5y - 16 + 3y + 20 = 180 degrees
Combine like terms on the left side
8y + 4 = 180
Subtract 4 from both sides.
8y = 176
Divide both sides by 8
y = 22
Plug 22 into y for each angle.
3y + 20 = 3(22) + 20 = 66 + 20 = 86 degrees
5y - 16 = 5(22) - 16 = 110 - 16 = 94
The angle opposite 5y - 16 also equals 94 because they are vertical angles.
The angle opposite of 3y + 20 also equals 86 because they are vertical angles
Answer:The scale is 1:12, and the actual car has 16-in wheels.
Step-by-step explanation:
So the setup would be a proportion:
1 over 12=x over 16
Then cross-multiply
12x=16
Divide by 12
X=16/12
=4/3 or 1 and 1/3 in. wheels on the model car.
Answer:
depends on data provided
Step-by-step explanation:
can explain the dat youd been given
Answer:
i think it is c but not sure
Step-by-step explanation:
What is the outlier in the following data set:<br>
15,11,10,8,9,1,8,7,5,4,2,3, and 37?
NeTakaya
Step-by-step explanation:
The steps to find an outlier:
1. Put the data in numerical order.
2. Find the median.
3. Find the medians for the top and bottom parts of the data. This divides the data into 4 equal parts.
The median with the smallest value is called Q1. The median for all the values - usually just called the median is also called Q2. The median with the largest value is Q3.
4. Subtract...Q3 - Q1. This value is the InterQuartileRange or IQR. Remember that the range means taking the largest minus the smallest. This is a special range having to do with the quartiles.
5. Multiply...1.5 * IQR
6. Take your answer from #5 and do 2 things with it. A). Subtract it from Q1 and B) Additional to Q3.
7. Look at all your data points. If any are SMALLER than Q1 - 1.5 *IQR, they are outliers. If any are LARGER than Q3 + 1.5 *IQR, they are also outliers.
For your data....the median, Q2 is
(43+38)/2 = 40.5.
Q1 = (30+26)/2 = 28.
Q3 = (54+52)/2 = 53
The IQR is 53 - 28 = 25
1.5 * IQR = 37.5
Q1 - 37.5 = 28 - 37.5 = -9.5. There is no data value less than -9.5.
Q3 + 37.5 = 53 + 37.5 = 90.5. there is no data value greater than 90.5.
My conclusion is that there are no outliers in this data.
I hope this helps!
