Answer:
For the critical value we need to calculate the degrees of freedom given by:
And since we have a one tailed test we need to look in the t distribution with 9 degrees of freedom a quantile who accumulates 0.05 of the area on a tail and we got:
Step-by-step explanation:
Previous concepts
A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.
Let put some notation
x=test value with right arm , y = test value with left arm
The system of hypothesis for this case are:
Null hypothesis: 
Alternative hypothesis: 
The first step is calculate the difference 
The second step is calculate the mean difference
The third step would be calculate the standard deviation for the differences, and we got:
The 4 step is calculate the statistic given by :
For the critical value we need to calculate the degrees of freedom given by:
And since we have a one tailed test we need to look in the t distribution with 9 degrees of freedom a quantile who accumulates 0.05 of the area on a tail and we got:
Hello!
To find this ordered pair, we solve for x and y and put them together in an ordered pair.
-4+y=8
x-5y=17
Let's solve the first equation for y.
y= 12
Now, let's plug 12 into the second equation for y.
x-5(12)=17
x-60=17
x=77
Therefore, our ordered pair is (77,12)
I hope this helps!
Answer: y = 2x
Step-by-step explanation:
The equation of a straight line can be represented in the slope intercept form as
y = mx + c
Where
m = slope = (change in the value of y in the y axis) / (change in the value of x in the x axis)
The equation of the given line is
y = 2x - 8
Comparing with the slope intercept form, slope = 2
If two lines are parallel, it means that they have the same slope. Therefore, the slope of the line passing through (- 3,-6) is 2
To determine the intercept, we would substitute m = 2, x = - 3 and
y = - 6 into y = mx + c. It becomes
- 6 = 2 × - 3 + c
- 6 = - 6 + c
c = - 6 + 6 = 0
The equation becomes
y = 2x
Answer: ![3ab\sqrt[3]{b^4}](https://tex.z-dn.net/?f=3ab%5Csqrt%5B3%5D%7Bb%5E4%7D)
Step-by-step explanation:
Given the following expression:
![\sqrt[3]{27a^3b^7}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27a%5E3b%5E7%7D)
You need to apply the Product of powers property, which states that:
Then, you can rewrite the expression as following:
![=\sqrt[3]{27a^3b^4b^3}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B27a%5E3b%5E4b%5E3%7D)
The next step is to descompose 27 into its prime factors:
Now you must substitute 
inside the given root. Then:
![=\sqrt[3]{3^3a^3b^4b^3}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B3%5E3a%5E3b%5E4b%5E3%7D)
You need to remember that, according to Radicals properties:
![\sqrt[n]{a^n}=a^{\frac{n}{n}}=a^1=a](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5En%7D%3Da%5E%7B%5Cfrac%7Bn%7D%7Bn%7D%7D%3Da%5E1%3Da)
Therefore, the final step is to apply this property in order to finally get the expression is its simplest form. This is:
![=3^{\frac{3}{3}}a^{\frac{3}{3}}b^{\frac{4}{3}}b^{\frac{3}{3}}=3ab^{\frac{4}{3}}b=3ab\sqrt[3]{b^4}](https://tex.z-dn.net/?f=%3D3%5E%7B%5Cfrac%7B3%7D%7B3%7D%7Da%5E%7B%5Cfrac%7B3%7D%7B3%7D%7Db%5E%7B%5Cfrac%7B4%7D%7B3%7D%7Db%5E%7B%5Cfrac%7B3%7D%7B3%7D%7D%3D3ab%5E%7B%5Cfrac%7B4%7D%7B3%7D%7Db%3D3ab%5Csqrt%5B3%5D%7Bb%5E4%7D)
Weight of an object=J
x=J/(379.2/150)
x=J/2.528
y=12.64/2.528
y=5
Hope this helps :)
