C
.) The - 3 at the end of the function means the function is shifted down 3 units. This also shifts the asymptote down 3 units.
Answer:
<h2>The flat pay $25 represents the intercept</h2>
Step-by-step explanation:
To answer this question we need to first understand and compare it with the equation of straight line.
i.e
which is the equation of line
where m= slope
y= dependent variable
x= independent variable
c= intercept
Given

comparing both expression we can see that
25 corresponds to c which is the intercept
A)<span> How much would it cost to ship a package weighing 3.2 pounds? Explain how you arrived at your answer.
Answer:Shipping cost of 3.2 package is $4.13.
</span>We can't use the cost assigned to 3 lbs because the package is over 3 lbs but it is not over 4 lbs so we use the cost of 4lbs. (check the attachment for price)
b)<span>What type of graph is needed to represent the Media Mail shipping prices as a function of the weight of the books shipped? Explain your thinking.
Answer:Line graph </span>is needed to represent the Media Mail shipping prices as a function of the weight of the books shipped.
In line graph, x values represent the weight in pounds and the y values represent the cost.
c)<span>Graph the Media Mail shipping prices as a function of the weight of the books shipped.
Answer;When we g</span>raph the Media Mail shipping prices as a function of the weight of the books shipped, it will be;
f(x) = 2.69 + 0.48(x-1)
She earned $3 more per hour for gardening than for office work.
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)