Во една продавница има повеќе видови на сокови. 16 л пепси, 18 л фанта, 26 л кока кола и 10 л џус од портокал. Колкав дел од сок
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1 answer:
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Answer:
1000 times
Step-by-step explanation:
Given:
The Sun is roughly 10^2 times as wide as the Earth.
The Star KW Sagittarii is roughly 10^5 times as wide as the Earth.
Question asked:
About how many times as wide as the Sun is KW Sagittarii?
Solution:
Let the width of the earth =
As the Sun is roughly 10^2 times as wide as the Earth, hence the width of the sun =
And as the Star KW Sagittarii is roughly 10^5 times as wide as the Earth, hence the width of the Star =
Now, to find that how many times width of the Star KW Sagittarii is as respect to the width of the Sun, we will simply divide:
Width of the Star KW Sagittarii =
Width of the Sun =
x canceled by x
Therefore, Star KW Sagittarii is 1000 times wider than Sun.
<em>First of all we calculated width of Sun in terms of width of earth and then calculated the width of the Star in terms of earth and for comparison we did simple division that showed that the Star KW Sagittarii is 1000 times wider than the Sun.</em>
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Answer:
The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.
Step-by-step explanation:
To solve this problem, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are 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.
Central limit Theorem
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
In this problem, we have that:
What is the level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L ?
This is the value of X when Z has a pvalue of 1-0.01 = 0.99. So X when Z = 2.33.
By the Central Limit Theorem
The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.