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Lelechka [254]
3 years ago
5

You are wrapping a present for your puppy the box is rectangular prism with a length of 15 in height of 13 in and width of 8 in.

if you have 500 square inches of wrapping paper do you have enough paper if know how much more do you need if yes how much extra
Mathematics
1 answer:
diamong [38]3 years ago
4 0

Answer:

1560 square inches, you'll need 1060 more square inches of wrapping  paper

Step-by-step explanation:

15 x 13 x 8 (L x W x H) = 1560

1560 - 500 = 1060

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Eva8 [605]

.25(4) = 1 gallon

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Last month for work Mr. Wheelboogie traveled 5150 miles. This month that total increased to 7895 miles. What is the percent incr
makvit [3.9K]

Answer:

53%

Step-by-step explanation:

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3 years ago
he time between arrivals of small aircraft at a county airport is exponentially distributed with a mean of one hour. Round the a
Eva8 [605]

Answer:

The answer is below

Step-by-step explanation:

The time between arrivals of small aircraft at a county airport is exponentially distributed with a mean of one hour. Round the answers to 3 decimal places.

(a) What is the probability that more than three aircraft arrive within an hour?

(b) If 30 separate one-hour intervals are chosen, what Is the probability that no interval contains more than three arrivals?

(c) Determine the length of an interval of time (in hours) such that the probability that no arrivals occur during the interval is 0.1.

Solution:

a) A poisson distribution is given by the formula:

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

λ = 1 hour

Therefore:

P(X > 3) = 1 - P(X < 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(x = 3)]

P(X=0)=\frac{e^{-1}*1^0}{0!}=0.3679

P(X=1)=\frac{e^{-1}*1^1}{1!}=0.3679

P(X=2)=\frac{e^{-1}*1^2}{2!}=0.1839

P(X=3)=\frac{e^{-1}*1^3}{3!}=0.0613

P(X > 3) = 1 - [0.3679+0.3679 + 0.1839 + 0.0613] = 0.019

b) Assuming 30 1 hour intervals, hence:

P(X \leq 3)^{30}=[1-P(X\geq 30)]^{30}=(1-0.019)^{30}=0.5624

c) mean = 1 hour

mean = 1 / λ

1 = 1 / λ

λ = 1

The cumulative distribution function of a continuous variable is:

F(x)=1-e^{-\lambda x}

0.1=1-F(x)\\\\0.1=1-(1-e^{- x})\\\\0.1=e^{- x}\\\\-x=ln(0.1)\\\\-x = -2.3\\\\x=2.3\ hours

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