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3 years ago

Which expression is equivalent to 4/5?

Mathematics

1 answer:

Sonja [21]3 years ago

7 0

Step-by-step explanation:

4/5=0.8

whatever expressions that equals 0.8

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Suppose a student carrying a flu virus returns to an isolated college campus of 5000 students. Determine a differential equation

zloy xaker [14]

Answer:

x(t) = 5000*( 1 - e^-kt)

Step-by-step explanation:

Given:

- Total number of students n = 5000

Find:

Differential equation governing the number of students x(t) who have contracted the flu.

Solution:

- Number of non-affected students = (5000 - x)

Hence,

- Rate at which students are infected:

dx / dt = k*(5000 - x )

- separate variables:

dx / (5000 - x ) = k*dt

- Integrate both sides:

- Ln(5000 - x) = kt + C

- Evaluate C for x = 0 @ t = 0

- Ln(5000 - 0) = k*0 + C

C = - Ln(5000)

- The solution to ODE is:

Ln(5000 - x) = -k*t + Ln(5000)

5000 - x = 5000*e^-kt

x(t) = 5000*( 1 - e^-kt)

4 0

3 years ago

The set of all points in a plane that are equidistant from two points is a(n): A. point B. line c. ray O D. angle

qwelly [4]

Answer:

Line

Step-by-step explanation:

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4 years ago

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Solve this fast please… and thank you so much :)

natita [175]

Answer:

Step-by-step explanation:

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3 years ago

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Find the equation of the line that is parallel to y = -7x + 2 and contains the point (-5,32)

photoshop1234 [79]

Answer:

Y=-7x-3

Step-by-step explanation:

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3 years ago

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In a region, 90% of the population have brown eyes. If 15 people are randomly selected, find the probability that at least 13

Andrew [12]

Answer:

a) $P(X \geq 13) = P(X=13) +P(X=14) +P(X=15)$

$P(X=13)=(15C13)(0.9)^{13} (1-0.9)^{15-13}=0.267$

$P(X=14)=(15C14)(0.9)^{14} (1-0.9)^{15-14}=0.343$

$P(X=15)=(15C15)(0.9)^{15} (1-0.9)^{15-15}=0.206$

And replacing we got:

$P(X \geq 13) = P(X=13) +P(X=14) +P(X=15)=0.267+0.343+0.206=0.816$

b) D. No, because the probability of this occurring is not small.

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Part a

If 15 people are randomly selected, find the probability that at least 13 of them have brown eyes.

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=15, p=0.90)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

And we want to find this probability:

$P(X \geq 13) = P(X=13) +P(X=14) +P(X=15)$

$P(X=13)=(15C13)(0.9)^{13} (1-0.9)^{15-13}=0.267$

$P(X=14)=(15C14)(0.9)^{14} (1-0.9)^{15-14}=0.343$

$P(X=15)=(15C15)(0.9)^{15} (1-0.9)^{15-15}=0.206$

And replacing we got:

$P(X \geq 13) = P(X=13) +P(X=14) +P(X=15)=0.267+0.343+0.206=0.816$

Part b

Is it unusual to randomly select 15 people and find that at least 13 of them have brown eyes? Note that a small probability is one that is less than 0.05.

Since our calculated probability is too higher compared to 0.05 we can conclude this:

D. No, because the probability of this occurring is not small.

7 0

4 years ago