THEOREM:
- Angle sum property —: The sum of interior angles of a triangle is 180°.
ANSWER:
Let the missing angle be x.
By angle sum property,
- ∠x + 20° + 80° = 180°
- ∠x + 100 = 180°
- ∠x = 180° – 100°
- ∠x = 80°
So, <u>Correct choice</u> - [C] 80°.
OptionAStep-by-step explanation:95% confidence interval is obtained using the following formulaCOnfidence interval lower bound = mean - critical value * sigma/sq rt nand upper bound = mean + critical value*sigma/sq rt nThus margin of error = critical value*sigma/rt nvaries indirectly as n provided others are remaining the same.Hence lower sample indicates higher margin of errorOut of 4 options given we find that 1 option has maximum margin of error as 21%Hence option. A is most likely coming from a small sample.
Answer:
no
Step-by-step explanation:
4 is greater than 0
<- Distributive Property
<- Combine Like Terms
If you're trying to solve for 0:
<- Subtracted 18 from both sides
<- Divided both sides by 60 and then simplified.
<- Fraction Form
<- Decimal Form
Give Brainliest for simple answer plz :P
Let us formulate the independent equation that represents the problem. We let x be the cost for adult tickets and y be the cost for children tickets. All of the sales should equal to $20. Since each adult costs $4 and each child costs $2, the equation should be
4x + 2y = 20
There are two unknown but only one independent equation. We cannot solve an exact solution for this. One way to solve this is to state all the possibilities. Let's start by assigning values of x. The least value of x possible is 0. This is when no adults but only children bought the tickets.
When x=0,
4(0) + 2y = 20
y = 10
When x=1,
4(1) + 2y = 20
y = 8
When x=2,
4(2) + 2y = 20
y = 6
When x=3,
4(3) + 2y= 20
y = 4
When x = 4,
4(4) + 2y = 20
y = 2
When x = 5,
4(5) + 2y = 20
y = 0
When x = 6,
4(6) + 2y = 20
y = -2
A negative value for y is impossible. Therefore, the list of possible combination ends at x =5. To summarize, the combinations of adults and children tickets sold is tabulated below:
Number of adult tickets Number of children tickets
0 10
1 8
2 6
3 4
4 2
5 0
