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Fudgin [204]
2 years ago
13

Eliza Savage received a statement from her bank showing a checking account balance of $324.18 as of January 18. Her own checkboo

k shows a balance of $487.38 as of January 29. The bank returned all of the cancelled checks but three. The amounts of these three checks are $15.00, $77.49, and $124.28. How much did Eliza deposit in her account between January 18 and January 29?
a. $201.12
b. $54.44
c. $197.24
d. $379.97
Mathematics
1 answer:
Lera25 [3.4K]2 years ago
8 0
Eliza Savage received a statement from her bank showing a checking account balance of $324.18 as of January 18. Her own checkbook shows a balance of $487.38 as of January 29. The bank returned all of the cancelled checks but three. The amounts of these three checks are $15.00, $77.49, and $124.28. How much did Eliza deposit in her account between January 18 and January 29?
     <span><span> <span> <span>   </span></span></span><span><span><span>Its D. $379.97</span></span></span></span>


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The total operating cost to run the organic strawberry farm varies directly with its number of acres. What is the constant of va
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Answer:

16\ acres

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form k=\frac{y}{x} or y=kx

In a proportional relationship the constant of proportionality k is equal to the slope m of the line and the line passes through the origin

Let

x ----> the number of acres

y ---> total operating costs

we have the ordered pair (20,551,520)

For x=20 acres, y=$551,520

<em>Find the value of the constant of proportionality k</em>

k=\frac{y}{x}

substitute the values of x and y

k=\frac{551,520}{20}=\$27,576\ per\ acre

The linear equation is equal to

y=27,576x

Find out how many acres are on a farm with a total operating cost of $441,216

so

For y=$441,216

substitute in the linear equation the value of y and solve for x

441,216=27,576x

x=441,216/27,576

x=16\ acres

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The article "Expectation Analysis of the Probability of Failure for Water Supply Pipes"† proposed using the Poisson distribution
oksian1 [2.3K]

Answer:

a. P(X ≤ 5) = 0.999

b. P(X > λ+λ) = P(X > 2) = 0.080

Step-by-step explanation:

We model this randome variable with a Poisson distribution, with parameter λ=1.

We have to calculate, using this distribution, P(X ≤ 5).

The probability of k pipeline failures can be calculated with the following equation:

P(k)=\lambda^{k} \cdot e^{-\lambda}/k!=1^{k} \cdot e^{-1}/k!=e^{-1}/k!

Then, we can calculate P(X ≤ 5) as:

P(X\leq5)=P(0)+P(1)+P(2)+P(4)+P(5)\\\\\\P(0)=1^{0} \cdot e^{-1}/0!=1*0.3679/1=0.368\\\\P(1)=1^{1} \cdot e^{-1}/1!=1*0.3679/1=0.368\\\\P(2)=1^{2} \cdot e^{-1}/2!=1*0.3679/2=0.184\\\\P(3)=1^{3} \cdot e^{-1}/3!=1*0.3679/6=0.061\\\\P(4)=1^{4} \cdot e^{-1}/4!=1*0.3679/24=0.015\\\\P(5)=1^{5} \cdot e^{-1}/5!=1*0.3679/120=0.003\\\\\\P(X\leq5)=0.368+0.368+0.184+0.061+0.015+0.003=0.999

The standard deviation of the Poisson deistribution is equal to its parameter λ=1, so the probability that X exceeds its mean value by more than one standard deviation (X>1+1=2) can be calculated as:

P(X>2)=1-(P(0)+P(1)+P(2))\\\\\\P(0)=1^{0} \cdot e^{-1}/0!=1*0.3679/1=0.368\\\\P(1)=1^{1} \cdot e^{-1}/1!=1*0.3679/1=0.368\\\\P(2)=1^{2} \cdot e^{-1}/2!=1*0.3679/2=0.184\\\\\\P(X>2)=1-(0.368+0.368+0.184)=1-0.920=0.080

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