We would need to look over the z table to find the area under the standard normal distribution curve to the left of z = 1.04. Then we'll subtract it from 1 to get the proportion of a normal distribution corresponding to z scores greater than 1.04.
By looking at the z table, we can see that the area to the left of z = 1.04 is 0.8508. So the proportion of a normal distribution to the right of z = 1.04 is 1 – 0.8508 = 0.1492.
The answer is 0.1492.
It should be C. Equilateral triangles
A=10x(15x)-p(4x)^2
A=150x^2-16px^2
A=(150-16p)x^2
correct Option is B i.e, 
whose simplified form is: 
Step-by-step explanation:
We need to solve the expression:
![2[(6x)(6x)]+4[(12x - 9)(6x)]](https://tex.z-dn.net/?f=2%5B%286x%29%286x%29%5D%2B4%5B%2812x%20-%209%29%286x%29%5D)
Solving:
![2[(6x)(6x)]+4[(12x - 9)(6x)]\\=2[36x^2]+4[(12x*6x)-(9*6x)]\\=72x^2+4[72x^2-54x]\\=72x^2+288x^2-216x\\=360x^2-216x](https://tex.z-dn.net/?f=2%5B%286x%29%286x%29%5D%2B4%5B%2812x%20-%209%29%286x%29%5D%5C%5C%3D2%5B36x%5E2%5D%2B4%5B%2812x%2A6x%29-%289%2A6x%29%5D%5C%5C%3D72x%5E2%2B4%5B72x%5E2-54x%5D%5C%5C%3D72x%5E2%2B288x%5E2-216x%5C%5C%3D360x%5E2-216x)
So, correct Option is B i.e, whose simplified form is:
Keywords: Surface area of the rectangular prism
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