Mr. Crow, the head groundskeeper at High Tech Middle School, mows the lawn along the side of the gym. The lawn is rectangular, a
nd the length is 5 feet more than twice the width. The perimeter of the lawn is 250 feet. Homework Help ✎
Use the 5-D Process to find the dimensions of the lawn.
Use the dimensions you calculated in part (a) to find the area of the lawn.
What are the answers and how do i solve it thank you
Use the 5-D Process to find the dimensions of the lawn.
Use the dimensions you calculated in part (a) to find the area of the lawn.
What are the answers and how do i solve it thank you
1 answer:
Answer: The dimension of the lawn = 85 × 40
Area of the lawn = 3250 square feet
Step-by-step explanation:
Let x represents the width of the rectangle,
Then According to the question,
The length of the rectangle = 2 x + 5
Also it is given that the perimeter of the rectangle = 250 feet.
But we know that the perimeter of the rectangle = 2 × ( length + width)
Thus, the perimeter of the rectangle = 2( 2x+5+x) = 2(3x+5) = 6x+10
If x = 30, the perimeter of the rectangle = 6 × 30 + 10 = 190 < 250
Thus, x ≠ 10.
If x = 50, the perimeter of the rectangle = 6 × 50 + 10 = 310 > 250
Thus, x ≠ 50
If x = 40, the the perimeter of the rectangle = 6 × 40 + 10 = 250 = 250
Thus, x = 40
Therefore, the width of the rectangle, x = 40 feet.
And, the length of the rectangle, 2 x + 5 = 2 × 40 + 5 = 85 feet.
1) The dimension of the rectangle = 85 × 40
2)The area of the rectangle = length × width = 85 × 40 = 3250 square feet.
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Answer:
a) 0.0016 = 0.16% probability that the sample mean is at least $30.00.
b) 0.8794 = 87.94% probability that the sample mean is greater than $26.50 but less than $30.00
c) 90% of sample means will occur between $26.1 and $28.9.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question, we have that:
a. What is the likelihood the sample mean is at least $30.00?
This is 1 subtracted by the pvalue of Z when X = 30. So
By the Central Limit Theorem, we have that:
has a pvalue of 0.9984
1 - 0.9984 = 0.0016
0.0016 = 0.16% probability that the sample mean is at least $30.00.
b. What is the likelihood the sample mean is greater than $26.50 but less than $30.00?
This is the pvalue of Z when X = 30 subtracted by the pvalue of Z when X = 26.50. So
From a, when X = 30, Z has a pvalue of 0.9984
When X = 26.5
has a pvalue of 0.1190
0.9984 - 0.1190 = 0.8794
0.8794 = 87.94% probability that the sample mean is greater than $26.50 but less than $30.00.
c. Within what limits will 90 percent of the sample means occur?
Between the 50 - (90/2) = 5th percentile and the 50 + (90/2) = 95th percentile, that is, Z between -1.645 and Z = 1.645
Lower bound:
Upper Bound:
90% of sample means will occur between $26.1 and $28.9.