2330 - 0540 milatary time

Which of the following is a trinomial with a constant term?

Leto [7]

I think the answer is b

4 0

3 years ago

A square root is categorized as what kind of number? Choices: Irrational Rational

anyanavicka [17]

Rational i believe is the answer i ain't for sure though

6 0

3 years ago

What is the solution to the equation e^3x=12 Round your answer to the nearest hundredth

Lunna [17]

Keywords:

equation, variable, clear, round, centesima, neperian logarithm, exponential

For this case we have the following equation $e ^ {3x} = 12$ , from which we must clear the value of the variable "x" and round to the nearest hundredth. To do this, we must apply properties of neperian and exponential logarithms. By definition:

$ln (e ^ x) = x$

So:

$e ^ {3x} = 12$ We apply Neperian logarithm to both sides:

$ln (e^{3x}) = ln (12)\\3x = ln (12)$

We divide between "3" both sides of the equation:

$\frac {3x} {3} = \frac {ln (12)} {3}\\x = \frac {ln (12)} {3}\\x = 0.828302217$

Rounding out the nearest hundredth we have:

$x = 0.83$

Answer:

$x = 0.83$

3 0

3 years ago

Read 2 more answers

Jayda take 4 hours to deliver 156 newspapers. How many newspapers can Jayda deliver in one hour?

Oksana_A [137]

Answer:

Step-by-step explanation:

39

8 0

3 years ago

Read 2 more answers

23% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and as

zlopas [31]

Answer:

a) There is a 29.42% probability that the number of college students who say they use credit cards because of the rewards program is exactly two.

b) There is a 41.37% probability that the number of college students who say they use credit cards because of the rewards program is more than two.

c) There is a 69.49% probability that the number of college students who say they use credit cards because of the rewards program is between two and five, inclusive.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this exercise using the binomial probability distribution.

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.

In this problem, we have that:

10 students are randomly selected, so $n = 10$ .

23% of college students say they use credit cards because of the rewards program. This means that $\pi = 0.23$

(a) exactly two

This is P(X = 2).

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

$P(X = 2) = C_{10,2}.(0.23)^{2}.(0.77)^{8} = 0.2942$

There is a 29.42% probability that the number of college students who say they use credit cards because of the rewards program is exactly two.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities of these events must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

So

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

$P(X = 0) = C_{10,0}.(0.23)^{0}.(0.77)^{10} = 0.0733$

$P(X = 1) = C_{10,1}.(0.23)^{1}.(0.77)^{9} = 0.2188$

$P(X = 2) = C_{10,2}.(0.23)^{2}.(0.77)^{8} = 0.2942$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0733 + 0.2188 + 0.2942 = 0.5863$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.5863 = 0.4137$

There is a 41.37% probability that the number of college students who say they use credit cards because of the rewards program is more than two.

(c) between two and five inclusive.

This is

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 2) = C_{10,2}.(0.23)^{2}.(0.77)^{8} = 0.2942$

$P(X = 3) = C_{10,3}.(0.23)^{3}.(0.77)^{7} = 0.2343$

$P(X = 4) = C_{10,4}.(0.23)^{4}.(0.77)^{6} = 0.1225$

$P(X = 5) = C_{10,3}.(0.23)^{5}.(0.77)^{5} = 0.0439$

So

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.2942 + 0.2343 + 0.1225 + 0.0439 = 0.6949$

There is a 69.49% probability that the number of college students who say they use credit cards because of the rewards program is between two and five, inclusive.

8 0

3 years ago