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12345 [234]
3 years ago
15

Please help me answer the question in the picture!

Mathematics
1 answer:
Natasha2012 [34]3 years ago
7 0

Answer:

x=75

Step-by-step explanation:

A( I mean angle)+B+C+D=360°

A=C

B=D

2A+2D=360°

A+D=180°

30°+x+x=180°

2x=150°

x=75

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Complete the equation of the line through (1,4) and (2,2)<br><br><br> Plz this is a Quiz!!
bagirrra123 [75]

Answer:

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Step-by-step explanation:

7 0
2 years ago
Evaluate the expression: a(b – c) if a = -8, b = 12, and c = 4
Llana [10]

Answer:

-64

Step-by-step explanation:

a(b - c)

a = -8

b = 12

c = 4

So, you'd plug in those numbers:

-8(12 - 4)

You'd start within the parenthesis's, so:

(12 - 4), which equals (8)

-8(8)

= -64

8 0
2 years ago
Read 2 more answers
Suppose a die is tossed 5 times. what is the probability of getting exactly 2 fours?
Tasya [4]

Answer:

the probability of getting exactly 2 fours is 0.16

Step-by-step explanation:

The probability of obtaining a number that is four = ¹/₆

The probability of obtaining a non 4 number = 1 - ¹/₆ = ⁵/₆

The number of ways 2 fours can be arrange in five numbers = ⁵C₂ = 10 ways

If the die is tossed five times, the probability of the events is calculated as;

P = 10 x (¹/₆)² x (⁵/₆)³

P = 10 x (¹/₃₆) x  (¹²⁵/₂₁₆)

P = 10 x 0.02778 x 0.5787

P = 0.16

Therefore, the probability of getting exactly 2 fours is 0.16

7 0
2 years ago
Please help immediately this is 2 days late
laiz [17]

Answer:

51

Step-by-step explanation:

8 0
3 years ago
Consider an urn containing 8 white balls, 7 red balls and 5 black balls.
weqwewe [10]

Answer + Step-by-step explanation:

1) The probability of getting 2 white balls is equal to:

=\frac{8}{20} \times \frac{7}{19}\\\\= 0.147368421053

2) the probability of getting 2 white balls is equal to:

=C^{2}_{5}\times (\frac{8}{20} \times \frac{7}{19}) \times (\frac{12}{18} \times \frac{11}{17} \times \frac{10}{16})\\=0.397316821465

3) The probability of getting at least 72 white balls is:

=C^{72}_{150}\times \left( \frac{8}{20} \right)^{72}  \times \left( \frac{7}{20} \right)^{78}  +C^{73}_{150}\times \left( \frac{8}{20} \right)^{73}  \times \left( \frac{7}{20} \right)^{77}  + \cdots +C^{149}_{150}\times \left( \frac{8}{20} \right)^{149}  \times \left( \frac{7}{20} \right)^{1}  +\left( \frac{8}{20} \right)^{150}

=\sum^{150}_{k=72} [C^{k}_{150}\times  \left( \frac{8}{15} \right)^{k}  \times \left( \frac{7}{15} \right)^{150-k}]

5 0
1 year ago
Read 2 more answers
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