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sesenic [268]
3 years ago
12

Hisham and Edgar are on the Baker High School wrestling team. Juan weighs 170 lbs. and wants to lose 4 pounds each week. Edgar w

eighs 135 lbs. and wants to gain 4 pounds each week. Write a system of equations that could be used to find out when Juan and Edgar would be the same weight. Then solve the system to find out how many weeks it will be until they are the same weight.
Mathematics
1 answer:
Ymorist [56]3 years ago
3 0

Answer:

If Jake loses 8 pounds, he will weigh twice as much as his sister. ... Together they now weigh 278 pounds. What is Jake's present weight, in pounds? (A) 131 (B) 135 ... Before we create our equations, we want to define some variables. ... To solve this equation, we can substitute 2S + 8 from Equation 1 for the variable J in ...

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I can try

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Dado el triángulo de vértices en A(-10, -13); B(-2,3) y C(2,1); Hallar la longitud de la perpendicular bajada desde el vértice B
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2 years ago
overnight a pipe cracked and started to leak into a basement. the graph shows the depth of the water in relation to time. how ma
Shkiper50 [21]

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if a water pipe bursts over night depends on the pressure of the water but i will say a foot or more an hour

Step-by-step explanation:

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3 years ago
Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie
ludmilkaskok [199]

Answer:

\lambda \geq 6.63835

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that X \sim Poisson(\lambda)

The probability mass function for the random variable is given by:

f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...

And f(x)=0 for other case.

For this distribution the expected value is the same parameter \lambda

E(X)=\mu =\lambda

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

P(X\geq 2)=1-P(X

Using the pmf we can find the individual probabilities like this:

P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}

P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}

And replacing we have this:

P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)

And we want this probability that at least of 99%, so we can set upt the following inequality:

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99

And now we can solve for \lambda

0.01 \geq e^{-\lambda}(1+\lambda)

Applying natural log on both sides we have:

ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)

ln(0.01) \geq -\lambda+ln(1+\lambda)

\lambda-ln(1+\lambda)+ln(0.01) \geq 0

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}

Where :

f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)

f'(x_n)=1-\frac{1}{1+\lambda}

Iterating as shown on the figure attached we find a final solution given by:

\lambda \geq 6.63835

4 0
3 years ago
Please answer this correctly
OlgaM077 [116]

Answer:

$21.57

Step-by-step explanation:

4.5 (1.22) + 0.8 (1.70) + 4.6 (3.20) =

5.49 + 1.36 + 14.72

21.57

4 0
3 years ago
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