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Arisa [49]
3 years ago
10

Suppose that y varies directly with x, and y=15 when x=24. Write a direct variation equation that relates x and y. Find y when x

=3
Mathematics
2 answers:
Mrrafil [7]3 years ago
5 0

Answer:

y=\frac{5}{8}x

y=\frac{15}{8}, when x=3.

Step-by-step explanation:

We have been given that  y varies directly with x, and y=15 when x=24.

We know that two direct proportional quantities are in form y=kx, where k is constant of proportionality.

Let us find constant of proportionality by substituting y=15 and x=24 in above equation.

15=k*24

\frac{15}{24}=\frac{k*24}{24}

\frac{3*5}{3*8}=k

\frac{5}{8}=k

Therefore, our required equation would be y=\frac{5}{8}x.

Let us substitute x=3 in the equation.

y=\frac{5}{8}(3)

y=\frac{15}{8}

Therefore, the value of y is \frac{15}{8}.

Lapatulllka [165]3 years ago
4 0

Just set up a proportion with the original numbers and then the numbers you want to find like so: (x1/y1)=(x2/y2) and since you want to find the second y value, just leave y as a variable like so: (15/24)=(3/y) after this just cross multiply and get the answer: 15y=72.....y=4.8


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A certain kind of sheet metal has, on average, 3 defects per 18 square feet. Assuming a Poisson distribution, find the probabili
Fofino [41]

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75.8% probability that a 31 square foot metal sheet has at least 4 defects.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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In this problem, we have that:

3 defects per 18 square feet.

So for 31 square feet, we have to solve a rule of three

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x = \frac{31*3}{18}

x = 5.17

So \mu = 5.17

Assuming a Poisson distribution, find the probability that a 31 square foot metal sheet has at least 4 defects.

Either it has three or less defects, or it has at least 4 defects. The sum of the probabilities of these events is decimal 1.

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P(X \leq 3) + P(X \geq 4) = 1

We want P(X \geq 4)

So

P(X \geq 4) = 1 - P(X \leq 3)

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P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

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P(X = 1) = \frac{e^{-5.17}*(5.17)^{1}}{(1)!} = 0.0294

P(X = 2) = \frac{e^{-5.17}*(5.17)^{2}}{(2)!} = 0.0760

P(X = 3) = \frac{e^{-5.17}*(5.17)^{3}}{(3)!} = 0.1309

So

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0057 + 0.0294 + 0.0760 + 0.1309 = 0.242

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P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.242 = 0.758

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