A school class went on a field trip to see a magician perform there were 17 females 20 males in the class the magicians randomly
2 answers:
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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.
Answer:
95% confidence interval for p, the true proportion of all women who will find success with this new treatment is [0.238 , 0.762].
Step-by-step explanation:
We are given that a pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome).
In the first stages of a clinical trial, it was successful for 7 out of the 14 women.
Firstly, the pivotal quantity for 95% confidence interval for the true proportion is given by;
P.Q. = ~ N(0,1)
where, = sample proportion of women who find success with this new treatment =
= 0.50
n = sample of women = 14
<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>
So, 95% confidence interval for the true proportion, p is ;
P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%
level of significance are -1.96 & 1.96}
P(-1.96 < < 1.96) = 0.95
P( <
<
) = 0.95
P( < p <
) = 0.95
<u />
<u>95% confidence interval for p</u> = [ ,
]
= [ ,
]
= [0.238 , 0.762]
Therefore, 95% confidence interval for p, the true proportion of all women who will find success with this new treatment is [0.238 , 0.762].