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Vsevolod [243]
2 years ago
15

Plzz help me i will give bairnlest frfr and i will give you a 10/10 and a thank you

Mathematics
1 answer:
Rus_ich [418]2 years ago
8 0

Answer:

the answer is 210,000

Step-by-step explanation:

30×7,000= 210,000

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Segments AB, DC and Ec intersect at point C. Angel DCE measures 148. Find the value of x.
Anna71 [15]

The image is attached.

Answer:

x = 16°

Step-by-step explanation:

Segment AB, segment DC and segment EC all intersect at point C.

Here, we are told angle DCE measures 148°.

This is a semi-circle, and the total angle of a semi-circle is 180°.

Which means, x+x+148 = 180

Solving for x, we have:

x+x+148 = 180

2x + 148 = 180

2x = 180 - 148

2x = 32

x = \frac{32}{2}

x = 16°

The value of x is 16°

6 0
3 years ago
Original amount = 80 dollars. 20 percent off is 64 dollars. 20 percent off is question mark.
lord [1]

Answer:

48

Step-by-step explanation:

80/100=0.8

0.8x60=48

7 0
2 years ago
Do I have a rare user? Just asking :)
Andre45 [30]

Answer:

lol yes?

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes
saul85 [17]

Answer:

a)0.08  , b)0.4  , C) i)0.84  , ii)0.56

Step-by-step explanation:

Given data

P(A) =  professor arrives on time

P(A) = 0.8

P(B) =  Student aarive on time

P(B) = 0.6

According to the question A & B are Independent  

P(A∩B) = P(A) . P(B)

Therefore  

{A}' & {B}' is also independent

{A}' = 1-0.8 = 0.2

{B}' = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B')  (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

P(\frac{B'}{A}) = \frac{P(B'\cap A)}{P(A)} = \frac{0.4\times 0.8}{0.8} = 0.4

Part c)

Assume the events are not independent

Given Data

P(\frac{{A}'}{{B}'}) = 0.4

=\frac{P({A}'\cap {B}')}{P({B}')} = 0.4

P({A}'\cap {B}') = 0.4 x P({B}')

= 0.4 x 0.4 = 0.16

P({A}'\cap {B}') = 0.16

i)

The probability that at least one of them is on time

P(A\cup B) = 1- P({A}'\cap {B}')  

=  1 - 0.16 = 0.84

ii)The probability that they are both on time

P(A\cap  B) = 1 - P({A}'\cup {B}') = 1 - [P({A}')+P({B}') - P({A}'\cap {B}')]

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0
3 years ago
Might just use all of my 900 remaining points on my next random questions​
pychu [463]

Answer:

do what you want

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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