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2 years ago

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Sheila has x 20 dollar bills, and Tamsin has y 10 dollar bills. Sheila has a total of $50 more than Tamsin, but Tamsin has 1 mor

e bill than Sheila. How many bills does Sheila have?

1 answer:

natka813 [3]2 years ago

6 0

Answer:

Sheila have 6 bills

Step-by-step explanation:

Suppose Sheila has x '20' dollar bills, so the total cost of her bill = $20x

Suppose Tamsin has y '10' dollar bills, so the total cost of her bill = $10y

<h3>Equation 1</h3>

<em>Sheila has a total of $50 more than Tamsin</em>

<h2><em>10y + 50 = 20x</em></h2><h2 /><h3>Equation 2</h3>

<em>Tamsin has 1 more bill than Sheila.</em>

<h2>y = x + 1 </h2><h2 />

<em>Now solve them by putting value of x in equation </em>

10( x + 1 ) + 50 = 20x

10x + 10 + 50 = 20x

60 = 20x - 10x

10x = 60

x = 60/10

x = 6

and

y = 7

<em> </em>

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

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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.

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2 years ago

Allen is showing his work in simplifying –8.3 + 9.2 – 4.4 + 3.7. Identify and explain any errors in his work or in his reasoning

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Additive inverse; you can't just take -4.4 and make it positive.

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2 years ago

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Answer: B) 4.963±0.019.

Step-by-step explanation:

Confidence interval for population mean ( when population standard deviation is not given) is given by :-

$\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}$

, where $\overline{x}$ = Sample mean

n= Sample size

s= sample standard deviation

t* = critical t-value.

As per given:

n= 50

Degree of freedom = n-1 =49

$\overline{x}= 4.963\ lb$

s= 0.067 lb

For df = 49 and significance level of 0.05 , the critical two-tailed t-value ( from t-distribution table) is 2.010.

Now , substitute all values in the formula , we get

$4.963\pm (2.010)\dfrac{0.067}{\sqrt{50}}\\\\ 4.963\pm (2.010)(0.0094752)\\\\ 4.963\pm0.019045152\approx4.963\pm0.019$

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2 years ago

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Step-by-step explanation:

Given,

Cost of each adult ticket = $6.50

Cost of each child's ticket = $3.50

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Let,

x represent the number of adult tickets sold

y represent the number of child tickets sold

According to given statement;

x+y=548 Eqn 1

6.50x+3.50y=2881 Eqn 2

Multiplying Eqn 1 by 6.50

$6.50(x+y=548)\\6.50x+6.50y=3562\ \ \ Eqn\ 3$

Subtracting Eqn 2 from Eqn 3

$(6.50x+6.50y)-(6.50x-3.50y)=3562-2881\\6.50x+6.50y-6.50x-3.50y=681\\3y=681$

Dividing both sides by 3

$\frac{3y}{3}=\frac{681}{3}\\y=227$

Putting y=227 in Eqn 1

$x+227=548\\x=548-227\\x=321$

Regal theater sold 321 adult tickets and 227 children tickets.

Keywords: linear equation, elimination method

Learn more about elimination method at:

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2 years ago

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Answer:

The answer is explained below

Step-by-step explanation:

STEP 1

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STEP 2

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2 years ago

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