A lot cost $120,000 the house cost 3 times more than a lot how much did the house cost

Mrs Stokes bought 3 packages of fruit juice. Each pkg has 2 rows of 6 boxes in each row. How many boxes of fruit juice did Mrs.

uysha [10]

Do 2x6=12×3 since there is 3 packages 2 rows and 6 in each row

4 0

2 years ago

Read 2 more answers

For each of the questions below, a histogram is described. Indicate in each case whether, in view of the Central Limit Theorem,

denis-greek [22]

Solution :

a). This histogram is not a bell shaped.

Here we cannot use the theorem of the central limit as the histogram is being plotted by using 365 samples data values of the one sample.

b). This histogram may be bell shaped approximately since the sample size is large.

Here there are 50 sample each having a size of 500 and there is 50 sample proportions. The histogram shows sampling distribution of the sample proportion. Thus the central limit theorem can be used.

The histogram is approximately bell shaped as there is 50 samples of each having 500 size and an average of 50. Thus the central limit theorem can be used.

c). The histogram is not a bell shaped.

In this case central limit theorem cannot be used as the histogram is plotted by using 100 sample data values of one sample.

d). The histogram is a bell shaped since the sample size is large.

In this case 200 samples of each having a size of 50 and a data values of 50. The histogram here shows a sampling distribution of the sample proportion. Thus the central limit theorem can be used.

In the second also the histogram is of bell shaped as the sample size is large.

There are 200 samples of each sample having a size of 40 and they have 40 data values , i.e. sum of rolls.

e). It is not a bell shaped.

The histogram here is plotted using 100 sample data values having one sample. Therefore we cannot use the central limit theorem.

6 0

1 year ago

which equation illustrates how the expression 1.15^t can be rewritten to approximate the equivalent monthly interest rate if the

denis23 [38]

sorry if this answer wrong

8 0

2 years ago

Luca had a large container of yogurt with 1 and StartFraction 5 Over 8 EndFraction pounds of yogurt left in it. If a serving of

lisov135 [29]

Answer:

4 servings

Step-by-step explanation:

Yogurt in the container = 1 5/8 pounds

A serving = 3/8 pounds

how many servings of yogurt are in the container?

Servings in the yogurt container = total yogurt in the container / yogurt per servings

= 1 5/8 ÷ 3/8

= 13/8 × 8/3

= 13/3

= 4.33

To the nearest whole serving

= 4 servings

7 0

2 years ago

Imaginá que tenés 125 dados cúbicos del mismo tamaño ¿Cuantos dados de altura tiene el cubo de mayor tamaño que podés armar apil

kumpel [21]

Answer:

(i) Debemos apilar 5 dados para construir el cubo de mayor tamaño.

(ii) Se necesita 121 dados cuadrados para formar el cuadrado con la mayor cantidad de dados posibles, quedando 4 dados sobrantes.

Step-by-step explanation:

(i) Sabemos por la Geometría Euclídea del Espacio que un cubo es un sólido regular con 6 caras cuadradas y longitudes iguales. Cada dado tiene un volumen de 1 dado cúbico y 125 dados dan un volumen total de 125 dados cúbicos.

El volumen de un cubo está dado por la siguiente fórmula:

$V = L^{3}$

Donde:

$L$ - Longitud de la arista, medida en dados.

$V$ - Volumen del cubo, medido en dados cúbicos.

Ahora, necesitamos despejar la longitud de la arista para calcular la altura máxima posible:

$L = \sqrt[3]{V}$

Dado que $V = 125\,dados^{3}$ , encontramos que la altura del cubo de mayor tamaño sería:

$L =\sqrt[3]{125\,dados^{3}}$

$L = 5\,dados$

Debemos apilar 5 dados para construir el cubo de mayor tamaño.

(ii) El área cuadrada formada por cubos está determinada por la siguiente fórmula:

$A = L^{2}$

Donde:

$L$ - Longitud de arista, medida en dados.

$A$ - Área, medida en dados cuadrados.

Puesto que la longitud de arista se basa en un conjunto discreto, esto es, el número de dados disponibles, debemos encontrar el valor máximo de $L$ tal que no supere 125 y de un área entera. Es decir:

$L \leq 125\,dados$

Si cada cubo tiene un área de 1 dado cuadrado, entonces un cuadrado conformado por 125 dados tiene un área total de 125 dados cuadrados. Entonces:

$L^{2}< 125\,dados^{2}$

Esto nos lleva a decir que:

$L < 11.180\,dados$

Entonces, la longitud máxima del cuadrado con la mayor cantidad de cubos posible es de 11 dados. El número total requerido de cubos es el cuadrado de esa cifra, es decir:

$n = (11\,dados)^{2}$

$n = 121\,dados$

Se necesita 121 dados cuadrados para formar el cuadrado con la mayor cantidad de dados posibles, quedando 4 dados sobrantes.

4 0

2 years ago