How hot is the air in the top (crown) of a hot air balloon? Information from Ballooning: The Complete Guide to Riding the Winds, by Wirth and Young (Random House), claims that the air in the crown should be an average of 100°C for a balloon to be in a state of equilibrium. However, the temperature does not need to be exactly 100°C. What is a reasonable and safe range of temperatures? This range may vary with the size and (decorative) shape of the balloon. All balloons have a temperature gauge in the crown. Suppose that 55 readings (for a balloon in equilibrium) gave a mean temperature of x = 97°C. For this balloon, σ ≈ 22°C.(a) Compute a 95% confidence interval for the average temperature at which this balloon will be in a steady-state equilibrium. (Round your answers to one decimal place.)lower limit °Cupper limit °C

i believe the answer is letter c) 8

step-by-step explanation:

the slope of the line is 60

one point on the line is (40,10)

step-by-step explanation:

in the equation outside the parenthesis the equation always tells you the slope. in this case it is 60.

in the parenthesis is also says 40 which is x because it is near the x and 10 is y because it is near y.

answer: ay  

fonsi  

dy  

oh

oh no, oh no

oh yeah

diridiri, dirididi daddy  

go

sí, sabes que ya llevo un rato mirándote  

tengo que bailar contigo hoy (dy)  

vi que tu mirada ya estaba llamándome  

muéstrame el camino que yo voy (oh)

tú, tú eres el imán y yo soy el metal  

me voy acercando y voy armando el plan  

solo con pensarlo se acelera el pulso (oh yeah)

ya, ya me está gustando más de lo normal  

todos mis sentidos van pidiendo más  

esto hay que tomarlo sin ningún apuro

despacito  

quiero respirar tu cuello despacito  

deja que te diga cosas al oído  

para que te acuerdes si no estás conmigo

despacito  

quiero desnudarte a besos despacito  

firmo en las paredes de tu laberinto  

y hacer de tu cuerpo todo un manuscrito (sube, sube, sube)

(sube, sube)

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)