Step-by-step explanation:
You have to have tracing paper or patty paper (cause that's what my teacher uses). Next you draw the figure along with the x and y-axis on it as well. Then you mark the x and y-axis and rotate the paper 90 degrees clockwise. Lastly if the x-axis is on the y-axis, mark it as the y-axis and if the y-axis is on the x-axis, also mark it as the x-axis.
I hope this helped :)
Length: 2(x + 6); Width: 3.5x
A rectangle has 4 sides. 2 sides are lengths and 2 sides are widths.
The perimeter is the sum of the measures of all 4 sides.
perimeter = length + length + width + width
perimeter = 2(x + 6) + 2(x + 6) + 3.5x + 3.5x
Use the distributive property on 2(x + 6).
perimeter = 2x + 12 + 2x + 12 + 3.5x + 3.5x
Now let's group all terms with x first, then all the numbers.
perimeter = 2x + 2x + 3.5x + 3.5x + 12 + 12
Now we add like terms. Like terms have exactly the same variables and the same exponents. All terms with x are like terms and can be added together. All terms with no variable are like terms and can be added together.
perimeter = 11x + 24
11x and 24 are not like terms since 11x contains the variable x and 24 has no variable. Since 11x and 24 are not like terms, they cannot be added together. No more simplification can be done, and 11x + 24 is the answer.
Answer: 11x + 24
It will cost $90 and you will need 100 boards
Answer:
that confusing
Step-by-step explanation:
Answer:
Step-by-step explanation:
Hello!
The objective is to estimate the average time a student studies per week.
A sample of 8 students was taken and the time they spent studying in one week was recorded.
4.4, 5.2, 6.4, 6.8, 7.1, 7.3, 8.3, 8.4
n= 8
X[bar]= ∑X/n= 53.9/8= 6.7375 ≅ 6.74
S²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/7*[376.75-(53.9²)/8]= 1.94
S= 1.39
Assuming that the variable "weekly time a student spends studying" has a normal distribution, since the sample is small, the statistic to use to perform the estimation is the student's t, the formula for the interval is:
X[bar] ± 
* (S/√n)
6.74 ± 2.365 * (1.36/√8)
[5.6;7.88]
Using a confidence level of 95% you'd expect that the average time a student spends studying per week is contained by the interval [5.6;7.88]
I hope this helps!
