A university health services physician is concerned about how much sleep freshman are getting. She asks a simple random sample of 50 students if they got at least 8 hours of sleep the previous night. Then she constructs a 95% confidence interval for the true proportion of all freshman college students who got at least 8 hours of sleep the previous night. (a) Explain what would happen to the width of the interval if the confidence level were decreased to 90%.
(b) Explain what you would expect to happen to the width of the interval if the sample size was increased to 200 students.
(c) After calculating the interval, the physician realizes that the sample was drawn from only the 70% of freshman who had turned in their health forms by the time they arrived on campus. Can she adjust the confidence interval to take this undercoverage into account? If so, how? If not, why not?

step-by-step explanation:

this is not possible to solve with the info given.   you need at least an one angle somewhere.

28

step-by-step explanation:

because that's the answer

we have three similar triangles, because each has a right angle and shares an angle.   let's write the angles in order: opposite to short leg, long leg, hypotenuse.

cab similar to bad similar to cbd

or as ratios,

ca: ab: cb = ba: ad: bd = cb: bd: cd

we also know

ac = ad + cd

(ad+cd): ab: cb = ba: ad: bd = cb: bd: cd

(ad+cd)/cb=cb/cd

we have cb=6, ad=5 and seek x=cd.

we reject the negative root and conclude x=4

answer: 4

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

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step-by-step explanation: