Answer:
A table divided into cells by category with counts for each category in each cell. For example, let's say you were counting the number of cars and trucks that drove down a road each day over a 5-day week. Your categories would be vehicle and day. You could summarise this as a frequency table:
Step-by-step explanation:
Answer:
El ancho del río es 59.9 metros.
Step-by-step explanation:
El ancho del río lo podemos calcular con la siguiente relación trigonométrica asumiendo que la torre forma un triángulo rectángulo con el río:

En donde:
CA: es el cateto adyacente = Altura de la torre = 28.2 m
CO: es el cateto opuesto = ancho del río =?
θ: es el ángulo adyacente a CA
Dado que el ángulo de depresión (25.2°) está ubicado fuera de la parte superior de la hipotenusa del triángulo que forma la torre con la orilla opuesta del río, debemos calcular el ángulo interno (θ) como sigue:

Ahora, el ancho del río es:

Por lo tanto, el ancho del río es 59.9 metros.
Espero que te sea de utilidad!
Answer:
8
Step-by-step explanation:
simplemente encuentre el total de los 8 pagos. Hay 8 pagos ya que habrá 8 retiros totales: (2 años) × (cuatro retiros por año) = 8 retiros.
Answer:
Hi there!
Your answer is: 150 people prefer that school assemblies be held on Friday afternoon
Step-by-step explanation:
3 out of 4 students means that 75% of students prefer assemblies on Friday afternoon!
To find 75% of 200:
200 = 100%
/100 to find 1%
2 people = 1%
Multiply it by 75 to determine 75%
2×75= 150
150people is 75% of 200 people!
I hope this makes sense! Let me know if you have questions!
There is a relationship between confidence interval and standard deviation:

Where

is the mean,

is standard deviation, and n is number of data points.
Every confidence interval has associated z value. This can be found online.
We need to find the standard deviation first:

When we do all the calculations we find that:

Now we can find confidence intervals:

We can see that as confidence interval increases so does the error margin. Z values accociated with each confidence intreval also get bigger as confidence interval increases.
Here is the link to the spreadsheet with standard deviation calculation:
https://docs.google.com/spreadsheets/d/1pnsJIrM_lmQKAGRJvduiHzjg9mYvLgpsCqCoGYvR5Us/edit?usp=sharing