a.) 1/2n 4=3−5/n b.) 1/2(n 4)=5/n−3 c.) 1/2(n 4)=3−5/n d.) 1/2n 4=5/n−3
Let's assume the number be n.
So, sum of number and four can be written as n+4.
Now one-half of n+4 will be .
Quotient of five and the number can be written as .
Now three less \frac{5}{n} can be written as \frac{5}{n} -3.
So, one-half the sum of a number and four is three less than the quotient of five and the number can be converted into equation as follows:
So, b is the correct choice.
Answer:
Statement A is correct about Sam's box and whisker plot.
Step-by-step explanation:
We have been given a box plot and we are asked to find out true statement according to the box plot.
Since we know data represented by box plot is divided in four equal parts.
Upon looking at our box plot we can see that our data is symmetric. It's median is 15, which means half the math assignments have less than 15 problems and half of the math assignments have more than 15 problems.
Interquartile range represents 50% values of data and it is the difference between upper quartile and lower quartile. IQR is not affected by outliers.
Upon substituting given values from box plot we get,
From IQR we can conclude that half of the assignments contained 15 problems or fewer.Therefore, option A is the correct choice.
To inscribe a circle in the attached triangle, we must follow the following steps. The resultant figure is the first one below.
1. Draw the angle bisector of angle JKL. The straight line in the green one.
2. Draw the angle bisector of angle KLJ. The straight line in the blue one.
3. Draw point D at these two lines.
4. Draw a line through point D perpendicular to side JK. The straight line in the red one.
5. Mark point G at the red line and side JK.
6. Draw a circle with center at D (point of concurrency) and a radius of DG.
To circumscribe a circle in the attached triangle, we must follow the following steps:
1. Draw the perpendicular bisector of the side FG. The straight line in the red one. The resultant figure is the second one below.
2. Draw the perpendicular bisector of the side GH. The straight line in the blue one.
3. The center of the circle is the intersection of these two lines
4. Draw a circle with center at D (point of concurrency).
<h3>3. Construct the two lines tangents to circle</h3>A tangent to a circle is a straight line that touches the circle at an only point. If a point A is outside a circle, we can draw two lines that pass through this point and are tangent to the circle at two different points each. So this is shown in the third figure. Lines red and blue are tangent to the circle at two different points.
