To solve this, you’d multiply 50 by 1.15 to get 57.5 grams of crisps.
Answer:
6%, 49%, 16%, 21%, 80%, 93%
Step-by-step explanation:
add the percent sign to each each number (6, 49, 16, 21, 80, and 93).
Answer:
280 hours
Step-by-step explanation:
Amount paid + profit = cost of a lesson per hour
$2000 + $5000 = $25(h)
7000 = 25h
7000/25 = h
280 = h
(a)
![\displaystyle \frac{1}{2} \cdot \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \left(4^2 - (3 + 2\cos\theta)^2 \right) \, d\theta](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%20%5Cint_%7B%5Cfrac%7B%5Cpi%7D%7B3%7D%7D%5E%7B%5Cfrac%7B5%5Cpi%7D%7B3%7D%7D%20%5Cleft%284%5E2%20-%20%283%20%2B%202%5Ccos%5Ctheta%29%5E2%20%5Cright%29%20%5C%2C%20d%5Ctheta)
or, via symmetry
![\displaystyle\frac{1}{2} \cdot 2 \int_{\frac{\pi}{3}}^{\pi} \left(4^2 - (3 + 2\cos\theta)^2 \right) \, d\theta](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%202%20%5Cint_%7B%5Cfrac%7B%5Cpi%7D%7B3%7D%7D%5E%7B%5Cpi%7D%20%5Cleft%284%5E2%20-%20%283%20%2B%202%5Ccos%5Ctheta%29%5E2%20%5Cright%29%20%5C%2C%20d%5Ctheta)
____________
(b)
By the chain rule:
![\displaystyle \frac{dy}{dx} = \frac{ dy/ d\theta}{ dx/ d\theta}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%20%5Cfrac%7B%20dy%2F%20d%5Ctheta%7D%7B%20dx%2F%20d%5Ctheta%7D)
For polar coordinates, x = rcosθ and y = rsinθ. Since
<span>r = 3 + 2cosθ, it follows that
![x = (3 + 2\cos\theta) \cos \theta \\ y = (3 + 2\cos\theta) \sin \theta](https://tex.z-dn.net/?f=x%20%3D%20%283%20%2B%202%5Ccos%5Ctheta%29%20%5Ccos%20%5Ctheta%20%5C%5C%20%0Ay%20%3D%20%283%20%2B%202%5Ccos%5Ctheta%29%20%5Csin%20%5Ctheta)
Differentiating with respect to theta
![\begin{aligned} \displaystyle \frac{dy}{dx} &= \frac{ dy/ d\theta}{ dx/ d\theta} \\ &= \frac{(3 + 2\cos\theta)(\cos\theta) + (-2\sin\theta)(\sin\theta)}{(3 + 2\cos\theta)(-\sin\theta) + (-2\sin\theta)(\cos\theta)} \\ \\ \left.\frac{dy}{dx}\right_{\theta = \frac{\pi}{2}} &= 2/3 \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%0A%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%26%3D%20%5Cfrac%7B%20dy%2F%20d%5Ctheta%7D%7B%20dx%2F%20d%5Ctheta%7D%20%5C%5C%0A%26%3D%20%5Cfrac%7B%283%20%2B%202%5Ccos%5Ctheta%29%28%5Ccos%5Ctheta%29%20%2B%20%28-2%5Csin%5Ctheta%29%28%5Csin%5Ctheta%29%7D%7B%283%20%2B%202%5Ccos%5Ctheta%29%28-%5Csin%5Ctheta%29%20%2B%20%28-2%5Csin%5Ctheta%29%28%5Ccos%5Ctheta%29%7D%20%5C%5C%20%5C%5C%0A%5Cleft.%5Cfrac%7Bdy%7D%7Bdx%7D%5Cright_%7B%5Ctheta%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%0A%26%3D%202%2F3%0A%5Cend%7Baligned%7D)
2/3 is the slope
____________
(c)
"</span><span>distance between the particle and the origin increases at a constant rate of 3 units per second" implies dr/dt = 3
A</span>ngle θ and r are related via <span>r = 3 + 2cosθ, so implicitly differentiating with respect to time
</span><span />
Answer:
97.10% probability that five or more of the original 2000 components fail during the useful life of the product.
Step-by-step explanation:
For each component, there are only two possible outcomes. Either it works correctly, or it does not. The probability of a component falling is independent from other components. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And p is the probability of X happening.
In this problem we have that:
![n = 2000, p = 1-0.995 = 0.005](https://tex.z-dn.net/?f=n%20%3D%202000%2C%20p%20%3D%201-0.995%20%3D%200.005)
Approximate the probability that five or more of the original 2000 components fail during the useful life of the product.
We know that either less than five compoenents fail, or at least five do. The sum of the probabilities of these events is decimal 1. So
![P(X < 5) + P(X \geq 5) = 1](https://tex.z-dn.net/?f=P%28X%20%3C%205%29%20%2B%20P%28X%20%5Cgeq%205%29%20%3D%201)
We want ![P(X \geq 5)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%205%29)
So
![P(X \geq 5) = 1 - P(X < 5)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%205%29%20%3D%201%20-%20P%28X%20%3C%205%29)
In which
![P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)](https://tex.z-dn.net/?f=P%28X%20%3C%205%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29)
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 0) = C_{2000,0}.(0.005)^{0}.(0.995)^{2000} = 0.000044](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B2000%2C0%7D.%280.005%29%5E%7B0%7D.%280.995%29%5E%7B2000%7D%20%3D%200.000044)
![P(X = 1) = C_{2000,1}.(0.005)^{1}.(0.995)^{1999} = 0.000445](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20C_%7B2000%2C1%7D.%280.005%29%5E%7B1%7D.%280.995%29%5E%7B1999%7D%20%3D%200.000445)
![P(X = 2) = C_{2000,2}.(0.005)^{2}.(0.995)^{1998} = 0.002235](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20C_%7B2000%2C2%7D.%280.005%29%5E%7B2%7D.%280.995%29%5E%7B1998%7D%20%3D%200.002235)
![P(X = 3) = C_{2000,3}.(0.005)^{3}.(0.995)^{1997} = 0.007480](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20C_%7B2000%2C3%7D.%280.005%29%5E%7B3%7D.%280.995%29%5E%7B1997%7D%20%3D%200.007480)
![P(X = 4) = C_{2000,4}.(0.005)^{4}.(0.995)^{1996} = 0.018765](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20C_%7B2000%2C4%7D.%280.005%29%5E%7B4%7D.%280.995%29%5E%7B1996%7D%20%3D%200.018765)
![P(X < 5) = P(X = 0) + `P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.000044 + 0.000445 + 0.002235 + 0.007480 + 0.018765 = 0.0290](https://tex.z-dn.net/?f=P%28X%20%3C%205%29%20%3D%20P%28X%20%3D%200%29%20%2B%20%60P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29%20%3D%200.000044%20%2B%200.000445%20%2B%200.002235%20%2B%200.007480%20%2B%200.018765%20%3D%200.0290)
![P(X \geq 5) = 1 - P(X < 5) = 1 - 0.0290 = 0.9710](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%205%29%20%3D%201%20-%20P%28X%20%3C%205%29%20%3D%201%20-%200.0290%20%3D%200.9710)
97.10% probability that five or more of the original 2000 components fail during the useful life of the product.