Answer:
a) The equation is:
Confidence Interval = Mean ± Z score × Standard deviation/√Number of samples
b) The 98% confidence interval = (5.62784, 6.37216)
Step-by-step explanation:
a. Write down the equation you should use to construct the confidence interval for the average number of days absent per term for all the children. (10 points)
Confidence Interval = Mean ± Z score × Standard deviation/√Number of samples
b. Determine a 98% confidence interval estimate for the average number of days absent per term for all the children. (10 points)
Confidence Interval = Mean ± Z score × Standard deviation/√Number of samples
Mean = 6 days
Standard deviation = 1.6 days
Number of samples = 100
Z score of 98% confidence interval = 2.326
Confidence interval = 6 ± 2.326 × 1.6/√100
= 6 ± 2.326 × 1.6/10
= 6 ± 0.37216
= 6 - 0.37216
= 5.62784
6 + 0.37216
= 6.37216
Therefore, the 98% confidence interval = (5.62784, 6.37216)
<span>We have to calculate the length of the missing side.
Use the Pythagorean theorem:
I think. Your answers to choice are wrong.
</span>
Check the picture below, so the hyperbola looks more or less like so, so let's find the length of the conjugate axis, or namely let's find the "b" component.
![\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Ctextit%7Bhyperbolas%2C%20horizontal%20traverse%20axis%20%7D%20%5C%5C%5C%5C%20%5Ccfrac%7B%28x-%20h%29%5E2%7D%7B%20a%5E2%7D-%5Ccfrac%7B%28y-%20k%29%5E2%7D%7B%20b%5E2%7D%3D1%20%5Cqquad%20%5Cbegin%7Bcases%7D%20center%5C%20%28%20h%2C%20k%29%5C%5C%20vertices%5C%20%28%20h%5Cpm%20a%2C%20k%29%5C%5C%20c%3D%5Ctextit%7Bdistance%20from%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bcenter%20to%20foci%7D%5C%5C%20%5Cqquad%20%5Csqrt%7B%20a%20%5E2%20%2B%20b%20%5E2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
