The number of people who have heard a rumor is modeled by the logistic function f(x)=2601+23e−0.06x
, where x is the number of days since the rumor began.
Approximately how many days have passed when the rate at which the rumor spreads begins to decrease?
Answers
Answer:
C
Step-by-step explanation:
The number of people who have heard a rumor is modeled by the logistic function:
[tex]\displaystyle f(x)=\frac{260}{1+23e^{-0.06x}}[/tex]
And we want how many days has passed since the rate at which the rumor spread began to decrease.
First, the numerator tells us that the total number of people who will have heard the rumor by the end is 260.
Recall that the point of maximum growth is always half of the maximum capacity.
In this case, that point will be 260/2 = 130.
For a logistic function, the rate of all values to the left of the halfway point is increasing, (i.e. concave up), while the rate of all values to the right is decreasing (i.e. concave down). Therefore, we can solve for x when f(x) = 130:
[tex]\displaystyle 130=\frac{260}{1+23e^{-0.06x}}[/tex]
Cross-multiply:
[tex]130(1+23e^{-0.06x}})=260[/tex]
Distribute:
[tex]130+2990e^{-0.06x}=260[/tex]
Solving for x:
[tex]2990e^{-0.06x}=130[/tex]
Therefore:
[tex]\displaystyle e^{-0.06x}=\frac{130}{2990}=\frac{1}{23}[/tex]
Taking the natural log of both sides:
[tex]\displaystyle -0.06x=\ln\Big(\frac{1}{23}\Big)[/tex]
Therefore:
[tex]\displaystyle x=\frac{\ln(1/23)}{-0.06}\approx52.3[/tex]
After about 52.3 days, the rate at which the rumor started spreading will begin to decrease.
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Answers
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Step-by-step explanation:
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Answers
Answer:
Step-by-step explanation:
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We wish to modify the cellular telephone coding system in order to reduce the number of errors. In particular, if there are two or three zeroes in the received sequence of 5 bits, we will say that a deletion (event D) occurs. Otherwise, if at least 4 zeroes are received, then the receiver decides a zero was sent. Similarly, if at least 4 ones are received, then the receiver decides a one was sent. We say that an error occurs if either a one was sent and the receiver decides zero was sent or if a zero was sent and the receiver decides a one was sent. For this modified protocol, what is the probability P[E] of an error? What is the probability P[D] of a deletion?
Answers
Solution :
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Now if the 0 is being transmitted, then the 0 is sent 5 number of times and then we call the decoding a 0 success. Let the repeated transmission are independent. In this case, we have 5 trails. Let each repeated bit the receiver decodes as 1 a success, using the binomial probabilities. We using the [tex]$S_{k,5}$[/tex] to denote an event of [tex]$k$[/tex] successes in the 5 trails, then the probability k are decoded at the receiver is:
[tex]$P[S_{k,5}]=\binom{5}{k}p^k(1-p)^{5-k}$[/tex]
[tex]$=\binom{5}{k}(0.9)^k(1-0.9)^{5-k}$[/tex]
[tex]$=\binom{5}{k}(0.9)^k(0.1)^{5-k}$[/tex]
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[tex]$=\binom{5}{3}(0.9)^5(0.1)^{0}+\binom{5}{4}(0.9)^4(0.1)^1$[/tex]
[tex]$=(0.9)^5+5(0.9)^4(0.1)^1$[/tex]
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[tex]$P[D]=P[S_{3,5}]+P[S_{2,5}]$[/tex]
[tex]$=\binom{5}{3}(0.9)^3(0.1)^{2}+\binom{5}{2}(0.9)^2(0.1)^3$[/tex]
[tex]$=10(0.9)^3(0.1)^2+10(0.9)^2(0.1)^3$[/tex]
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The probability of an error is
[tex]$P[E]=P[S_{1,5}]+P[S_{0,5}]$[/tex]
[tex]$=\binom{5}{1}(0.9)^1(0.1)^{4}+\binom{5}{0}(0.9)^0(0.1)^5$[/tex]
[tex]$=5(0.9)^1(0.1)^4+(0.1)^5$[/tex]
= 0.00046
Anyone know this please I really need help
Answers
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Step-by-step:
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A rocket fired upward from some initial distance above the ground. Its height (in feet) h, above the ground t seconds after it is fired is given by h(t)=-16t^2+160t+384
What is the rockets maximum height?
After it is fired, The rocket reaches the ground at t=____ seconds
Answers
Answer:
Part A)
784 feet in the air (after five seconds).
Part B)
After 12 seconds.
Step-by-step explanation:
The height h (in feet) of a rocket t seconds after being fired is modeled by the function:
[tex]h(t)=-16t^2+160t+384[/tex]
Part A)
We want to find the rocket's maximum height.
Since our function is a quadratic, the maximum height will occur at its vertex. The vertex of a quadratic is given by:
[tex]\displaystyle \Big(-\frac{b}{2a}\, f\Big(-\frac{b}{2a}\Big)\Big)[/tex]
In this case, a = -16, b = 160, and c = 384.
So, the vertex occurs at:
[tex]\displaystyle t=-\frac{160}{2(-16)}=\frac{160}{32}=5\text{ seconds}[/tex]
The maximum height is reached after five seconds.
Then the maximum height is:
[tex]h(t)_\text{max}=h(5)=-16(5)^2+160(5)+384=784\text{ feet}[/tex]
Part B)
When the rocket reaches the ground, its height h above the ground will be 0. Hence:
[tex]h(t)_\text{ground}=0=-16t^2+160t+384[/tex]
Solve for t. We can first divide both sides by -16:
[tex]t^2-10t-24=0[/tex]
Factor:
[tex](t-12)(t+2)=0[/tex]
Zero Product Property:
[tex]t-12=0\text{ or } t+2=0[/tex]
Solve for each case:
[tex]t=12\text{ or } t=-2[/tex]
Time cannot be negative. Hence, our only solution is:
[tex]t=12\text{ seconds}[/tex]
The rocket reaches the ground 12 seconds after it is fired.
What is g(x)?
A. g(x) = -2
B. g(x) = -11
C. g(x) = -x
D. g(x) = -x
WILL MARK BRAINIEST!!
Answers
Step-by-step explanation:
g(x) = -X^2
which isn't an option.