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Westkost [7]
3 years ago
8

If two distinct lines intersect, which MUST be true?

Mathematics
1 answer:
emmainna [20.7K]3 years ago
7 0

Answer:

Answer options

A) the lines are parallel

B) the lines are perpendicular

C) the lines form angles at the intersection

D) the lines form right angles at the intersection

the answer is C

Step-by-step explanation:

I got it right on my test

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How far is 0, 7 from the origin
Delvig [45]

Answer:

7

Step-by-step explanation:

Count the numbers in between (0,0) and (0,7).

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3 years ago
A public health organization reports that 40%of baby boys 6-8 months old in the United
Sveta_85 [38]

Using the binomial distribution, the probabilities are given as follows:

  • 0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.
  • 0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.
  • Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

<h3>What is the binomial distribution formula?</h3>

The formula is:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

The values of the parameters for this problem are:

n = 10, p = 0.4.

The probability that more than 4 weigh more than 20 pounds is:

P(X > 4) = 1 - P(X \leq 4)

In which:

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

Then:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.4)^{0}.(0.6)^{10} = 0.0061

P(X = 1) = C_{10,1}.(0.4)^{1}.(0.6)^{9} = 0.0403

P(X = 2) = C_{10,2}.(0.4)^{2}.(0.6)^{8} = 0.1209

P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.2150

P(X = 4) = C_{10,4}.(0.4)^{4}.(0.6)^{6} = 0.2502

Hence:

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0061 + 0.0403 + 0.1209 + 0.2150 + 0.2502 = 0.6325

P(X > 4) = 1 - P(X \leq 4) = 1 - 0.6325 = 0.3675

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.

The probability that fewer than 3 weigh more than 20 pounds is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0061 + 0.0403 + 0.1209 = 0.1673

0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.

For more than 7, the probability is:

P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 8) = C_{10,8}.(0.4)^{8}.(0.6)^{2} = 0.0106

P(X = 9) = C_{10,9}.(0.4)^{9}.(0.6)^{1} = 0.0016

P(X = 10) = C_{10,10}.(0.4)^{10}.(0.6)^{0} = 0.0001

Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

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