Answer: The value of x in trapezoid ABCD is 15
Step-by-step explanation: The trapezoid as described in the question has two bases which are AB and DC and these are parallel. Also it has sides AD and BC described as congruent (that is, equal in length or measurement). These descriptions makes trapezoid ABCD an isosceles trapezoid.
One of the properties of an isosceles trapezoid is that the angles on either side of the two bases are equal. Since line AD is equal to line BC, then angle D is equal to angle C. It also implies that angle A is equal to angle B.
With that bit of information we can conclude that the angles in the trapezoid are identified as 3x, 3x, 9x and 9x.
Also the sum of angles in a quadrilateral equals 360. We can now express this as follows;
3x + 3x + 9x + 9x = 360
24x = 360
Divide both sides of the equation by 24
x = 15
Therefore, in trapezoid ABCD
x = 15
Answer:
3 miles
Step-by-step explanation:
First, we find the rate at which she drives, that is her speed.
Speed is given as:
speed = distance / time
When Cara drives to work, it takes her 30 minutes to drive 15 miles. Her speed is:
s = 15 / 30 = 0.5 miles per minute.
She drives at the same rate to The Dreamy Donut Shop and it takes her 6 minutes.
From the formula of speed, distance is given as:
distance = speed * time
Therefore, the distance of The Dreamy Donut Shop is:
distance = 0.5 * 6 = 3 miles
The Dreamy Donut Shop is 3 miles away.
<span>The expression given (27x^2)z/(-3x^2)(z^6), can be simplify as below:
By properties of powers, if you have the same base, you can substract the exponents. Then:
=(</span>27x^2)z/(-3x^2)(z^6)
=-9z/z^6 (As you can see: x^2-2= x^0=1)
=-9/z^5 (Then, z^6-1=z^5)
<span>
Therefore, the answer is: The exponent on the variable z is 5.</span>
Answer:
the expected value of Xn , E(Xn) = 0 and the variance σ²(Xn) = n*(1-2n)
Step-by-step explanation:
If X1= number of tails when n fair coins are flipped , then X1 follows a binomial distribution with E(X1) = n*p , p=0,5 and the number of heads obtained is X2=n-X1
therefore
Xn =X1-X2 = X1- (n-X1) = 2X1-n
thus
E(Xn) =∑ (2*X1-n) p(X1) = 2*∑[X1 p(X1)] -n∑p(X1) = 2*E(X1)-n = 2*n*p--n= 2*n*1/2 -n = n-n =0
the variance will be
σ²(Xn) = ∑ [Xn - E(Xn)]² p(Xn) = ∑ [(2X1-n) - 0 ]² p(X1) = ∑ (4*X1²-4*X1*n+n²) p(X1) = = 4*∑ X1²p(X1) - 4n ∑X1 p(X1) - n²∑p(X1) = 2*E(X1²) -4n*E(X1)- n²
since
σ²(X1) = n*p*(1-p) = n*0,5*0,5=n/4
and
σ²(X1) = E(X1²) - [E(X1)]²
n/4 = E(X1²) - (n/2)²
E(X1²) = n(n+1)/4
therefore
σ²(Xn) = 4*E(X1²) -4n*E(X1)- n² = 4*n(n+1)/4 - 4*n*n/2 - n² = n(n+1) - 2n² - n²
= n - 2n² = n(1-2n)
σ²(Xn) = n(1-2n)
I think it’s 1.7444921e+26
