We use the chi-square distribution when making inferences about a single population variance.
Short Description of Chi-Square Distribution
The continuous probability distribution known as the chi-square distribution. The number of degrees of freedom (k) a chi-square distribution has determines its shape. This type of sampling distribution has a variance of 2k and a mean equal to its number of degrees of freedom (k). The range is of a chi-square distribution is from 0 to ∞.
Variance plays a key role in the analysis of risk and uncertainty. The sample variance, an unbiased estimator of population variance, is expressed by the following formula of core statistic for a sample size 'n' and Y' as the sample mean:
S² = ∑(Yₓ - Y') / (n-1)
The formula, (n-1)S² / σ² has the central chi-square distribution as χ²ₙ₋₁. Here (n-1) represents the degrees of freedom.
Learn more about chi-square distribution here:
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Multiply 3 by both sides answer x=27/16 or in decimal 1.6875
If you need the geometry and trigonometry I'll post those. Instead I found an answer at this link: http://jwilson.coe.uga.edu/EMAT6450/Class%20Projects/Scarpelli/Scarpelli_MathematicsBaseballActivity.....
The distance from home plate to the pitcher's mound is 60.5 feet and from home plate to second base is <span>127.28 feet.
Pitcher's Mound to 2nd base = </span><span>
<span>
<span>
66.78</span> </span>f</span>eet
Given those distances, we KNOW a 50 foot sprinkler will NOT reach home plate and second base from the pitcher's mound.
I didn't figure out the pitcher's mound to 1st or to 3rd, since the question is already answered.
<h2>
Answer:</h2>
<h2>
Step-by-step explanation:</h2>
As the question states,
John's brother has Galactosemia which states that his parents were both the carriers.
Therefore, the chances for the John to have the disease is = 2/3
Now,
Martha's great-grandmother also had the disease that means her children definitely carried the disease means probability of 1.
Now, one of those children married with a person.
So,
Probability for the child to have disease will be = 1/2
Now, again the child's child (Martha) probability for having the disease is = 1/2.
Therefore,
<u>The total probability for Martha's first child to be diagnosed with Galactosemia will be,</u>
(Here, we assumed that the child has the disease therefore, the probability was taken to be = 1/4.)
<em><u>Hence, the probability for the first child to have Galactosemia is 
</u></em>
