First you solve for the area of the unshaded triangle. Use the formula of area of triangle which is 1/2bh to find. 1/2(9)(4) = 18 cm^2. The area of the unshaded triangle is 18cm^2. Now we need to find the area of the rectangle. We can find that by using the formula for area of rectangle which is (b)(h). 12*4 = 48 cm ^2. Now that we have the area of the triangle and the area of the rectangle, we just need to subtract the area of the triangle from the area of the rectangle to get the area of the shaded region. 48 cm^2 - 18 cm^2 = 30 cm^2.
So our answer is: The area of the shaded region is 30 cm^2. In this case it’s option C or the third option for your quiz.
ANSWER
9
EXPLANATION
We want to find the distance between the points (3, -5) and (-6, -5).
The given points have the same y-coordinates .
This means it is a horizontal line.
We use the absolute value method to find the distance between the two points.
We find the absolute value of the distance between the x-values.
The distance between the two points is
|3--6|=|3+6|=|9|=9
John decided to look at new and used cars. John found a used car for $6000. A new car is $18000, so what percent of the price of a new car does John pay for a used car? Round your answer to the nearest tenth if necessary.
Answer:
216 ft
Step-by-step explanation:
you multiply 12 and 18
Answer:
There is a 24.3% probability that one of the calculators will be defective.
Step-by-step explanation:
For each calculator, there are only two possible outcomes. Either it is defective, or it is not. So we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which 
is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
The probability of a defective calculator is 10 percent.
This means that 
If 3 calculators are selected at random, what is the probability that one of the calculators will be defective
This is P(X = 1) when n = 3. So
There is a 24.3% probability that one of the calculators will be defective.
