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yulyashka [42]
3 years ago
11

What is 0.2 repeating as a simplified fraction

Mathematics
1 answer:
Lisa [10]3 years ago
6 0
The answer is 2/9 because <span>Since </span>x<span> is recurring in 1 decimal places, we multiply it by 10.</span>
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A<br> 100 cm²<br> b<br> 9 cm²<br> c<br> 225 cm²<br> d<br> 28.26 cm²
nasty-shy [4]

Answer:

28.26 cm^2

Step-by-step explanation:

A = (pi)r^2

A = 3.14 * (3 cm)^2

A = 3.14 * 9 cm^2

A = 28.26 cm^2

6 0
3 years ago
What is the answer to, - (-x - 3)?
bekas [8.4K]

Step-by-step explanation:

(+)(+)=(-)(-)=(+)\\(+)(-)=(-)(+)=(-)\\\\-(-x-3)=-1(-x-3)\qquad\text{use the distributive property}\\\\=(-1)(-x)+(-1)(-3)=x+3

3 0
3 years ago
1/2+1/4+3/4=<br> pls help this my last question on my quiz!
galben [10]

Answer: 1 1/2 or 6/4

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Please help me and I’ll give you brainiest!!!!!!!!!
Aleksandr-060686 [28]

Answer:

Graph it on a graphing calculator

Step-by-step explanation:

6 0
3 years ago
Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving
lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

P(x=0) = ^nC_x P^x (1-P)^n^-^x

P(x=0) = ^5C_0  (0.4)^0 (1-0.4)^5^-^0

P(x=0) = ^5C_0 (0.4)^0 (0.60)^5

P(x=0) = 0.778

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - ^5C_0 (0.4)^0 * (0.6)^5

= 1 - 0.0778

= 0.9222

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2  (0.4)^2  (0.6)^3

= 0.6826

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

= 1 - [^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2 * (0.4)^2  (0.6)^3]

= 1 - 0.6826

= 0.3174

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

7 0
3 years ago
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