At a small coffee shop, the distribution of the number of seconds it takes for a cashier to process an order is approximately no
1 answer:
Answer:
The correct option is;
A. 0.17
Step-by-step explanation:
The given information are;
The mean duration it takes a cashier to process an order, μ = 276 seconds
The standard deviation from the mean, σ = 38 seconds
The z-score of the order processing time of x = 240 seconds is given as follows;
Therefore;
The probability
P(z = -0.9474) = 0.17361
Therefore, the proportion of orders which are processed in less than 240 is approximately 0.17361 or 0.17 to two places of decimal.
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Since the number of bird decreases 3% per year then for year 1 there will be 29 100 birds left. This value is 97% of 30000. From this the number of birds for year 1 is N = 30000(1-0.03)
A. N = 30,000
B. To solve this we use the equation we established in letter A. We get the answer by substituting T with 20. N = 16,000. <span />
A. N = 30,000
B. To solve this we use the equation we established in letter A. We get the answer by substituting T with 20. N = 16,000. <span />
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.