Factor
in form
ax²+bx+c form
use the ac method
1. mulitply a and c
2. factor the result such that the 2 factors add to b
3. split b into those 2 factors
4. group the terms
5. undistribute common factors
6. undistribute again
first multiply ac
4 times 5=20
now
what 2 numbers multiply to 20 and add to 12
2 and 10
4x²+2x+10x+5
group
(4x²+2x)+(10x+5)
undistribute
2x(2x+1)+5(2x+1)
undistribute the (2x+1)
(2x+1)(2x+5) is factored form whih you have there, nice
![a\cdot a\cdot a=216\\\\a^3=216\to a=\sqrt[3]{216}\\\\\boxed{a=6}\\\\\text{Substitute}\ b\cdot c=52\ \text{to the second expression}\ a\cdot b\cdot c=96:\\\\abc96\ \wedge\ bc=52\to a(52)=96\qquad\text{divide both sides by 52}\\\\a=\dfrac{96}{52}\to a=\dfrac{24}{13}\neq6](https://tex.z-dn.net/?f=a%5Ccdot%20a%5Ccdot%20a%3D216%5C%5C%5C%5Ca%5E3%3D216%5Cto%20a%3D%5Csqrt%5B3%5D%7B216%7D%5C%5C%5C%5C%5Cboxed%7Ba%3D6%7D%5C%5C%5C%5C%5Ctext%7BSubstitute%7D%5C%20b%5Ccdot%20c%3D52%5C%20%5Ctext%7Bto%20the%20second%20expression%7D%5C%20a%5Ccdot%20b%5Ccdot%20c%3D96%3A%5C%5C%5C%5Cabc96%5C%20%5Cwedge%5C%20bc%3D52%5Cto%20a%2852%29%3D96%5Cqquad%5Ctext%7Bdivide%20both%20sides%20by%2052%7D%5C%5C%5C%5Ca%3D%5Cdfrac%7B96%7D%7B52%7D%5Cto%20a%3D%5Cdfrac%7B24%7D%7B13%7D%5Cneq6)
a = 6 and a = 24/13 FALSE!!!
<h3>Answer: NO SOLUTION.</h3>
Answer:
a) H0:
H1:
b) 
And the critical values with
on each tail are:

c)
d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34
Step-by-step explanation:
Information provided
n = 10 sample size
s= 1.186 the sample deviation
the value that we want to test
represent the p value for the test
t represent the statistic (chi square test)
significance level
Part a
On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:
H0:
H1:
The statistic is given by:
Part b
The degrees of freedom are given by:

And the critical values with
on each tail are:

Part c
Replacing the info we got:
Part d
For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34
The answer is -166, your welcome
(8/9)^2 is 64/81. Exponent is 2. Hope this help