Approximate circumference of a semicircle with a radius of 30 centimeters is 94.2 centimeter
<em><u>Solution:</u></em>
Given semicircle with a radius of 30 centimeters
<em><u>To find: approximate circumference of a semicircle</u></em>
To find circumference of semicircle we can divide the circumference of circle by 2
circumference of semicircle = circumference of circle 
2
Substituting the given radius = 30 cm,
Thus approximate circumference of a semicircle with a radius of 30 centimeters is 94.2 centimeter
Answer:
Bivariate Frequency Distribution.
Cumulative Frequency Distribution.
Relative Frequency Distribution.
The length of the side QR in the task content can be determined as; 12.22cm.
<h3>What is the length of the third side QR?</h3>
It follows from the task content that 2 lengths and one included angle are given in the task content. It therefore follows that the length of side QR can be evaluated by the cosine rule as follows;
QR² = 13² + 4² - (2×4×13×cos70)
QR = √149.4
QR = 12.22cm
Read more on cosine rule;
brainly.com/question/23720007
#SPJ1
3.06% Next time just google it my guy. (longer number 3.06499205451137%)
Answer: Exam A
Step-by-step explanation:
We must analyze how far are you from the mean in both cases, where the "step" that we will use to measure is the standard deviation.
In exam A, the mean is 20.5 and the standard deviation is 4.9.
If you scored a 27; then you need to see:
20.5 + 4.9 = 25.4
20.5 + 4.9 + 4.9 = 30.3
So you are within two times the standard deviation (more than the mean).
In the B exam, the mean is 1022 and the standard deviation is 214, where you scored 1209.
1022 + 214 = 1236
So in this exam, you are by one standard deviation away from the media.
With this, you can see that you did score better in exam A.
