Answer:
1.![\frac{dy}{dt}=ky](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%3Dky)
2.543.6
Step-by-step explanation:
We are given that
y(0)=200
Let y be the number of bacteria at any time
=Number of bacteria per unit time
![\frac{dy}{dt}\proportional y](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%5Cproportional%20y)
![\frac{dy}{dt}=ky](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%3Dky)
Where k=Proportionality constant
2.
,y'(0)=100
Integrating on both sides then, we get
![lny=kt+C](https://tex.z-dn.net/?f=lny%3Dkt%2BC)
We have y(0)=200
Substitute the values then , we get
![ln 200=k(0)+C](https://tex.z-dn.net/?f=ln%20200%3Dk%280%29%2BC)
![C=ln 200](https://tex.z-dn.net/?f=C%3Dln%20200)
Substitute the value of C then we get
![ln y=kt+ln 200](https://tex.z-dn.net/?f=ln%20y%3Dkt%2Bln%20200)
![ln y-ln200=kt](https://tex.z-dn.net/?f=ln%20y-ln200%3Dkt)
![ln\frac{y}{200}=kt](https://tex.z-dn.net/?f=ln%5Cfrac%7By%7D%7B200%7D%3Dkt)
![\frac{y}{200}=e^{kt}](https://tex.z-dn.net/?f=%5Cfrac%7By%7D%7B200%7D%3De%5E%7Bkt%7D)
![y=200e^{kt}](https://tex.z-dn.net/?f=y%3D200e%5E%7Bkt%7D)
Differentiate w.r.t
![y'=200ke^{kt}](https://tex.z-dn.net/?f=y%27%3D200ke%5E%7Bkt%7D)
Substitute the given condition then, we get
![100=200ke^{0}=200 \;because \;e^0=1](https://tex.z-dn.net/?f=100%3D200ke%5E%7B0%7D%3D200%20%5C%3Bbecause%20%5C%3Be%5E0%3D1)
![k=\frac{100}{200}=\frac{1}{2}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B100%7D%7B200%7D%3D%5Cfrac%7B1%7D%7B2%7D)
![y=200e^{\frac{t}{2}}](https://tex.z-dn.net/?f=y%3D200e%5E%7B%5Cfrac%7Bt%7D%7B2%7D%7D)
Substitute t=2
Then, we get ![y=200e^{\frac{2}{2}}=200e](https://tex.z-dn.net/?f=y%3D200e%5E%7B%5Cfrac%7B2%7D%7B2%7D%7D%3D200e)
![y=200(2.718)=543.6=543.6](https://tex.z-dn.net/?f=y%3D200%282.718%29%3D543.6%3D543.6)
e=2.718
Hence, the number of bacteria after 2 hours=543.6
(13 + x)/(50 - x ) = (2/1)
Cross Multiply
2(50-x) = 1(13 + x)
100 - 2x = 13 + x
add 2x to both sides
100 -2x + 2x = 13 + x + 2x
100 = 13 + 3x
Subtract 13 from both sides
100 - 13 = 13 - 13 + 3x
87 = 3x
divide both sides by 3
29 = x
(13 + 29)/(50-29) = 42/21 = 2 to 1 ratio
Answer:
(D)109
Step-by-step explanation:
Mean = 450 seconds
Standard deviation = 50
First, we determine the probability that the expected response time is between 400 seconds and 500 seconds, P(400<x<500)
Using the Z-Score,
![P(\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=P%28%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%20%3Cx%20%3C%20%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%29%5C%5C%3DP%28%5Cfrac%7B400-450%7D%7B50%7D%20%3Cx%20%3C%20%5Cfrac%7B500-450%7D%7B50%7D%29%5C%5C%3DP%28-1%3Cx%3C1%29)
From the Z-Score table
P(-1<x<1) = 0.68269
The probability that the expected response time is between 400 seconds and 500 seconds is 0.68269.
Since there are 160 Emergencies
Number whose expected time is between 400 seconds and 500 seconds
![=160 X 0.68269 \approx 109](https://tex.z-dn.net/?f=%3D160%20X%200.68269%20%5Capprox%20109)
Answer:
B. The maximum occurs at the function's x-intercept.
Step-by-step explanation:
Given table:
![\large\begin{array}{| c | c | c | c | c | c | c |}\cline{1-7} x & -5 & -4 & -3 & -2 & -1 & 0\\\cline{1-7} g(x) & -1 & 0 & -1 & -4 & -9 & -16\\\cline{1-7}\end{array}](https://tex.z-dn.net/?f=%5Clarge%5Cbegin%7Barray%7D%7B%7C%20c%20%7C%20c%20%7C%20c%20%7C%20c%20%7C%20c%20%7C%20c%20%7C%20c%20%7C%7D%5Ccline%7B1-7%7D%20x%20%26%20-5%20%26%20-4%20%26%20-3%20%26%20-2%20%26%20-1%20%26%200%5C%5C%5Ccline%7B1-7%7D%20g%28x%29%20%26%20-1%20%26%200%20%26%20-1%20%26%20-4%20%26%20-9%20%26%20-16%5C%5C%5Ccline%7B1-7%7D%5Cend%7Barray%7D)
From inspection of the table, we can see that:
and![g(-3) = -1](https://tex.z-dn.net/?f=g%28-3%29%20%3D%20-1)
This indicates <u>symmetry</u>.
The line of symmetry is the mid-point between the two x-values.
Therefore, the <u>line of symmetry</u> is x = -4
The vertex (minima/maxima) is on the line of symmetry, therefore the vertex is at (-4, 0). As the function decreases as x → 0, the vertex is a <u>maximum</u>.
As the y-value of the vertex is 0, the maximum occurs at the function's <u>x-intercept</u>.
In a positively skewed distribution, the extreme scores are larger, thus the mean is larger than the median. Since the mean is 122, the only possible median is 118. Therefore, D.
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