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3 years ago

5

) Jenna works at a full time job where she earns $13.75 per hour. She is saving up for a vacation trip and already has $213.50 i

n her savings account. Write an equation modeling this story, then find out how much money she will have saved up if she works 10 hours.

1 answer:

almond37 [142]3 years ago

8 0

Answer:

She will have $227.25

Step-by-step explanation:

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The rectangle has an area of x^2-7x+10 square meters and a width of x−5 meters. What expression represents the length of the rec

Katyanochek1 [597]

Answer:

length = x - 2

Step-by-step explanation:

Area (A) of rectangle = lw ( l is the length and w the width )

A = x² - 7x + 10 = (x - 2)(x - 5) ← in factored form

l = $\frac{A}{w}$ = $\frac{(x-2)(x-5)}{x-5}$

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3 years ago

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

Fofino [41]

Answer:

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:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

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

3 defects - 18 square feet

x defects - 31 square feet

$18x = 3*31$

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

$x = 5.17$

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.

So

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

We want $P(X \geq 4)$

So

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

In which

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

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-5.17}*(5.17)^{0}}{(0)!} = 0.0057$

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

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

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

So

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

Finally

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

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

8 0

3 years ago