I wonder if you mean to write
in place of
...
If you meant what you wrote, then we have


If you meant to write
(the cube root of 256), then we could go on to have
![\sqrt[3]{256}=\sqrt[3]{16^2}=\sqrt[3]{(4^2)^2}=\sqrt[3]{4^4}=\sqrt[3]{4^3\cdot4}=4\sqrt[3]4](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B256%7D%3D%5Csqrt%5B3%5D%7B16%5E2%7D%3D%5Csqrt%5B3%5D%7B%284%5E2%29%5E2%7D%3D%5Csqrt%5B3%5D%7B4%5E4%7D%3D%5Csqrt%5B3%5D%7B4%5E3%5Ccdot4%7D%3D4%5Csqrt%5B3%5D4)
Answer: p = 6
Step-by-step explanation:
HI !
px + 3y - ( p-3)=0
a₁ = p , b₁ = 3 , c₁ = - (p-3)
=====================
12x + py - p=0
a₂ = 12 , b₂ = p , c₂ = -p
since , the equations have infinite solutions ,
a₁/a₂ = b₁/b₂ = c₁/c₂
p/12 = 3/p = p-3/p
--------------------------------------
p/12 = 3/p
cross multiply ,
p² = 36
p = √36
p = 6
for the value of p = 6 , the equations will have infinitely many solutions
Working conditions in the United States in the early 1900’s can best be described as atrocious.
Answer:
<h3>
X = 2 , X = -3</h3>
Option C is the correct option.
Step-by-step explanation:

Discriminant = 
Given d = 15
When discriminant D > 0 --> roots are real and distinct. ( unequal )
So ,
option has real and distinct roots.
Hope this helps..
Best regards!!
We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:
![CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack%20x-Z_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%2Cx%2BZ_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%5Crbrack)
Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:
![CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack30.0-Z_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%2C30.0%2BZ_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%5Crbrack)
Where (from tables):

Finally, the interval at 98% confidence level is: