Mathematics, 11.04.2021 04:40 abbie722

Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. Assume that the population has a normal distribution. n = 10, x=13.8, s =2.4, 95 percent

A. 12.11, 15.49
B. 12.08, 15.52
C. 12.09, 15.51
D. 12.41, 15.19

Answers: 2

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Answer from: Quest

The linear equation is: y=ax+b

Answer from: Quest

Ok here we go ..there are plenty of ways to solve such math problems but we will go for the smartest and easiest ..5x+2y=7 -2x+6y=9we rearrange them into the following ..by multiplying the 1st equation by 2 it becomes 10x + 4y = 14we rearrange it and so it becomes 10x = 14 - 4yas for the second equation we multiply it by 5 and rearrange it so it becomes as to the following ..-10x + 30y = 45rearranging so 10x = 30y -45now 10x equals both 14 - 4y and 30y -45so 14 - 4y = 30y -4531 = 26yy= 31/26 = 1.19 or around 1.2so by now we can calculate x easily as -2x+6y=9so 2x = 6y -9 = 6 * 31/26 - 9 = -1.84 then x = - 0.92 or around -0.9best luck!

Answer from: Quest

answer:

some responsibilities that come with having a checking account -

1. we have to maintain check register with exactness to avoid checks being returned for insufficient funds which is called bouncing a check.

2. we need to update our check register each time we make a transaction using a convenience card. updating check register will save us from overdrawing of checking account.

3. read the agreement the bank gave you when you signed up for our checking account. this includes all the fees and the minimum balance we need to keep in our account to avoid maintenance charges.

explanation:

note : - checking account is a transactional deposit account. we can use it to deposit and withdraw funds. money in a checking account is very liquid. it can be withdrawn using electronic debits, automated cash machines, or checks among other methods.

Answer from: Quest

let's first notice, the center of the hyperbola is at the origin, since the focus points are equidistantly 3 units away from the origin, that also means that the distance "c" from the center to either focus is 3.

it should also be notable that, the traverse axis for this hyperbola is vertical, or namely over the y-axis and therefore, the positive fraction will be the one with y-variable.

the traverse "a" component or distance from the center to either vertex is not 2, is more than 2, so we can safely put it at √5 ≈ 2.24.

$\bf \textit{hyperbolas, vertical traverse axis } \cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h, k\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} [-0.35em] \rule{34em}{0.25pt} \begin{cases} h=0\\ k=0\\ a=\sqrt{5}\\ c=3 \end{cases}\implies c=\sqrt{a^2+b^2}\implies 3=\sqrt{(\sqrt{5})^2+b^2}\implies 3^2=5+b^2$

$\bf 3^2-5=b^2\implies 9-5=b^2\implies 4=b^2\implies \sqrt{4}=b\implies \boxed{2=b} [-0.35em] ~\dotfill \cfrac{(y-0)^2}{(\sqrt{5})^2}-\cfrac{(x-0)^2}{2^2}=1\implies \cfrac{y^2}{5}-\cfrac{x^2}{4}=1$

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