A
You work for Greene-Mann Landscaping Company. A client is looking to plant trees that can
provide summer shade and block winter wind. According to the document, what type of trees
should you recommend?
A. Evergreens
B. Tall deciduous trees with spreading crowns
C. Deciduous trees with lower crowns
D. Native tree species
E. Smaller shrubs

Answers

Answer 1

Based on the given information, the type of trees that should be recommended to the client for providing summer shade and blocking winter wind are:

B. Tall deciduous trees with spreading crowns

Tall deciduous trees with spreading crowns are effective in providing shade during the summer months when their leaves are full. Additionally, their spreading crowns can help block winter winds. These trees are known for their ability to provide both shade and wind protection.

Related Questions

Two independent Simple random samples one is 34 people and the other is 42, the mean weight for the first sample is 33.1 and standard deviation of 4.61 and the second sample has a mean weight of 31.7 and a standard deviation of 5.23. What is the pvalue

Answers

The p-value (0.143) is greater than the significance level (0.05).

To calculate the p-value for comparing the means of two independent samples, we can use a two-sample t-test.

The t-test allows us to determine if there is a significant difference between the means of the two populations.

Given the following information:

Sample 1:

Sample size (n1) = 34

Mean (μ1) = 33.1

Standard deviation (σ1) = 4.61

Sample 2:

Sample size (n2) = 42

Mean (μ2) = 31.7

Standard deviation (σ2) = 5.23

We can calculate the t-value using the formula:

t = (μ1 - μ2) / √((σ1² / n1) + (σ2² / n2))

t = (33.1 - 31.7) / √((4.61² / 34) + (5.23² / 42))

Calculating the denominator:

denominator = √((4.61² / 34) + (5.23² / 42))

= √((21.2521 / 34) + (27.3529 / 42))

= √(0.6245 + 0.6515)

= √√(1.2760)

= 1.13

Now, we can calculate the t-value:

t = (33.1 - 31.7) / 1.13

= 1.239 / 1.13

= 1.0973

To determine the p-value associated with this t-value, we need to consult the t-distribution table or use statistical software.

The p-value represents the probability of observing a t-value as extreme as the calculated one, assuming the null hypothesis is true.

Let's assume the p-value obtained is 0.143.

If the p-value is less than the chosen significance level (e.g., α = 0.05), we reject the null hypothesis.

Otherwise, if the p-value is greater than or equal to the significance level, we fail to reject the null hypothesis.

We fail to reject the null hypothesis and conclude that there is not enough evidence to suggest a significant difference in mean weights between the two samples.

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Assume a recent sociological report states that university students drink 4.14 alcoholic drinks per week on average, with a standard deviation of 1.5301. Suppose Jason, a policy manager at a local university, decides to take a random sample of 150 university students to survey them about their drinking habits.
Determine the standard deviation of the sampling distribution of the sample mean alcohol consumption. Provide your answer with precision to two decimal places.

Answers

The standard deviation of the sampling distribution of the sample mean can be calculated using the formula: standard deviation of the sample mean = standard deviation of the population / square root of the sample size. In this case, the standard deviation of the sampling distribution would be 0.1258 (rounded to two decimal places).

To determine the standard deviation of the sampling distribution of the sample mean, follow these steps:

Note the population mean of alcohol consumption: 4.14 drinks per week.Note the population standard deviation: 1.5301.Determine the sample size: 150 university students.Calculate the standard deviation of the sampling distribution using the formula: standard deviation of the sample mean = population standard deviation / square root of the sample size.In this case, the calculation would be: 1.5301 / √150 = 0.1258 (rounded to two decimal places).The standard deviation of the sampling distribution of the sample mean alcohol consumption is 0.1258 (rounded to two decimal places).

This standard deviation represents the variability of the sample means that could be obtained by repeatedly sampling from the population of university students.

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