distance between them is:

Where:
--- 
--- 
So:





Solving (b): The radius;
In (a), we have:
--- circle P
--- circle Q
By comparison

<em>Hence, circle Q has a smaller radius</em>
Answer:
a) The level of significance is large because here z is not appropriate.
b) The true critical value is larger than z because the level of significance is large.
Step-by-step explanation:
a) Based on the question under consideration, the true level of significance is larger than .05 level of significance is large, the reason being that z is not appropriate in this case.
b) The true critical value is larger than that of the critical z, this is because the level of significance is large.
t* in P(T > t*) = 0.975 is more than z* = 1.96 in P(z> z*) =0.975