Since the probability of a student being in either tenth, eleventh, or twelfth grade = 1 = 100% (that is, certainty), then the probability of a randomly drawn student being in twelfth grade is (100-33.7-35.3)% = 31.0%.

When randomly choosing one student from the whole school, it is most likely (35.3%) that the student is in the eleventh grade.

4 0

3 years ago

What is the solution to the equation 3/m+3-m/3-m=m^2+9/m^2-9

Bas_tet [7]

Answer:

no solution

Step-by-step explanation:

If we subtract the left side of the equation, we get ...

$\dfrac{m^2+9}{m^2-9}-\dfrac{3}{m+3}+\dfrac{3}{3-m}=0\\\\\dfrac{m^2+9-3(m-3)-3(m+3)}{(m-3)(m+3)}=0\\\\\dfrac{m^2-6m+9}{(m-3)(m+3)}=0=\dfrac{(m-3)^2}{(m-3)(m+3)}\\\\\dfrac{m-3}{m+3}=0\quad\text{m$\ne$3}$

This equation will equal zero only if m=3, which is disallowed because it makes the denominator zero. Thus, there is no solution.

7 0

3 years ago

The new building is 22.5 inches from the gymnasium in the model. What will be the actual

Len [333]

Answer:

67.5feet

Step-by-step explanation:

Given parameters:

Model distance between building and gymnasium = 22.5 inches

Scale of model : 1 inch = 3 feet

Unknown:

Actual ground distance = ?

To solve this problem, we first must understand the concept of scale. A scale is a relationship that represents a dimension on a map/model compared to the true ground expression. In order to visualize or represent some real life objects on paper or in a computer, we use models. These models are an abstraction of the real world based on scales. There are different ways of representing a scale.

In this problem;

the scale is given as;

1 inch on model represents 3 feet on ground

Now, to find 22.5 inches, simply cross multiply and solve;

If 1 inch on model represents 3 feet on ground

22.5 inches on a model will be = $\frac{22. 5 inches x 3 feet }{1 inches}$

= 67.5feet

Therefore, the actual distance is 67.5feet

8 0

3 years ago