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Anuta_ua [19.1K]
3 years ago
9

Carolina y su mamá compraron tres tortas de sabores diferentes para la reunión con sus amigos del colegio. De la torta de naranj

a se comieron 0,7 partes; de chocolate 9/12 partes y de vainilla 6/8 partes. Si todas las tortas tienen la misma forma y tamaño y cada torta la partieron en diferentes porciones, ¿ en total cuánta torta se comieron carolina y sus amigos?
Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
5 0

Answer: 2 1/5 partes  de torta

Step-by-step explanation:

Para contestar esta pregunta debemos sumar cada parte de torta:

0.7+9/12+6/8 =

Convertimos las fracciones en decimales, para tener todos los valores expresados de la misma forma. Dividiendo:

9/12=0.75

6/8=0.75

Sumamos :

0.7+0.75+0.75 = 2.2 partes = 2 1/5 partes

En conclusión, Carolina y sus amigos comieron 2 1/5 partes de torta.

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Each day a commuter takes a bus to work, the transportation system has a phone app that tells her what time the bus will arrive.
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Answer:

Step-by-step explanation:

Hello!

The commuter is interested in testing if the arrival time showed in the phone app is the same, or similar to the arrival time in real life.

For this, she piked 24 random times for 6 weeks and measured the difference between the actual arrival time and the app estimated time.

The established variable has a normal distribution with a standard deviation of σ= 2 min.

From the taken sample an average time difference of X[bar]= 0.77 was obtained.

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H₀:μ=0

H₁:μ≠0

b. This test is a one-sample test for the population mean. To be able to do it you need the study variable to be at least normal. It is informed in the test that the population is normal, so the variable "difference between actual arrival time and estimated arrival time" has a normal distribution and the population variance is known, so you can conduct the test using the standard normal distribution.

c.

Z_{H_0}= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } }

Z_{H_0}= \frac{0.77-0}{\frac{2}{\sqrt{24} } }= 1.89

d. This hypothesis test is two-tailed and so is the p-value.

p-value: P(Z≤-1.89)+P(Z≥1.89)= P(Z≤-1.89)+(1 - P(Z≤1.89))= 0.029 + (1 - 0.971)= 0.058

e. 90% CI

Z_{1-\alpha /2}= Z_{0.95}= 1.645

X[bar] ± Z_{1-\alpha /2}* (\frac{Sigma}{\sqrt{n} } )

0.77 ± 1.645 * (\frac{2}{\sqrt{24} } )

[0.098;1.442]

I hope this helps!

4 0
3 years ago
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