Answer and Step-by-step explanation:
According to the given situation, The r-value associated with the ordered pairs for the linear function is very nearest to zero which does not results in adequate presentation of the outcome.
A quadratic model could properly comprise of a combination of data, as the set of data has a turning point.
This result in data rises and falls which represents the graph of quadratic's graph
First, divide the original measurements with the scaled ones. You'll get the same answer meaning that is the scale used by the model.
1500 ft. ÷7.5 ft = 200
600 ft. ÷ 3 ft. = 200
To simply check, if you divide 1500 with 200, the answer is 7.5 and 600 divided by 200 is 3.
To find the scaled dimension of the tennis court with actual dimension of 120 ft by 60 ft, divide both values with the scale used in the model which is 200.
120 ft. ÷ 200 = 0.6 ft
60 ft. ÷ 200 = 0.3 ft
The dimension of the tennis court in the scaled model is 0.6 ft long and 0.3 ft. wide.
We are given that the elevator is descending, so the coefficient of t must be negative. The elevator's initial position is 500 feet above the ground, so this is a positive value of +500. Therefore the only choice that fits these is h(t) = -5t + 500.
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
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)
