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2 years ago

Which statements are true? Check all that apply.

Mathematics

2 answers:

Colt1911 [192]2 years ago

6 0

Answer:

A. mArc C B = 120°

C. m∠COB = 2(m∠CAB)

D. m∠COB = 120°

german2 years ago

4 0

Answer:

A,C,D

Step-by-step explanation:

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Could someone help me out?

worty [1.4K]

Since the grade of the numerator and the denominator is the same, then the limit exists and is distinct from 0. The limit of the expression is 4/7.

<h3>How to determine the limit of a rational expression when x tends to infinite</h3>

In this problem we must apply some algebraic handling and some known limits to determine whether the limit exists or not. The limit exists if and only if the result exists.

$\lim_{x \to \infty} \frac{4\cdot x - 1}{7\cdot x + 3}$

$\lim_{x \to \infty} \frac{4\cdot x - 1}{7\cdot x + 3} \cdot \frac{x}{x}$

$\lim_{x \to \infty} \frac{4 - \frac{1}{x} }{7 + \frac{3}{x} }$

$\lim_{x \to \infty} \frac{4}{7}$

4/7

Since the grade of the numerator and the denominator is the same, then the limit exists and is distinct from 0. The limit of the expression is 4/7.

To learn more on limits: brainly.com/question/12207558

#SPJ1

8 0

1 year ago

A machine making computer chips isn't working correctly, and 5% of the computer chips it makes are defective. If an inspector ch

Rainbow [258]

Answer:

0.25% probability that they are both defective

Step-by-step explanation:

For each computer chip, there are only two possible outcomes. Either they are defective, or they are not. The probability of a computer chip being defective is independent of other chips. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

5% of the computer chips it makes are defective.

This means that $p = 0.05$

If an inspector chooses two computer chips randomly (meaning they are independent from each other), what is the probability that they are both defective?

This is P(X = 2) when n = 2. So

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

$P(X = 2) = C_{2,2}.(0.05)^{2}.(0.95)^{0} = 0.0025$