All of the following materials handled in vehicle maintenance and recycling can be toxic to fish, wildlife and humans, EXCEPT:

London bought a bag containing 95 small candies. She starts eating the

Savatey [412]

Answer: 19

Step-by-step explanation: let the number of candies left in the bag be y and the number of minutes eating be x.

Because she eats 5 candies per minute

8 0

2 years ago

How do you solve this problem using substitution or elimination?

saw5 [17]

3x=4y
x+y=21

x = 21-y
3(21-y)=4y

63-3y=4y
63=7y
y=9

x+y=21
x+9=21
x=12

answer is (12, 9)

6 0

3 years ago

Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen

k0ka [10]

Let $a,b,c$ be the randomly selected lengths. Without loss of generality, suppose $a[tex]P(A + B \ge C) = P(A + B - C \ge 0)$

where $A,B,C$ are independent random variables with the same uniform distribution on [0, 1].

By their mutual independence, we have

$P(A=a,B=b,C=c) = P(A=a) \times P(B=b) \times P(C=c)$

so that the joint density function is

$P(A=a,B=b,C=c) = \begin{cases}1 & \text{if }(a,b,c)\in[0,1]^3 \\ 0 & \text{otherwise}\end{cases}$

where $[0,1]^3=[0,1]\times[0,1]\times[0,1]$ is the cube with vertices at (0, 0, 0) and (1, 1, 1).

Consider the plane

$a + b - c = 0$

with $(a,b,c)\in\Bbb R^3$ . This plane passes through (0, 0, 0), (1, 0, 1), and (0, 1, 1), and thus splits up the cube into one tetrahedral region above the plane and the rest of the cube under it. (see attached plot)

The point (0, 0, 1) (the vertex of the cube above the plane) does not belong the region $a+b-c\ge0$ , since $0+0-1=-1$ . So the probability we want is the volume of the bottom "half" of the cube. We could integrate the joint density over this set, but integrating over the complement is simpler since it's a tetrahedron.

Then we have

$\displaystyle P(A+B-C\ge0) = 1 - P(A+B-C < 0) \\\\ ~~~~~~~~ = 1 - \int_0^1\int_0^{1-a}\int_{a+b}^1 P(A=a,B=b,C=c) \, dc\,db\,da \\\\ ~~~~~~~~ = 1 - \int_0^1 \int_0^{1-a} (1 - a - b) \, db \, da \\\\ ~~~~~~~~ = 1 - \int_0^1 \frac{(1-a)^2}2\,da \\\\ ~~~~~~~~ = 1 - \frac16 = \boxed{\frac56}$

5 0

1 year ago

Mrs.smith has 5 times many markers as colored pencils. The total number of markers and colored pencils is 54. How many markers d

Slav-nsk [51]

Answer:

270 Markers

Step-by-step explanation:

5 x 54 = 270

(sorry if I'm wrong)

6 0

2 years ago

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2500 individ

Ne4ueva [31]

Answer:

The approximate distribution of the number who carry this gene is approximately normal with mean $\mu = 5$ and standard deviation $\sigma = 2.23$

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they carry the defective gene that causes inherited colon cancer, or they do not. The probability of a person carrying this gene is independent from other people. So the binomial probability distribution is used to solve this question.

A sample of 2500 individuals is quite large, so we use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

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

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer.

This means that $p = \frac{1}{500} = 0.002$

In a sample of 2500 individuals, what is the approximate distribution of the number who carry this gene?

$n = 2500$

So

$\mu = E(X) = np = 2500*0.002 = 5$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2500*0.002*0.998} = 2.23$

So the approximate distribution of the number who carry this gene is approximately normal with mean $\mu = 5$ and standard deviation $\sigma = 2.23$

4 0

3 years ago