Find the distance between two points (-5,8) and (19,53)
1 answer:
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Answer:
10.57% probability that the mean contents of the 625 sample cans is less than 9.995 ounces.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
What is the probability that the mean contents of the 625 sample cans is less than 9.995 ounces?
This is the pvalue of Z when X = 9.995. So
has a pvalue of 0.1057
So there is a 10.57% probability that the mean contents of the 625 sample cans is less than 9.995 ounces.
Answer:
Al estudiante le falta caminar de kilómetro.
Step-by-step explanation:
Un estudiante debe recorrer dos tercios de kilómetro para llegar al colegio. Dos tercios equivale a la fracción .
El estudiante avanza medio kilómetro en bus. Un medio equivale a la fracción , por lo tanto un medio es la mitad de una cantidad. Para calcular la distancia recorrida en bus se realiza la siguiente multiplicación:
Resolviendo:
Entonces, la distancia recorrida en bus es de kilómetro.
Para calcular la distancia recorrida a pie se calcula la diferencia entre la distancia total a recorrer por el estudiante y la distancia recorrida en bus. Esto es:
Resolviendo:
<u><em>Al estudiante le falta caminar </em></u><u><em> de kilómetro.</em></u>
Let's say we use "x" of the 25% OJ... ok... well, how much juice is really in that "x" liters? well, 25% of it is OJ, the rest is water, how much is 25% of x? well, (25/100) * x, or 0.25x.
likewise, we'll use "y" of the 10% OJ, and it will end up with (10/100) *y or 0.10y of juice in it.
whatever "x" and "y" are, they must add up to 15 Liters, and the juice concentration on that mixture, must also be the sum of their sum concentration.
how much will be it of the 10% juice? well, y = 15 - x.
likewise, we'll use "y" of the 10% OJ, and it will end up with (10/100) *y or 0.10y of juice in it.
whatever "x" and "y" are, they must add up to 15 Liters, and the juice concentration on that mixture, must also be the sum of their sum concentration.
how much will be it of the 10% juice? well, y = 15 - x.