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Vladimir79 [104]
3 years ago
12

Por una calculadora y un cuaderno habríamos pagado, hace tres días, 10,80 €. El precio de la calcula-dora ha aumentado un 8%, y

el cuaderno tiene una rebaja del 10%. Con estas variaciones, los dos artículos nos cuestan 11,34 €. ¿Cuánto costaba cada uno de los artículos hace tres días?
Mathematics
1 answer:
Inessa05 [86]3 years ago
4 0

Answer:

La calculadora y el cuaderno costaba 9 y 1,8 euros hace tres días.

Step-by-step explanation:

A partir del enunciado podemos construir el sistema de ecuaciones lineales:

Hace tres días:

x+y = 10.80 (1)

Actualmente:

1.08\cdot x + 0.9\cdot y = 11.34 (2)

Donde:

x - Precio de la calculadora.

y - Precio del cuaderno.

Por métodos algebraicos, tenemos que el sistema de ecuaciones tiene la siguiente solución:

x = 9, y = 1.8

La calculadora y el cuaderno costaba 9 y 1,8 euros hace tres días.

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4 years ago
Write down the solution to the simultaneous equations.
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x = 5/2; y = 7

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2x + y = 12

y = 2x + 2

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4x + 2 = 12

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x = 5/2

y = 2x + 2

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Researchers studying the effect of diet on growth would like to know if a vegetarian diet affects the height of a child. Twelve
Eva8 [605]

Answer:

a)  t_{0.035},11=\pm2.201

b)  t=-2.963

c)  Reject\ H_0\ when\ \alpha=0.05

d)   t

   -2.963

Step-by-step explanation:

From the question we are told that:

Sample size n=12

Mean \=x=42.5

Standard deviation \sigma=3.8

Population mean \mu=45.75

Significance \alpha=0.05

 

Generally the hypothesis given by

H_0;\mu=45.75\\H_1:\neq =45.75

Generally the equation for test statistics is mathematically given by

t=\frac{\=x-\mu}{\sigma/\sqrt{n} }

t=\frac{42.5-45.75}{3.8/\sqrt{12} }

t=-2.963

Generally the Critical value is mathematically given by

t_{\alpha/2},d_t

\alpha=0.05 \\\alpha/2=0.025\\d_t=n-1=11

t_{0.035},11

From table

t_{0.035},11=\pm2.201

Therefore

t

-2.963

Reject\ H_0\ when\ \alpha=0.05

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3 years ago
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