83.95% ? 83.95%
<
=
Submit
Answer: =
Step-by-step explanation:
83.95% is equal to 83.95% (look at it)
Answer:
=
Step-by-step explanation:
umm 83.95 = 83.95?
Answer:
a) For this case we can use the definition of weighted average given by:
And if we replace the values given we have:
b)
c)
Step-by-step explanation:
Assuming the following question: "One sample has a mean of M=8 and a second sample has a mean of M=16 . The two samples are combined into a single set of scores.
a) What is the mean for the combined set if both of the original samples have n=4 scores"
For this case we can use the definition of weighted average given by:
And if we replace the values given we have:
b) what is the mean for the combined set if the first sample has n=3 and the second sample has n=5
Using the definition we have:
c) what is the mean for the combined set if the first sample has n=5 and the second sample has n=3
Using the definition we have:
The required sum of -3x, 5x + y, and 7x - 8 - 3y in descending order is 9x - 2y - 8.
Simplification involves applying rules of arithmetic and algebra to remove unnecessary terms, factors, or operations from an expression. The goal is to obtain an expression that is easier to work with, manipulate, or solve.
Here,
To add -3x, 5x + y, and 7x - 8 - 3y, we can first combine like terms.
Group the x-terms together:
-3x + 5x + 7x = 9x
Group the y-terms together:
y - 3y = -2y
Group the constants together:
-8
Putting it all together, we get:
9x - 2y - 8
Therefore, the sum of -3x, 5x + y, and 7x - 8 - 3y in descending order is 9x - 2y - 8.
Learn more about simplification here:
#SPJ7
Answer:
The 80% confidence interval for the average net change in a student's score after completing the course is (15.4, 26.3).
Step-by-step explanation:
The net change in 7 students' scores on the exam after completing the course are:
S = {37 ,12 ,12 ,17 ,13 ,32 ,23}
Compute the sample mean and sample standard deviation as follows:
As the population standard deviation is not known, a t-interval will be formed.
Compute the critical value of t for 80% confidence interval and 6 degrees of freedom as follows:
*Use a t-table.
Compute the 80% confidence interval for the average net change in a student's score after completing the course as follows:
Thus, the 80% confidence interval for the average net change in a student's score after completing the course is (15.4, 26.3).
Answer:
In a quadrilateral ABCD, ∠A = 100°, ∠B = 105° and ∠C = 70°, find ∠D.
Solution:
Here the sum of the four angles
or, ∠A + ∠B + ∠C + ∠D = 360°
We know, ∠A = 100°, ∠B = 105° and ∠C = 70°
or, 100° + 105° + 70° + ∠D = 360°
or, 275° + ∠D = 360°
∠D = 360° - 275°
Therefore, = 85°
i hope this was helpful. <3
crown?