Answer:
The 95% confidence interval estimate of the population mean rating for Miami is (6.0, 7.5).
Step-by-step explanation:
The (1 - <em>α</em>)% confidence interval for the population mean, when the population standard deviation is not provided is:
The sample selected is of size, <em>n</em> = 50.
The critical value of <em>t</em> for 95% confidence level and (<em>n</em> - 1) = 49 degrees of freedom is:
*Use a <em>t</em>-table.
Compute the sample mean and sample standard deviation as follows:
![\bar x=\frac{1}{n}\sum X=\frac{1}{50}\times [1+5+6+...+10]=6.76\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{49}\times 31.12}=2.552](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5Cfrac%7B1%7D%7Bn%7D%5Csum%20X%3D%5Cfrac%7B1%7D%7B50%7D%5Ctimes%20%5B1%2B5%2B6%2B...%2B10%5D%3D6.76%5C%5C%5C%5Cs%3D%5Csqrt%7B%5Cfrac%7B1%7D%7Bn-1%7D%5Csum%20%28x-%5Cbar%20x%29%5E%7B2%7D%7D%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B49%7D%5Ctimes%2031.12%7D%3D2.552)
Compute the 95% confidence interval estimate of the population mean rating for Miami as follows:
Thus, the 95% confidence interval estimate of the population mean rating for Miami is (6.0, 7.5).
Answer:
3 is the correct answer
u can find common ratio by dividing a specific term in GP by its preceding term. ...just as here
4/(4/3) = 3
also 12/4 = 3
also 36 /12 = 3
hence the common ratio for this GP is 3
hope it helped
Step-by-step explanation:
Answer:
see the explanation
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
m∠PQS+m∠SQR=m∠PQR ----> equation A (by Addition Angle Postulate)
we have that
m∠PQR=90° ----> equation B given problem (because is a right angle)
substitute equation B in equation A
m∠PQS+m∠SQR=90°
Remember that
Two angles re complementary is their sum is equal to 90 degrees (Definition of complementary angles)
therefore
m∠PQS and m∠SQR are complementary angles
When you have 3 choices for each of 6 spins, the number of possible "words" is
3^6 = 729
The number of permutations of 6 things that are 3 groups of 2 is
6!/(2!×2!×2!) = 720/8 = 90
A) The probability of a word containing two of each of the letters is 90/729 = 10/81
The number of permutations of 6 things from two groups of different sizes is
(2 and 4) : 6!/(2!×4!) = 15
(3 and 3) : 6!/(3!×3!) = 20
(4 and 2) : 15
(5 and 1) : 6
(6 and 0) : 1
B) The number of ways there can be at least 2 "a"s and no "b"s is
15 + 20 + 15 + 6 + 1 = 57
The probability of a word containing at least 2 "a"s and no "b"s is 57/729 = 19/243.
_____
These numbers were verified by listing all possibilities and actually counting the ones that met your requirements.
Answer: y=mx+b
Step-by-step explanation:
