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3 years ago

Ok this is not related to school but can any Christians tell me about the "mark of the beast "

Mathematics

2 answers:

Ghella [55]3 years ago

6 0

Answer:

People usally call mark of the beast 666

Step-by-step explanation:

"And that no man might buy or sell, save he that had the mark, or the name of the beast, or the number of his name. Here is wisdom. Let him that hath understanding count the number of the beast: for it is the number of a man; and his number is 666."

KatRina [158]3 years ago

4 0

Answer:

its an evil thing where the government makes u have some sort of identification to be able to buy stuff and pretty much live

Step-by-step explanation:

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3 years ago

The proportion of high school seniors who are married is 0.02. Suppose we take a random sample of 300 high school seniors; a.) F

MrRissso [65]

Answer:

a) Mean 6, standard deviation 2.42

b) 10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c) 14.85% probability that we find less than 4 of the seniors are married.

d) 99.77% probability that we find at least 1 of the seniors are married

Step-by-step explanation:

For each high school senior, there are only two possible outcomes. Either they are married, or they are not. The probability of a high school senior being married is independent from other high school seniors. 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.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

In this problem, we have that:

$n = 300, p = 0.02$

a.) Find the mean and standard deviation of the sample count X who are married.

Mean

$E(X) = np = 300*0.02 = 6$

Standard deviation

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.02*0.98} = 2.42$

b.) What is the probability that, in our sample of 300, we find that 8 of the seniors are married?

This is P(X = 8).

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

$P(X = 8) = C_{300,8}.(0.02)^{8}.(0.98)^{292} = 0.1040$

10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c.) What is the probability that we find less than 4 of the seniors are married?

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

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

$P(X = 0) = C_{300,0}.(0.02)^{0}.(0.98)^{300} = 0.0023$

$P(X = 1) = C_{300,1}.(0.02)^{1}.(0.98)^{299} = 0.0143$

$P(X = 2) = C_{300,2}.(0.02)^{2}.(0.98)^{298} = 0.0436$

$P(X = 3) = C_{300,3}.(0.02)^{3}.(0.98)^{297} = 0.0883$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0023 + 0.0143 + 0.0436 + 0.0883 = 0.1485$

14.85% probability that we find less than 4 of the seniors are married.

d.) What is the probability that we find at least 1 of the seniors are married?

Either no seniors are married, or at least 1 one is. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

From c), we have that $P(X = 0) = 0.0023$ . So

$0.0023 + P(X \geq 1) = 1$

$P(X \geq 1) = 0.9977$

99.77% probability that we find at least 1 of the seniors are married

5 0

3 years ago