16x0 + 2x2 • y –1 Evaluate the expression for x = 2 and y = 4.
1 answer:
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A sample size of 60 is required.
We use the formula
We first find the z-score associated with this level of confidence:
Convert 99% to a decimal: 99/100 = 0.99
Subtract from 1: 1-0.99 = 0.01
Divide by 2: 0.01/2 = 0.005
Subtract from 1: 1-0.005 = 0.995
Using a z-table (http://www.z-table.com) we see that this value is equally distant from 2.57 and 2.58; therefore we will use 2.575:
We use the formula
We first find the z-score associated with this level of confidence:
Convert 99% to a decimal: 99/100 = 0.99
Subtract from 1: 1-0.99 = 0.01
Divide by 2: 0.01/2 = 0.005
Subtract from 1: 1-0.005 = 0.995
Using a z-table (http://www.z-table.com) we see that this value is equally distant from 2.57 and 2.58; therefore we will use 2.575:
Answer:
<h2>It must be shown that both j(k(x)) and k(j(x)) equal x</h2>Step-by-step explanation:
Given the function j(x) = 11.6 and k(x) =
, to show that both equality functions are true, all we need to show is that both j(k(x)) and k(j(x)) equal x,
For j(k(x));
j(k(x)) = j[(ln x/11.6)]
j[(ln (x/11.6)] = 11.6e^{ln (x/11.6)}
j[(ln x/11.6)] = 11.6(x/11.6) (exponential function will cancel out the natural logarithm)
j[(ln x/11.6)] = 11.6 * x/11.6
j[(ln x/11.6)] = x
Hence j[k(x)] = x
Similarly for k[j(x)];
k[j(x)] = k[11.6e^x]
k[11.6e^x] = ln (11.6e^x/11.6)
k[11.6e^x] = ln(e^x)
exponential function will cancel out the natural logarithm leaving x
k[11.6e^x] = x
Hence k[j(x)] = x
From the calculations above, it can be seen that j[k(x)] = k[j(x)] = x, this shows that the functions j(x) = 11.6 and k(x) =
are inverse functions.