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Mkey [24]
3 years ago
7

Flor tiene una papelería y vende los lápices en $12 y los bolígrafos en $20. ¿Qué expresión algebraica permite saber cuánto debe

cobrar Flor por 10 bolsas que contengan x cantidad de lápices y "y" cantidad de bolígrafos?
Mathematics
1 answer:
stiks02 [169]3 years ago
7 0

Answer:

c=120x+200y

Step-by-step explanation:

La expresión algebraica debería indicar que el valor total que debe cobrar Flor es igual a 10 que representa el número de bolsas por el resultado de la suma del precio de cada item por la cantidad, lo que se puede expresar de la siguiente forma:

c=10(12x+20y)

c=120x+200y

De acuerdo a esto, la expresión algebraica que permite saber cuánto debe cobrar Flor por 10 bolsas que contengan x cantidad de lápices y "y" cantidad de bolígrafos es c=120x+200y.

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For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to les
kaheart [24]

Answer:

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

The margin of error for this case is given by:

ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

And replacing we got:

ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259

And replacing into the confidence interval formula we got:

0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941

0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459

And the 95% confidence interval would be given (0.4941;0.5459).

Step-by-step explanation:

Data given and notation  

n=1000 represent the random sample taken    

\hat p=0.52 estimated proportion of of U.S. employers were likely to require higher employee contributions for health care coverage

\alpha=0.05 represent the significance level (no given, but is assumed)    

Solution to the problem

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

The margin of error for this case is given by:

ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

And replacing we got:

ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259

And replacing into the confidence interval formula we got:

0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941

0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459

And the 95% confidence interval would be given (0.4941;0.5459).

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