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Minchanka [31]
2 years ago
6

HELPPPP!!!PLEASE!!!!!

Mathematics
2 answers:
nadezda [96]2 years ago
6 0
The two angles are vertically opposite meaning that 156 is equal to 3x + 36.
So, 3x + 36 = 156.
Now, solve it like you would.
1. Bring 36 to the other side.
3x = 156 - 36
3x = 120
2. Now, divide each side by 3
120/3=40
3x/3 = 1
x = 40
lbvjy [14]2 years ago
3 0

Answer:

x=40

Step-by-step explanation:

Both angles are equal. So the equation needs to equal 156.

Just plug in points for x until the answer is 156.

In this case, when we plug in 40, it works!

(3*40)+36=156

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Answer:

The correct answer is B

Step-by-step explanation:

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5 0
3 years ago
Let f(x)=5x+5 find f(-1)<br><br>I need your help with this. please help me​
Alisiya [41]

f(x) = 5x+5\\\\f(-1) = 5(-1) +5  = -5+5=0

5 0
2 years ago
If town A had 35 people, 12 left, 13 joined, 39 joined, 25 got killed, how many people were there left?
Makovka662 [10]

Answer:

50

Step-by-step explanation:

35 - 12 + 13 + 39 - 25 = 50

6 0
3 years ago
What is y=x+2and 2x+y=8
Leya [2.2K]

Answer:

1. x=y−2

2. -1/2y + 4

Step-by-step explanation:

5 0
3 years ago
An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What
galina1969 [7]

Answer:

Part A:

The probability that all of the balls selected are white:

P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\      P(A)=\frac{5}{66}=0.075757576

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516

Step-by-step explanation:

A is the event all balls are white.

D_i is the dice outcome.

Sine the die is fair:

P(D_i)=\frac{1}{6} for i∈{1,2,3,4,5,6}

In case of 10 black and 5 white balls:

P(A|D_1)=\frac{5_{C}_1}{15_{C}_1} =\frac{5}{15}=\frac{1}{3}

P(A|D_2)=\frac{5_{C}_2}{15_{C}_2} =\frac{10}{105}=\frac{2}{21}

P(A|D_3)=\frac{5_{C}_3}{15_{C}_3} =\frac{10}{455}=\frac{2}{91}

P(A|D_4)=\frac{5_{C}_4}{15_{C}_4} =\frac{5}{1365}=\frac{1}{273}

P(A|D_5)=\frac{5_{C}_5}{15_{C}_5} =\frac{1}{3003}=\frac{1}{3003}

P(A|D_6)=\frac{5_{C}_6}{15_{C}_6} =0

Part A:

The probability that all of the balls selected are white:

P(A)=\sum^6_{i=1} P(A|D_i)P(D_i)

P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\      P(A)=\frac{5}{66}=0.075757576

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

We have to find P(D_3|A)

The data required is calculated above:

P(D_3|A)=\frac{P(A|D_3)P(D_3)}{P(A)}\\ P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516

7 0
3 years ago
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