the y axis is on the top and the x axis is acroos
:]]]]
Answer:
The value of t test statistics is -2.25.
Step-by-step explanation:
We are given that a manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 402 gram setting. It is believed that the machine is under filling the bags.
A 25 bag sample had a mean of 393 grams with a standard deviation of 20.
<u><em>Let </em></u>
<u><em> = mean filling of bags by the machine.</em></u>
So, Null Hypothesis,
:
402 gram {means that the the machine is not under filling the bags}
Alternate Hypothesis,
:
< 402 gram {means that the the machine is under filling the bags}
The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;
T.S. =
~ 
where,
= sample mean bag filling = 393 grams
s = sample standard deviation = 20 grams
n = sample of bags = 25
So, <u><em>test statistics</em></u> =
~ 
= -2.25
The value of t test statistics is -2.25.
Answer:
Para resolver correctamente el ejercicio debemos seguir los pasos que se muestran en la explicación.
Step-by-step explanation:
El ejercicio enunciado de manera completa es el siguiente:
7.Sabiendo que el segmento DE es paralelo a la base del triángulo, las medidas de los segmentos a y b son:
(la gráfica del ejercicio se guardo como archivo adjunto)
-a = 8 cm y b = 10
-a = 9 cm y b = 11
-Ninguna de las respuestas anteriores es correcta.
<em>SOLUCIÓN:</em>
<em>Pasos:</em>
-Primero debemos hallar el valor de a, la gráfica nos indica que debemos hacer,tenemos que restar 15 de 6; entonces:
a= 15cm -6cm=9 cm
-Ahora debemos aplicar el teorema de thales ;como los segmentos son proporcionales podemos establecer la siguiente relación de esa manera podemos hallar el valor del segmento b;
=
( para resolver debemos multiplicar en cruz)
(9x7)=6xb
63=6b (aquí debemos hallar el valor de b,para eso b pasa a dividir a 63)
= b
b= 10.5 cm
La respuesta correcta es: a= 9cm y b= 10.5 cm
(De las opciones de respuesta dadas en el ejercicio ,ninguna corresponde a la respuesta hallada)
We have to assume that the speed before being stuck was sufficient to get to the destination on time had there been no delay. Call that speed "s" in km/h.
Since 200 km is "halfway", the total distance must be 400 km.
time = distance / speed
total time = (time for first half) + (delay) + (time for second half)
400/s = 200/s + 1 + 200/(s+10) . . . .times are in hours, distances in km
200/s = 1 + 200/(s+10) . . . . . . . . . . subtract 200/s
200(s+10) = s(s+10) +200s . . . . . . .multiply by s(s+10)
0 = s² +10s - 2000 . . . . . . . . . . . . . .subtract the left side
(s+50)(s-40) = 0
Solutions are s = -50, s = 40
The speed of the bus before the traffic holdup was 40 km/h.