3 years ago

How to get my answer

1 answer:

4 0

We know that rectangle EFGD has a width of 3. When we compare this 3 unit width to rectangle ABCD's width, we notice EFGC is 1/10 of ABCD. This means we are dividing by 10. ABCD's length happens to be 10. 10 divided by 10 is 1. EFGD's length is 1. Since we are solving perimeter, the expression would be 2l + 2w (w = 3, l = 1). 1 + 1 + 3 + 3 = 8 units.

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Suppose that W1 is a random variable with mean μ and variance σ21 and W2 is a random variable with mean μ and variance σ2. From

vivado [14]

Answer:

Step-by-step explanation:

The concept of variance in random variable is applied in solving for the value of c for the estimator cW1 + (1 − c)W2 to be most efficient. Appropriate differentiation of the estimator with respect to c will give the value of c when the result is at minimum.

The detailed analysis and step by step approach is as shown in the attachment.