The answer is nine pounds of grapes.
X=0
Y=9
X=0
Y=9
sunita has to ribbons of lengths 25 inches and 35 inches he want to cut the Ribbon into strips of equal length what is the longe
1 answer:
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The complete question in the attached figure
we know that
find the square of the speeds of individual particles
2.8²----> 7.84
3.2²---->10.24
5.8²----> 33.64
7.3²----> 53.29
7.4²---> 54.76
average=[7.84+10.24+33.64+53.29+54.76]/5-----> 159.77/5----> 31.954
step 2
find the square root of the average
√31.954=5.65
therefore
the answer is
the option b) 5.7 m/s
we know that
the root mean square speed, is the square root of the average (mean) of all of the square of the speeds of individual particles in a gas.
step 1find the square of the speeds of individual particles
2.8²----> 7.84
3.2²---->10.24
5.8²----> 33.64
7.3²----> 53.29
7.4²---> 54.76
average=[7.84+10.24+33.64+53.29+54.76]/5-----> 159.77/5----> 31.954
step 2
find the square root of the average
√31.954=5.65
therefore
the answer is
the option b) 5.7 m/s
Answer:
The 90th percentile for the distribution of the total contributions is $6,342,525.
Step-by-step explanation:
To solve this question, 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 normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For sums of size n, the mean is and the standard deviation is
In this question:
The 90th percentile for the distribution of the total contributions
This is X when Z has a pvalue of 0.9. So it is X when Z = 1.28. Then
By the Central Limit Theorem
The 90th percentile for the distribution of the total contributions is $6,342,525.