Answer:
The correct option is f.
Step-by-step explanation:
The (1 - <em>α</em>)% confidence interval for the population proportion is:
![CI=\hat p\pm z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}](https://tex.z-dn.net/?f=CI%3D%5Chat%20p%5Cpm%20z_%7B%5Calpha%2F2%7D%5Ccdot%5Csqrt%7B%5Cfrac%7B%5Chat%20p%281-%5Chat%20p%29%7D%7Bn%7D%7D)
It is provided that an experimenter flips a coin 100 times and gets 58 heads.
That is the sample proportion of heads is,
.
The critical value of <em>z </em> for 90% confidence level is, <em>z</em> = 1.645.
*Use a <em>z</em>-table.
Compute the 90% confidence interval for the probability of flipping a head with this coin as follows:
![CI=\hat p\pm z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}](https://tex.z-dn.net/?f=CI%3D%5Chat%20p%5Cpm%20z_%7B%5Calpha%2F2%7D%5Ccdot%5Csqrt%7B%5Cfrac%7B%5Chat%20p%281-%5Chat%20p%29%7D%7Bn%7D%7D)
![=0.58\pm 1.645\cdot\sqrt{\frac{0.58\times(1-0.58)}{100}}\\\\=0.58\pm 0.0812\\\\=(0.4988, 0.6612)\\\\\approx (0.499, 0.661)](https://tex.z-dn.net/?f=%3D0.58%5Cpm%201.645%5Ccdot%5Csqrt%7B%5Cfrac%7B0.58%5Ctimes%281-0.58%29%7D%7B100%7D%7D%5C%5C%5C%5C%3D0.58%5Cpm%200.0812%5C%5C%5C%5C%3D%280.4988%2C%200.6612%29%5C%5C%5C%5C%5Capprox%20%280.499%2C%200.661%29)
Thus, the 90% confidence interval for the probability of flipping a head with this coin is (0.499, 0.661).
The correct option is f.
Answer:
$11,991.60
Step-by-step explanation:
An appropriate formula is ...
A = P(1 +r/n)^(nt)
where r is the annual rate, n is the number of time per year interest is compounded, and t is the number of year. P is the principal invested.
Filling in the given numbers, we have ...
A = $2000(1 +0.12/12)^(12·15) = $2000(1.01^180) ≈ $11,991.60
The account balance after 15 years will be $11,991.60.
Step-by-step explanation:
the answer will be 122,653 in a bank account
Answer:
yes, you can form a triangle with the side lengths given
<span>The line will shift vertically down by 4 tons. </span>