Answer: 313
Step-by-step explanation:
Let x be the attendance before the drop.
Given: An article reports, "attendance dropped 4% this year, to 300.
Wec can write 4% = 0.04
Then, 4% of x = 0.04x
Now according to the question we have:-
![x-0.04x=300\\\\\Rightarrow\ x(1-0.04)=300\\\\\Rightarrow\ 0.96x=300\\\\\Rightarrow\ x=\dfrac{300}{0.96}=312.5\approx313\ \ \ \ \text{[Rounded to the nearest whole numbers.]}](https://tex.z-dn.net/?f=x-0.04x%3D300%5C%5C%5C%5C%5CRightarrow%5C%20x%281-0.04%29%3D300%5C%5C%5C%5C%5CRightarrow%5C%200.96x%3D300%5C%5C%5C%5C%5CRightarrow%5C%20x%3D%5Cdfrac%7B300%7D%7B0.96%7D%3D312.5%5Capprox313%5C%20%5C%20%5C%20%5C%20%5Ctext%7B%5BRounded%20to%20the%20nearest%20whole%20numbers.%5D%7D)
Hence, the the attendance before the drop = 313
x+7=4
x=-3
x+7=-4
x=-11
A is correct
for absolute value problems, solve for x when the total is positive and negative
The assumption here is that, both triangles are similar. Now, if that's the case,
![\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\ -------------------------------\\\\ \cfrac{small}{large}\qquad \cfrac{25}{35}\implies \stackrel{simplified}{\cfrac{5}{7}}\qquad \qquad \cfrac{5}{7}=\cfrac{\sqrt{270}}{\sqrt{x}}\implies \cfrac{5}{7}=\sqrt{\cfrac{270}{x}} \\\\\\ \left( \cfrac{5}{7} \right)^2=\cfrac{270}{x}\implies \cfrac{5^2}{7^2}=\cfrac{270}{x}\implies x=\cfrac{7^2\cdot 270}{5^2}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5Ctextit%7Bsimilar%20shape%7D%7D%7B%5Ctextit%7Bsimilar%20shape%7D%7D%5Cqquad%20%5Ccfrac%7Bs%7D%7Bs%7D%3D%5Ccfrac%7B%5Csqrt%7Bs%5E2%7D%7D%7B%5Csqrt%7Bs%5E2%7D%7D%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bs%5E3%7D%7D%7B%5Csqrt%5B3%5D%7Bs%5E3%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Ccfrac%7Bsmall%7D%7Blarge%7D%5Cqquad%20%5Ccfrac%7B25%7D%7B35%7D%5Cimplies%20%5Cstackrel%7Bsimplified%7D%7B%5Ccfrac%7B5%7D%7B7%7D%7D%5Cqquad%20%5Cqquad%20%5Ccfrac%7B5%7D%7B7%7D%3D%5Ccfrac%7B%5Csqrt%7B270%7D%7D%7B%5Csqrt%7Bx%7D%7D%5Cimplies%20%5Ccfrac%7B5%7D%7B7%7D%3D%5Csqrt%7B%5Ccfrac%7B270%7D%7Bx%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cleft%28%20%5Ccfrac%7B5%7D%7B7%7D%20%5Cright%29%5E2%3D%5Ccfrac%7B270%7D%7Bx%7D%5Cimplies%20%5Ccfrac%7B5%5E2%7D%7B7%5E2%7D%3D%5Ccfrac%7B270%7D%7Bx%7D%5Cimplies%20x%3D%5Ccfrac%7B7%5E2%5Ccdot%20270%7D%7B5%5E2%7D)
<span>0.3046% chance of damaging at least one of five payloads.
First, determine how many standard deviations it takes for the parachute to open at 100 meters. So
200 m - 100 m = 100 m
100 m / 31 m = 3.2258
Looking up on a standard normal table, the probability of having that deviation or higher is 0.00061, or 0.061%
Since you're looking for the probability of damaging at least one of five, the easiest way of calculating that is to subtract from 1 the probability of having all five parachutes successfully deploy. So
1 - (1 - 0.00061)^5 = 1 - 0.99939^5 = 1 - 0.996953719 = 0.003046281 = 0.3046%</span>
