The answer is 7. If you follow the order of PEMDAS this is an easy question.
Answer:
1/18
Step-by-step explanation:
1/6 * 1/3
1 *1 = 1
6 * 3 = 18
1/18
Answer:
m∠RQS = 72°
m∠TQS = 83°
Step-by-step explanation:
m∠RQS +m ∠TQS = m∠RQT
The two angles combine to make a larger angle
So
m∠RQS = (4x - 20)
m∠TQS = (3x + 14)
(4x - 20) + (3x + 14) = 155
Group the Xs and the constants
4x + 3x - 20 + 14 = 155
Combine like terms
7x - 6 = 155
Add 6 to both sides
7x = 161
Divide by 7 on both sides
x = 23
Check:
4(23) - 20 + 3(23) + 14 = 155
92 - 20 + 69 + 14 = 155
155 = 155
But we need to find m∠RQS and m∠TQS. So plug in x = 23 to the values.
m∠RQS = 4(23) - 20 = 72°
m∠TQS = 3(23) + 14 = 83°
Checking:
72 + 83 = 155
Answer:
The solution is the point (0.5,-3)
Step-by-step explanation:
we have
----> equation A
----> equation B
Solve the system by substitution
Substitute equation B in equation A
Solve for x
Adds 5 both sides
Divide by 4 both sides
therefore
The solution is the point (0.5,-3)
<em>Verify your answer using the graph</em>
using a graphing tool
Remember that the solution of the system of equations is the intersection point both graphs
The intersection point is (0.5,-3)
therefore
The solution is the point (0.5,-3)
see the attached figure
Answer:
18
Step-by-step explanation:
Given the data:
180kg, 250kg, 200kg, 209kg, 195kg, 205kg, 190kg, 188kg, 192kg
The interquartile range (IQR) = Q3 - Q1
Reordering the data:
180, 188, 190, 192, 195, 200, 205, 209, 250
Q3 = 3/4(n+1)th term
Q3 = 3/4(10) = 7.5 th term = (205+209)/2 = 207
Q1 = 1/4(n+1)th term
Q1 = 1/4(10) = 2.5th term = (188+190)/2 = 189
Q3 - Q1 = 207 - 189 = 18
