Roy wants to make a path from one corner of his yard to the other as shown below. The path will be 4 feet wide. He wants to find
the area of lawn that remains.
Roy claims that the area of the lawn is 300 square feet since it covers exactly one-half of the yard. Which statement about his claim is correct?
He is incorrect. The path will have an area of (4)(40)=160 sq ft. The yard has an area of 600 sq ft. The area of the lawn will be the difference of the yard and path, so it is 440 sq ft.
He is incorrect. The path will have an area of 1/2(4)(40)=80 sq ft. The yard has an area of 300 sq ft. The area of the lawn will be the difference of the yard and path, so it is 220 sq ft.
He is incorrect. The path will have an area of (4)(40)=160 sq ft. The yard has an area of 300 sq ft. The area of the lawn will be the difference of the yard and path, so it is 140 sq ft.
He is incorrect. The path will have an area of (9)(40)=360 sq ft. The yard has an area of 600 sq ft. The area of the lawn will be the difference of the yard and path, so it is 240 sq ft.
Roy claims that the area of the lawn is 300 square feet since it covers exactly one-half of the yard. Which statement about his claim is correct?
He is incorrect. The path will have an area of (4)(40)=160 sq ft. The yard has an area of 600 sq ft. The area of the lawn will be the difference of the yard and path, so it is 440 sq ft.
He is incorrect. The path will have an area of 1/2(4)(40)=80 sq ft. The yard has an area of 300 sq ft. The area of the lawn will be the difference of the yard and path, so it is 220 sq ft.
He is incorrect. The path will have an area of (4)(40)=160 sq ft. The yard has an area of 300 sq ft. The area of the lawn will be the difference of the yard and path, so it is 140 sq ft.
He is incorrect. The path will have an area of (9)(40)=360 sq ft. The yard has an area of 600 sq ft. The area of the lawn will be the difference of the yard and path, so it is 240 sq ft.
1 answer:
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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:
Then, we can calculate P(X ≤ 5) as:
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:
Answer:
95% of the text messages have length between 23 units and 47 units.
Step-by-step explanation:
We are given the following in the question:
The lengths of text messages are normally distributed.
95% confidence interval:
(23,47)
Thus, we could interpret the confidence interval as:
About 95% of the text messages have length between 23 units and 47 units.
By Empirical rule for a normally distributed data, about 95% of data lies within 2 standard deviations of mean , thus we can write:
Thus, the mean length of text messages is 23 units and standard deviation is 6 units.