What is the sum of the areas of circle C and circle D?
1 answer:
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Answer: There are 5 grasshoppers, 15 crickets and 20 ladybugs
Step-by-step explanation:
The question in english is:
Jose has found several insects. There are 10 less grasshoppers than crickets. There are 5 less crickets than ladybugs. If Jose has found 5 grasshoppers, how many ladybugs did he find and how many crickets? Write two equations to solve the problem.
Let's tag grasshoppers with , crickets with
and ladybugs with
.
So, we are tolde there are 10 less grasshoppers than crickets:
(1) We have the first equation
Then, we are told there are 5 less crickets than ladybugs:
(2) This is the second equation
If and we substitute this in both equations we will have:
(3)
Isolating :
(4) There are 15 crickets
Substituting (4) in (2):
(5)
Isolating :
There are 20 ladybugs
Therefore:
There are 15 crickets and 20 ladybugs
Answer:
There is a 5.5% probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.
Step-by-step explanation:
To solve this problem, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
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:
The weights of a random sample of 48 commercial boat passengers were recorded. The sample mean was determined to be 177.6 pounds. Find the probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.
The probability of an extreme value below the mean.
This is the pvalue of Z when X = 177.6.
So
By the Central Limit Theorem
has a pvalue of 0.0274.
So there is a 2.74% of having a sample mean as extreme than that and lower than the mean.
The probability of an extrema value above the mean.
Measures above the mean have a positive z score.
So this probability is 1 subtracted by the pvalue of
has a pvalue of 0.9726.
So there is a 1-0.9726 = 0.0274 = 2.74% of having a sample mean as extreme than that and above than the mean.
Total:
2*0.0274 = 0.0548 = 0.055
There is a 5.5% probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.