Answer:
87.471
Step-by-step explanation:
Given that scores for test, quiz, final exam , average quizzes.
There are weight given as 55% for quiz, 30% for homework, 5% Final Exam 10% Tests:
Now to calculate the student final grade, we can use weighted average method.
Final exam average
67
90
44
29
100
100
100
75.71428571
Use the averages to find:
Score Weight Score*weight
Quizzes 90 55% 49.5
Homework 85 30% 25.5
Tests 75.71 10% 7.571
final exam 98 5% 4.9
total 87.471
Hence answer is 87.471
DUDE THESE WERE SO EASY I DID THIS IN 4TH GRADE YOYR TOO YOUNG OK OK
Answer: do L x W x H = a
Answer: The common number is 26.
Step-by-step explanation:
We know that the n-th term of a sequence is:
aₙ = 3*n^2 - 1
And the n-th term of another sequence is:
bₙ = 30 - n^2
Remember that in a sequence n is always an integer number.
We want to find a number that belongs to both sequences, then we want to find a pair of integers x and n, such that:
aₙ = bₓ
This is:
3*n^2 - 1 = 30 - x^2
Let's isolate one of the variables, i will isolate n.
3*n^2 = 30 - x^2 + 1 = 31 - x^2
n^2 = (31 - x^2)/3
n = √( (31 - x^2)/3)
Now we can try with different integer values of x, and see if n is also an integer.
if x = 1
n = √( (31 - 1^2)/3) = √10
We know that √10 is not an integer, so we need to try with another value of x.
if x = 2:
n = √( (31 - x^2)/3) = √(27/3) = √9 = 3
Then if we have x= 2, n is also an integer, n = 3.
Then we have:
a₃ = b₂
The common number between both sequences is:
a₃ = 3*(3)^2 - 1 = 26
b₂ = 30 - 2^2 = 26
Answer:
1) no
2)yes
3)yes
4)yes
Step-by-step explanation:
Answer:
a. CI=[128.79,146.41]
b. CI=[122.81,152.39]
c. As the confidence level increases, the interval becomes wider.
Step-by-step explanation:
a. -Given the sample mean is 137.6 and the standard deviation is 20.60.
-The confidence intervals can be constructed using the formula;
where:
is the sample standard deviation
is the s value of the desired confidence interval
we then calculate our confidence interval as:
![\bar X\pm z\frac{s}{\sqrt{n}}\\\\=137.60\pm z_{0.05/2}\times\frac{20.60}{\sqrt{21}}\\\\=137.60\pm1.960\times \frac{20.60}{\sqrt{21}}\\\\=137.60\pm8.8108\\\\\\=[128.789,146.411]](https://tex.z-dn.net/?f=%5Cbar%20X%5Cpm%20z%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%5C%5C%5C%5C%3D137.60%5Cpm%20z_%7B0.05%2F2%7D%5Ctimes%5Cfrac%7B20.60%7D%7B%5Csqrt%7B21%7D%7D%5C%5C%5C%5C%3D137.60%5Cpm1.960%5Ctimes%20%5Cfrac%7B20.60%7D%7B%5Csqrt%7B21%7D%7D%5C%5C%5C%5C%3D137.60%5Cpm8.8108%5C%5C%5C%5C%5C%5C%3D%5B128.789%2C146.411%5D)
Hence, the 95% confidence interval is between 128.79 and 146.41
b. -Given the sample mean is 137.6 and the standard deviation is 20.60.
-The confidence intervals can be constructed using the formula in a above;
![\bar X\pm z\frac{s}{\sqrt{n}}\\\\=137.60\pm z_{0.01/2}\times\frac{20.60}{\sqrt{21}}\\\\=137.60\pm3.291\times \frac{20.60}{\sqrt{21}}\\\\=137.60\pm 14.7940\\\\\\=[122.806,152.394]](https://tex.z-dn.net/?f=%5Cbar%20X%5Cpm%20z%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%5C%5C%5C%5C%3D137.60%5Cpm%20z_%7B0.01%2F2%7D%5Ctimes%5Cfrac%7B20.60%7D%7B%5Csqrt%7B21%7D%7D%5C%5C%5C%5C%3D137.60%5Cpm3.291%5Ctimes%20%5Cfrac%7B20.60%7D%7B%5Csqrt%7B21%7D%7D%5C%5C%5C%5C%3D137.60%5Cpm%2014.7940%5C%5C%5C%5C%5C%5C%3D%5B122.806%2C152.394%5D)
Hence, the variable's 99% confidence interval is between 122.81 and 152.39
c. -Increasing the confidence has an increasing effect on the margin of error.
-Since, the sample size is particularly small, a wider confidence interval is necessary to increase the margin of error.
-The 99% Confidence interval is the most appropriate to use in such a case.
