Answer:
Yes. Have the same distribution.
Step-by-step explanation:
Given that X is a random day of the week, coded so that Monday is 1, Tuesday is 2, etc. (so X takes values 1, 2,..., 7, with equal probabilities).
Y is the next day after X
So Y can be written as X+1
But since highest value is 7, we can write
Y = X+1 mod 7
i.e. whenever X=7, Y= 8= 1 again.
Thus Y can take values as
Y 1 2 3 4 5 6 7
P(Y) = P(X+1) = 1/7
i.e. Y also has the same distribution as X
The definition of a coefficient is the number BEFORE the variable. So in 76x^2, the coefficient is 76.<span>
</span><span>
</span>
F(x) = 3x + 1
let f(x) = y
y = 3x + 1
y - 1 = 3x
3x = y - 1
x = (y - 1)/3
From y = f(x), x = f⁻¹(y)
<span>x = (y - 1)/3
</span>
f⁻¹(y) = (y - 1)/3
f⁻¹(7) = <span>(7 - 1)/3 = 6/3 = 2
</span>f⁻¹(7)<span> = 2</span>
Answer:
the answer is the top choice
Answer:
∠1 is 33°
∠2 is 57°
∠3 is 57°
∠4 is 33°
Step-by-step explanation:
First off, we already know that ∠2 is 57° because of alternate interior angles.
Second, it's important to know that rhombus' diagonals bisect each other; meaning they form 90° angles in the intersection. Another cool thing is that the diagonals bisect the existing angles in the rhombus. Therefore, 57° is just half of something.
Then, you basically just do some other pain-in-the-butt things after.
Since that ∠2 is just the bisected half from one existing angle, that means that ∠3 is just the other half; meaning that ∠3 is 57°, as well.
Next is to just find the missing angle ∠1. Since we already know ∠3 is 57°, we can just add that to the 90° that the diagonals formed at the intersection.
57° + 90° = 147°
180° - 147° = 33°
∠1 is 33°
Finally, since that ∠4 is just an alternate interior angle of ∠1, ∠4 is 33°, too.
