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irinina [24]
3 years ago
8

Which polygon appears to be regular?

Mathematics
2 answers:
wariber [46]3 years ago
6 0
Figure B. is correct
Andrei [34K]3 years ago
4 0
Figure D seems to be Regular. D. Figure D, is your correct answer.
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Evaluate |3 - 5 + 7|.<br><br> -5<br> 5<br> -15<br> 15
Mnenie [13.5K]
The right answer is 5
7 0
3 years ago
The average number of pounds of red meat a person consumes each year is 196 with a standard deviation of 22 pounds (source: amer
disa [49]
Given:
Mean, μ = 196
Std. dev., σ = 22
A sample size of 50 (> 30) is large enough to provide meaningful data.

The random variable is x = 200.
The z-score is
z = (x - μ)/σ = (200 - 196)/22 = 0.1818

From normal distribution tables, obtain
Prob(X < 200) = 0.572 = 57.2%

Answer: 57.2%
7 0
3 years ago
I need help on number 29
choli [55]

x = All real numbers or 0.

First you should distribute both sides of the equation. This gives you 42x+42=42x+42.

Then, subtract 42 from both sides. This gives you 42x=42x.

Now, divide 42 on both sides. This gives you x=0 or all real numbers.

7 0
3 years ago
Read 2 more answers
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
Create two whole number division problems
AlekseyPX
98 divided by 2 equals 49 55 divided by 5 equals 11
3 0
3 years ago
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