Answer:
See below
Step-by-step explanation:
The function
oscillates between -1 and 1, so the range is
.
Answer:
the axis of symmetry is X = 1
Step-by-step explanation:
make sure to ask if you need any further guidance.
The increasing order of the horizontal widths of their asymptote rectangles is dependent on the values gotten from y = ± x.
<h3>What is a Hyperbola?</h3>
This is defined as a two-branched open curve formed by the intersection of a plane perpendicular to the bases of a double cone.
The rectangular hyperbola has two asymptotes which are defined as y = ± x in this scenario.
Read more about Hyperbola here brainly.com/question/3351710
#SPJ1
.06*120=$7.20
.005*44=$0.22
Answer:
Bias for the estimator = -0.56
Mean Square Error for the estimator = 6.6311
Step-by-step explanation:
Given - A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and X2. The mean is estimated using the formula (3X1 + 4X2)/8.
To find - Determine the bias and the mean squared error for this estimator of the mean.
Proof -
Let us denote
X be a random variable such that X ~ N(mean = 4.5, SD = 7.6)
Now,
An estimate of mean, μ is suggested as

Now
Bias for the estimator = E(μ bar) - μ
= 
= 
= 
= 
= 
= 3.9375 - 4.5
= - 0.5625 ≈ -0.56
∴ we get
Bias for the estimator = -0.56
Now,
Mean Square Error for the estimator = E[(μ bar - μ)²]
= Var(μ bar) + [Bias(μ bar, μ)]²
= 
= 
= ![\frac{1}{64} ( [{3Var(X_{1}) + 4Var(X_{2})] }) + 0.3136](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B64%7D%20%28%20%5B%7B3Var%28X_%7B1%7D%29%20%2B%204Var%28X_%7B2%7D%29%5D%20%20%7D%29%20%2B%200.3136)
= ![\frac{1}{64} [{3(57.76) + 4(57.76)}] } + 0.3136](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B64%7D%20%5B%7B3%2857.76%29%20%2B%204%2857.76%29%7D%5D%20%20%7D%20%2B%200.3136)
= ![\frac{1}{64} [7(57.76)}] } + 0.3136](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B64%7D%20%5B7%2857.76%29%7D%5D%20%20%7D%20%2B%200.3136)
= ![\frac{1}{64} [404.32] } + 0.3136](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B64%7D%20%5B404.32%5D%20%20%7D%20%2B%200.3136)
= 
= 6.6311
∴ we get
Mean Square Error for the estimator = 6.6311