Answer:
The lower limit of 95% confidence interval is 99002.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = $100,000
Sample mean,
= $111,000
Sample size, n = 36
Alpha, α = 0.05
Population standard deviation, σ = $36,730
First, we design the null and the alternate hypothesis

We have to find the lower limit of the 95% confidence interval.
95% Confidence interval:

Putting the values, we get,


The lower limit of 95% confidence interval is 99002.
The volume of figure B is 15
Hey!
So what you can do is:
1 x 1.5 = 1.5(2nd term)
1.5 x 1.5=2.25(3rd term) x 1.5=3.375(4th term) x 1.5 = 5.0625(5th term)
5.0625 x 1.5=7.59375(6th term)x1.5=<span> 11.390625
</span>11.390625 x1.5= 17.0859375(8th term)x1.5=<span> 25.62890625</span>(9th term)
The answer is D!
Hope this helps!
It says use the table you have to send a pic
A) x=-2,1,3(x+2)(x-1)(x-3)=0(x+2)(x2-4x+3)=0x3-4x2+3x+2x2-8x+6=0x3-2x2-5x+6=0 b) x=-3,3,i and also -i(x-3)(x+3)(x-i)(x+i)=0(x2-9)(x2-i2)=0(x2-9)(x2+1)=0x4-9x2+x2-9=0x4-8x2-9=0 c) x=-2,-2,2-3i (also 2+3i), 4+√2 (also 4-√2)(x+2)(x+2)(x-[2-3i])(x-[2+3i])(x-[4+√2])(x-[4-√2])(x2+4x+4)(x2-[2-3i]x-[2+3i]x+4-9i2)(x2-[4+√2]x-[4-√2]x+16-2)(x2+4x+4)(x2-4x+13)(x2-8x+14)you now multiply all three trinomials together