Help easy math I think

Which of the following is correct about environmental policy?a. The national forests are protected from the exploitive ""dual us

Andrew [12]

Answer:

Option C is correct.

Environmental regulation has led to dramatic improvements in air and water quality.

Step-by-step explanation:

- The national forests are protected from the exploitive ""dual use policy.

The dual use policy on any idea, technology, breakthrough etc. means such concept can be used for its intended purpose and for military use.

Natural forests are not excluded from this dual use policy & plenty forest exploiters hide under this policy to literally exploit natural forests.

- The Environmental Protection Agency was elevated to cabinet status in 1998 by President Clinton and the Republican Congress

This is not true as the EPA is still not a Cabinet department, but the Administrator is normally given cabinet rank.

- Environmental regulation has led to dramatic improvements in air and water quality.

Regulations have led plenty industries at the forefront of pollution to sit up and fix up. Strict laws and hefty fines have discouraged various water and air pollution means since those regulations came into place. This is the only correct statement about environmental policy amongst the options.

- Policymakers always give more consideration to environmental protection than to economic development when the two conflict.

This is true sometimes, but 'always'!? No, policymakers do not always give more consideration to environmental protection than to economic development when the two conflict.

6 0

3 years ago

a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-

Tresset [83]

Answer:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

$M = \frac{1}{2}(Lower + Upper)$

Where

$Lower \to$ Lower class interval

$Upper \to$ Upper class interval

So, we have:

Class 63-65:

$M = \frac{1}{2}(63 + 65) = 64$

Class 66 - 68:

$M = \frac{1}{2}(66 + 68) = 67$

When the computation is completed, the frequency distribution will be:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

$\bar x = 70.2903$