The formula to calculate standard deviation from probability is \sqrt(n*p*(1-p)). n is the sample size, and 200 in this case (number of putts for practice). p is 80% or 0.8, the probability that he can make it. So the standard deviation is \sqrt(200*0.8*(1-0.8)=\sqrt(200*0.8*0.2)=\sqrt(16)=4.
Answer:
A. 217
Step-by-step explanation:
3,472/16 = 217
Answer:
20%
Step-by-step explanation:
In the graph, there is a cycle of increasing and decreasing. First, the original circle increases by 25% into the new, larger circle. Then, the circle does back to its original size. So, to solve we can assign a value to the smaller circle, increase it by 25%, and then find what percentage brings this back to the original value.
It is easiest to assign the value of 100 because it is easy to work with. So, let's say the smaller circle is 100; this increased by 25% is 125. The next step is to find what percentage of 125 brings us back to 100. There are multiple ways to do this such as guess and check or proportions/fractions. Either way, the correct percentage is 20%. If you decrease 125 by 20% you get the correct value of 100. This means that the missing number must be 20%.
When two parallel lines are cut by a transversal, due to the natures of the different relationships among the angles, all of them will be either equal to the measure of the first angle or equal to that angle subtracted from 180. If the transversal is perpendicular to the parallel lines, forms a right angle, then both the angle given and the angle subtracted from 180 will be 90. This means all of the angles will be 90.
The new area will be 320 in²
<em><u>Explanation</u></em>
Lin has a drawing with an area of 20 in² and she increases all sides by a scale factor of 4.
<u>The general rule</u> we need to use here.......
"<em>If the lengths of the sides in a shape are all increased by a scale factor of 
, then the area will be increased by a scale factor of 
"</em>
Here the sides are increased by a scale factor of 4. So, the area will be increased by a scale factor of 
Thus, the new area will be: 
