Jamie took 20 pieces of same-sized colored paper and put them in a hat. Four pieces were red, three pieces were blue, and the re
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Answer:
The cost is $9.70 per kilogram.
Step-by-step explanation:
This can be solved by a rule of three.
In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.
When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too. In this case, the rule of three is a cross multiplication.
When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease. In this case, the rule of three is a line multiplication.
In this problem, the measures are the weight of the cheese and the price. As the weight increases, so does the price. It means that this is a direct rule of three.
Solution:
The problem states that cheese costs $4.40 per pound. Each kg has 2.2 pounds. How many kg are there in 1 pound. So:
1 pound - xkg
2.2 pound - 1 kg
kg
Since cheese costs $4.40 per pound, and each pound has 0.45kg, cheese costs $4.40 per 0.45kg. How much does is cost for 1kg?
$4.40 - 0.45kg
$x - 1kg
The cost is $9.70 per kilogram.
Answer:
Each adult male has a 5.05% probability of having a hip width greater than 16.2 inches.
There is a 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.
Step-by-step explanation:
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
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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.
In this problem
Assume that adult men have hip breadths that are normally distributed with a mean of 14.4 inches and a standard deviation of 1.1 inches. This means that .
What is the probability that any one of those adult male will have a hip width greater than 16.2 inches?
For each one of these adult males, the probability that they have a hip width greater than 16.2 inches is 1 subtracted by the pvalue of Z when X = 16.2. So:
has a pvalue of 0.9495.
This means that each male has a 1-0.9495 = 0.0505 = 5.05% probability of having a hip width greater than 16.2 inches.
For the average of the sample
What is the probability that the 110 adult men will have an average hip width greater than 16.2 inches?
Now, we need to find the standard deviation of the sample before using the zscore formula. That is:
.
Now
has a pvalue of 0.9999.
This means that there is a 1-0.9999 = 0.0001 = 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.