Total of angles in 180 so:
46+90+8x+4=180
8x=40
x=5
Question:
1. The females worked less than the males, and the female median is close to Q1.
2. There is a high data value that causes the data set to be asymmetrical for the males.
3. There are significant outliers at the high ends of both the males and the females.
4. Both graphs have the required quartiles.
Answer:
The correct option is;
1. The females worked less than the males, and the female median is close to Q1
Step-by-step explanation:
Based on the given data, we have;
For males
Minimum = 0
Q1 = 1
Median or Q2 = 20
Q3 = 25
Maximum = 50
For females;
Minimum = 0
Q1 = 5
Median or Q2 = 6
Q3 = 10
Maximum = 18
Therefore, the values of data that affect the statistical measures of spread and center are that
The females worked less than the males as such the statistical data for the females have less variability than the males in terms of interquartile range
Also the female median is very close to Q1, therefore it affects the definition of a measure of center.
Answer:
1) 35+5w=110
2) the starting heart rate is 73 beats per minute
3)B
4) 6mins
Step-by-step explanation:
1) since she deposits 5 dollars each week we can multiply 5 and w to find out how much money hes depositing after all the weeks are done. We need to add that with 35 to get the TOTAL money in the bank account.
2) 73 is the y intercept. The y intercept is the number when "x" is 0. So when hes just starting out/not excersizing his heart beats 73 beats per min.
3) we know that 10% of her tips is given back to her hence x(number of tips) is multiplied by 0.1, and added to 50 to find the total amount of money she receives
4) 6 mins for 1 order
