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weqwewe [10]
2 years ago
7

Identify any zeros of y = x2 + 4x + 5

Mathematics
2 answers:
pav-90 [236]2 years ago
6 0

Answer:

<h2>B. There are no zeros</h2>

Step-by-step explanation:

\text{The zeros for y = 0:}

x^2+4x+5=0\qquad\text{subtract 5 from both sides}\\\\x^2+2(x)(2)=-5\qquad\text{add}\ 2^2\ \text{to both sides}\\\\x^2+2(x)(2)+2^2=-5+2^2\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\(x+2)^2=-5+4\\\\(x+2)^2=-1

Andrei [34K]2 years ago
4 0

Answer:

B. There are no zeros

Step-by-step Explanation:

None of the given point even lie on the parabola (curved line) of the graph, so they can't possibly be zeros. The only explanation is that there simply aren't any zeros.

So your answer is...B. There are no zeros.

Hope this helps!

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JulijaS [17]

a. Your answer is correct.

b. Divide the given value of sigma by √n where n = sample size = 7. Then perform the same calculation you did for part a. The probabiilty is 0.4427.

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3 years ago
Which of the following are coefficient
malfutka [58]

Answer:

2 and -3

Step-by-step explanation:

because a coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression (e.g. 4 in 4x).

so in 2x-3y+5, 2 and -3 are the coefficients.

4 0
3 years ago
In a classroom, 4/6of the students are wearing
Airida [17]
The answer is
4/6-3/6=1/6
6 0
1 year ago
When grading an exam, 90% of a professor's 50 students passed. If the professor randomly selected 10 exams, what is the probabil
Damm [24]

Using the binomial distribution, it is found that there is a:

a) 0.9298 = 92.98% probability that at least 8 of them passed.

b) 0.0001 = 0.01% probability that fewer than 5 passed.

For each student, there are only two possible outcomes, either they passed, or they did not pass. The probability of a student passing is independent of any other student, hence, the binomial distribution is used to solve this question.

<h3>What is the binomial probability 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.

In this problem:

  • 90% of the students passed, hence p = 0.9.
  • The professor randomly selected 10 exams, hence n = 10.

Item a:

The probability is:

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)

In which:

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

P(X = 8) = C_{10,8}.(0.9)^{8}.(0.1)^{2} = 0.1937

P(X = 9) = C_{10,9}.(0.9)^{9}.(0.1)^{1} = 0.3874

P(X = 10) = C_{10,10}.(0.9)^{10}.(0.1)^{0} = 0.3487

Then:

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1937 + 0.3874 + 0.3487 = 0.9298

0.9298 = 92.98% probability that at least 8 of them passed.

Item b:

The probability is:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

Using the binomial formula, as in item a, to find each probability, then adding them, it is found that:

P(X < 5) = 0.0001

Hence:

0.0001 = 0.01% probability that fewer than 5 passed.

You can learn more about the the binomial distribution at brainly.com/question/24863377

3 0
2 years ago
N − 5 ≥− 2 please show your work to how u got your answer.
kondaur [170]

Answer:

n ≥ 3

Step-by-step explanation:

add (-5) to both sides of equations. Solve for n.

3 0
3 years ago
Read 2 more answers
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