Use synthetic division and the Remainder Theorem to find P(a).

Given }(x) = x and g(x) = 1 - 5x, find ( o 3)(x) and its domain

Citrus2011 [14]

I think it’s b buhh im really not sure

7 0

3 years ago

Literally if somebody will do my math test I will give you all of my points

Vika [28.1K]

Answer:

how many points u got

Step-by-step explanation:

7 0

3 years ago

How many magazines at most can the company print daily,

Dvinal [7]

Answer:

The most they can print is 800

Step-by-step explanation:

.5 N + 2.5M ≤ 6000

We are given N = 8000

Substitute this into the inequality

.5 (8000) + 2.5M ≤ 6000

4000 +2.5M≤ 6000

Subtract 4000 from each side

4000-4000 +2.5M≤ 6000 -4000

2.5M≤ 2000

Divide each side by 2.5

2.5M/2.5≤ 2000/2.5

M ≤800

6 0

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

Joaquin buys 3 packs light bulbs. Each pack contains 12 lights bulbs. After changing the bulbs in his house he has 15 light bulb

denpristay [2]

Answer:

Step-by-step explanation:

3x12=36 light bulbs total

36-15=21 used light bulbs with 15 left unused

8 0

3 years ago