Which of the following representations are functions?

Help i,m suck and i cant find the answer

Anestetic [448]

The answer is 2x
3
+5x
2
−x−6

7 0

2 years ago

Given that the measure of ∠x is 149°, and the measure of ∠y is 110°, find the measure of ∠z.

Rom4ik [11]

Answer:

∠z is 79°

Step-by-step explanation:

8 0

3 years ago

Jane must get at least three of the four problems on the exam correct to get an A. She has been able to do 80% of the problems o

NISA [10]

Answer:

a) There is n 81.92% probability that she gets an A.

b) If she gets the first problem correct, there is an 89.6% probability that she gets an A.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the answer is correct, or it is not. This means that we can solve this problem using binomial distribution probability concepts.

Binomial probability distribution

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

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

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

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

And $\pi$ is the probability of X happening.

For this problem, we have that:

The probability she gets any problem correct is 0.8, so $\pi = 0.8$ .

(a) What is the probability she gets an A?

There are four problems, so $n = 4$

Jane must get at least three of the four problems on the exam correct to get an A.

So, we need to find $P(X \geq 3)$

$P(X \geq 3) = P(X = 3) + P(X = 4)$

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

$P(X = 3) = C_{4,3}.(0.80)^{3}.(0.2)^{1} = 0.4096$

$P(X = 4) = C_{4,4}.(0.80)^{4}.(0.2)^{0} = 0.4096$

$P(X \geq 3) = P(X = 3) + P(X = 4) = 2*0.4096 = 0.8192$

There is n 81.92% probability that she gets an A.

(b) If she gets the first problem correct, what is the probability she gets an A?

Now, there are only 3 problems left, so $n = 3$

To get an A, she must get at least 2 of them right, since one(the first one) she has already got it correct.

So, we need to find $P(X \geq 2)$

$P(X \geq 3) = P(X = 2) + P(X = 3)$

$P(X = 2) = C_{3,2}.(0.80)^{2}.(0.2)^{1} = 0.384$

$P(X = 4) = C_{3,3}.(0.80)^{3}.(0.2)^{0} = 0.512$

$P(X \geq 3) = P(X = 2) + P(X = 3) = 0.384 + 0.512 = 0.896$

If she gets the first problem correct, there is an 89.6% probability that she gets an A.

3 0

3 years ago

Round 29.315 to the nearest hundredth

castortr0y [4]

Answer:

29.32

Step-by-step explanation:

sense 1 is the hundredth you just round now so 5 is big enough so it would make the hundreth number go up by 1.

6 0

3 years ago

A production manager at a wall clock company wants to test their new wall clocks. The designer claims they have a mean life of 1

Harman [31]

The probability that the mean clock life would differ from the population mean by greater than 12.5 years is 98.30%.

Given mean of 14 years, variance of 25 and sample size is 50.

We have to calculate the probability that the mean clock life would differ from the population mean by greater than 1.5 years.

μ=14,

σ= $\sqrt{25}$ =5

n=50

s orσ =5/ $\sqrt{50}$ =0.7071.

This is 1 subtracted by the p value of z when X=12.5.

So,

z=X-μ/σ

=12.5-14/0.7071

=-2.12

P value=0.0170

1-0.0170=0.9830

=98.30%

Hence the probability that the mean clock life would differ from the population mean by greater than 1.5 years is 98.30%.

Learn more about probability at brainly.com/question/24756209

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There is a mistake in question and correct question is as under:

What is the probability that the mean clock life would differ from the population mean by greater than 12.5 years?

5 0

2 years ago