Answer:
see explanation
Step-by-step explanation:
If there is a solution between x = 1.1 and x = 1.2 then there will be a change in sign when the equation is evaluated at the points, indicating the graph has crossed the x- axis, where the solution lies.
Given
x³ + 4x = 6 ( subtract 6 from both sides )
x³ + 4x - 6 = 0 ← in standard form
Evaluating for x = 1.1
(1.1)³ + 4(1.1) - 6
= 1.331 + 4.4 - 6 = - 0.269 ← < 0
Evaluating for x = 1.2
(1.2)³ + 4(1.2) - 6
= 1.728 + 4.8 - 6 = 0.528 ← > 0
Since there is a change in sign the graph has crossed the x-axis from below / indicating a solution between x = 1.1 and x = 1.2
5x-5=-7y
3x+2y=-8
First, you need to rearrange the first equation so it is in the same format as the second one.
5x-5=-7y
Add 5 to both sides
5x-5+5=-7y+5
Add 7y to both sides
5x+7y=-7y+7y+5
So you have 5x+7y=5
Now you need to multiply that equation by 3
3(5x+7y=5)
15x+21y=15
Multiply the second equation by -5
-5(3x+2y=-8)
-15x-10y=40
Now add them together
15x+21y=15
+(-15x-10y=40)
---------------------
11y=55
y=5
Now plug y=5 into one of the original equations and solve for x.
3x+2(5)=-8
3x+10=-8
3x=-8-10
3x=-18
x=-6
To check the solution plug them both into the other equation:
5(-6)-5=-7(5)
-30-5=-35
-35=-35
It checks.
Hope that helps.
The difference is 180 is the correct answer for the problems
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.
Answer:
the first one is 3 second is 1 third is 4 and the last is 2
